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Maths MCQ

Functions · Maharashtra Board STD 11 Science (MSBSHSE - English Medium). 143 questions.

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143 questions · 142 auto-graded MCQ + 1 self-marked written.

MCQ 11 Mark
The domain and range of $f(x)=2-|x-5|$ are
  • A
    $R+,(-\infty, 1]$
  • ✓
    $R,(-\infty, 2]$
  • C
    $R,(-\infty, 2)$
  • D
    $R+,(-\infty, 2]$
Answer
Correct option: B.
$R,(-\infty, 2]$
(B) $R_t(-\infty, 2]$
Image
Hint:
$f(x)=2-|x-5| $
$=2-(5-x), x<5$
$=2-(x-5), x \geq 5 $
$\therefore f(x)=x-3, x<5 $
$=7-x, x \geq 5$
Domain $=R$,
Range (from graph) $=(-\infty, 2]$
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MCQ 21 Mark
The domain of $\frac{1}{|x|-x}$, where $[\mathrm{x}]$ is greatest integer function is
  • A
    $\mathrm{R}$
  • B
    $\mathrm{Z}$
  • ✓
    $R-Z$
  • D
    $Q-\{0\}$
Answer
Correct option: C.
$R-Z$
(C) $R-Z$
Hint:
$
f(x)=\frac{1}{[x]-x}=\frac{1}{-\{x\}}
$
For $f$ to be defined, $\{x\} \neq 0$
$\therefore \mathrm{x}$ cannot be integer.
$\therefore$ Domain $=\mathrm{R}-\mathrm{Z}$
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MCQ 31 Mark
If $f(x)=2 x^2+b x+c$ and $f(0)=3$ and $f(2)=1$, then $f(1)$ is equal to
  • A
    $-2$
  • ✓
    $0$
  • C
    $1$
  • D
    $2$
Answer
Correct option: B.
$0$
(B) $0$
Hint:
$f(x)=2 x^2+b x+c $
$f(0)=3 $
$\therefore 2(0)+b(0)+c=3 $
$\therefore c=3 \ldots \ldots .(i) $
$\therefore f(2)=1$
$\therefore 2(4)+2 b+c=1 $
$\therefore 2 b+c=-7$
$\therefore 2 b+3=-7 \ldots . .[\text { From (i)] }$
$\therefore b=-5 $
$\therefore f(x)=2 x^2-5 x+3 $
$\therefore f(1)=2(1)^2-5(1)+3=0$
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MCQ 41 Mark
Let the function $f$ be defined by $f(x)=\frac{2 x+1}{1-3 x}$ then $f^{-1}(x)$ is:
  • ✓
    $\frac{x-1}{3 x+2}$
  • B
    $\frac{x+1}{3 x-2}$
  • C
    $\frac{2 x+1}{1-3 x}$
  • D
    $\frac{3 x+2}{x-1}$
Answer
Correct option: A.
$\frac{x-1}{3 x+2}$
(A)
$\frac{x-1}{3 x+2}$
Hint:
$f(x)=\frac{2 x+1}{1-3 x}=y$, say. then
$2 \mathrm{x}+1=\mathrm{y}(1-3 \mathrm{x})$
$\therefore \mathrm{y}-1=\mathrm{x}(2+3 \mathrm{y}) $
$\therefore \mathrm{x}=\frac{y-1}{2+3 y}=\mathrm{f}^{-1}(\mathrm{y})$
$\therefore \mathrm{f}^{-1}(\mathrm{x})=\frac{x-1}{2+3 x}$
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MCQ 51 Mark
If $f: R \rightarrow R$ is defined by $f(x)=x^3$, then $f^{-1}(8)$ is equal to:
  • ✓
    $\{2\}$
  • B
    $\{-2.2\}$
  • C
    $\{-2\}$
  • D
    $(-2.2)$
Answer
Correct option: A.
$\{2\}$
(A) $\{2\}$
Hint:
$
\begin{array}{ll}
& \mathrm{f}(x)=x^3=y, \text { say } \\
\therefore & x=y^{\frac{1}{3}}=\mathrm{f}^{-1}(y) \\
\therefore \quad & \mathrm{f}^{-1}(8)=(8)^{\frac{1}{3}}=\left(2^3\right)^{\frac{1}{3}} \\
\therefore & \mathrm{f}^{-1}(8)=2
\end{array}
$
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MCQ 61 Mark
If $f(x)=\frac{1}{1-x}$, then $f(f(f(x)))$ is
  • A
    $x-1$
  • B
    $1-x$
  • ✓
    $x$
  • D
    $-x$
Answer
Correct option: C.
$x$
(C) $\mathrm{x}$
Hint:
$f(x)=\frac{1}{1-x}$
$f(f(x))=f\left(\frac{1}{1-x}\right)=\frac{1}{1-\left(\frac{1}{1-x}\right)}$
$=\frac{1-x}{1-x-1}=\frac{x-1}{x}$
$f(f(f(x)))=f\left(\frac{x-1}{x}\right)=\frac{1}{1-\left(\frac{x-1}{x}\right)}=\frac{x}{x-x+1}=x $
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MCQ 71 Mark
Find $x$, if $2 \log _2 x=4$
  • A
    $4,-4$
  • ✓
    $4$
  • C
    $-4$
  • D
    not defined
Answer
Correct option: B.
$4$
(B) $4$
Hint:
$2 \log _2 x=4, x>0$
$\therefore \log _2\left(x^2\right)=4$
$\therefore x^2=16$
$\therefore x= \pm 4$
$\therefore x=4$
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MCQ 81 Mark
If $\log _{10}\left(\log _{10}\left(\log _{10} x\right)\right)=0$, then $x=$
  • A
    $1000$
  • ✓
    $10^{10}$
  • C
    $10$
  • D
    $0$
Answer
Correct option: B.
$10^{10}$
(B) $10^{10}$
Hint:
$\log _{10} \log _{10} \log _{10} x=0 $
$\therefore \log _{10}\left(\log _{10}(x)\right)=10^0=1$
$\therefore \log _{10} x=10^1=10$
$\therefore x=10^{10}$
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MCQ 91 Mark
If $\log (5 x-9)-\log (x+3)=\log 2$, then $x=$
  • A
    $3$
  • ✓
    $5$
  • C
    $2$
  • D
    $7$
Answer
Correct option: B.
$5$
(B) $5$
Hint:
$\log (5 x-9)-\log (x+3)=\log 2 $
$\therefore \frac{5 x-9}{x+3}=2 $
$\therefore 3 x=9+6 $
$\therefore x=5$
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MCQ 102 Marks
If $f(x)=\frac{2 x-3}{3 x-4}, x \neq \frac{4}{3}$, then the value of $f^{-1}(x)$ is
  • ✓
    $\frac{4 x-3}{3 x-2}$
  • B
    $\frac{3 x-2}{4 x+3}$
  • C
    $\frac{3 x-4}{4 x-2}$
  • D
    $\frac{2 x+3}{4 x-3}$
Answer
Correct option: A.
$\frac{4 x-3}{3 x-2}$
(a) $\frac{4 x-3}{3 x-2}$
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MCQ 112 Marks
The money invested in a company is compounded continuously. If ₹ 200 invested today becomes $₹ 400$ in 6 years, then at the end of 33 years it will become ₹
  • A
    $1600 \sqrt{2}$
  • B
    $3200 \sqrt{2}$
  • C
    $12800 \sqrt{2}$
  • ✓
    $6400 \sqrt{2}$
Answer
Correct option: D.
$6400 \sqrt{2}$
(d) : We know that, $A=P\left(1+\frac{R}{100}\right)^n$
Here, $400=200\left(1+\frac{R}{100}\right)^6$ ...(i)
$
\Rightarrow 2=\left(1+\frac{R}{100}\right)^6 \Rightarrow\left(1+\frac{R}{100}\right)=2^{1 / 6}
$
$\therefore$ At the end of 33 years, $A=200\left(1+\frac{R}{100}\right)^{33}$
$
\begin{aligned}
& =200\left(2^{1 / 6}\right)^{33}=20-2^{11 / 2} \quad \text { [Using (i)] } \\
& =200 \times\left(2^{11}\right)^{1 / 2}=200\left(2^{10+1}\right)^{1 / 2}=2002^5 \cdot 2^{1 / 2}=6400 \sqrt{2}
\end{aligned}
$
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MCQ 122 Marks
The equation $x^3+x-1=0$ has
  • A
    no real root.
  • B
    exactly two real roots.
  • ✓
    exactly one real roots.
  • D
    more than two real roots.
Answer
Correct option: C.
exactly one real roots.
(c) : $f(x)=x^3+x-1$ Now, $f(0)=0+0-1=-1<0$ and $f(1)=1+1-1=1>0$ So, $f(x)$ has a root in between 0 and 1 and $f^{\prime}(x)>0 \in R \forall$ $x>1 \therefore f(x)$ has exactly one real root.
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MCQ 132 Marks
The domain and range for the function $f(x)=e^{d x \sin x}$ are
  • ✓
    domain $=R$, range $(0, \infty)$
  • B
    domain $=R$, range $=R$
  • C
    domain $=R$, range $=[0, \infty)$
  • D
    domain $=R$, range $=[1, \infty$ )
Answer
Correct option: A.
domain $=R$, range $(0, \infty)$
(a) : $f(x)=e^{|x| \sin x}$
We know that domain of $|x|$ is from $(-\infty, \infty)$ and domain of $\sin x$ is also from $(-\infty, \infty)$
So, domain of $f(x)=e^{|x| \sin x}$ is from $(-\infty, \infty)$
We know that range of expononential function is from $(0, \infty)$. So, range of function $f(x)=e^{x \mid \sin x}$ is from $(0, \infty)$.
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MCQ 142 Marks
If $R$ denotes the set of all real numbers then the function $f: R \rightarrow R$ defined by $f(x)=|x|$ is
  • A
    surjective
  • B
    injective and surjective
  • ✓
    neither injective nor surjective
  • D
    injective
Answer
Correct option: C.
neither injective nor surjective
(c) : $f: R \rightarrow R, f(x)=|x|$
Now $_1 f(1)=|1|=1$ and $f(-1)=|-1|=1 \Rightarrow f(1)=f(-1)$, but $1 \neq-1$ $\therefore f$ is not one-one.
Also, $-1 \in R$, but $f(x)=|x|$ is non-negative.
$\Rightarrow f$ is not onto. $\Rightarrow f$ is neither injective nor surjective.
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MCQ 152 Marks
If $f(x)=3 x-2$ and $g(x)=x^2$, then $fog(x)=$
  • ✓
    $3 x^2-2$
  • B
    $3 x^2+2$
  • C
    $3 x-2$
  • D
    $2-3 x^2$
Answer
Correct option: A.
$3 x^2-2$
(a) : $f(x)=3 x-2, g(x)=x^2$
$fo g(x)=f(g(x))=3(g(x))-2=3 x^2-2$
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MCQ 162 Marks
If $f: R-\{2\} \rightarrow R$ is a function defined by $f(x)=\frac{x^2-4}{x-2}$, then its range is
  • A
    $R$
  • B
    $R-\{2\}$
  • ✓
    $R=\{4\}$
  • D
    $R=\{-2,2\}$
Answer
Correct option: C.
$R=\{4\}$
(c): We have, $y=f(x)=\frac{x^2-4}{x-2}, x \neq 2$ $=\frac{(x-2)(x+2)}{x-2}=x+2, x \neq 2$.
It follows from the above relation that $y$ take all real values except when $x$ takes values in the set $R-\{2\}$
$\therefore \quad$ Range $(f)=R-\{4\}$
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MCQ 172 Marks
$A=\{1,2,3,4\}, B=\{1,2,3,4,5,6\}$ are two sets and function $f : A \rightarrow B$ is defined by $f (x)=x+2 ; \forall x \in A$, then the function f is
  • A
    Bijective
  • B
    Onto
  • ✓
    One-one
  • D
    Many-one
Answer
Correct option: C.
One-one
(C)
$f (x)= f (y)$
$\Rightarrow x+2=y+2 \Rightarrow x=y$
∴ Function f is one-one
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MCQ 182 Marks
If $f: A \rightarrow B$ is a bijection and $g: B \rightarrow A$ is the inverse of $f$, then fog is equal to
Answer
Correct option: B.
$I _{ B }$
B
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MCQ 192 Marks
If $f : R \rightarrow R$ be defined by $f (x)= e ^x$ and $g : R \rightarrow R$ be defined by $g (x)=x^2$. The mapping gof : $R \rightarrow R$ be defined by
$( gof )(x)= g [ f (x)] \forall x \in R$. Then
  • A
    gof is bijective but f is not injective
  • B
    gof is injective but g is not injective
  • ✓
    gof is injective but g is not bijective
  • D
    gof is surjective and g is not surjective
Answer
Correct option: C.
gof is injective but g is not bijective
(C)
$g (x)$ is neither injective nor surjective
$(\operatorname{gof})(x)=\left( e ^{ x }\right)^2= e ^{2 x}$
This is an injective function.
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MCQ 202 Marks
Let $S , T , U$ be three non-void sets and $f : S \rightarrow T$, $g : T \rightarrow U$ be so that g of $: S \rightarrow U$ is surjective. Then
  • A
    g and f are both surjective
  • ✓
    g is surjective,f may not be so
  • C
    f is surjective,g may not be so
  • D
    f and g both may not be surjective
Answer
Correct option: B.
g is surjective,f may not be so
B
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MCQ 212 Marks
Let $f, g: R \rightarrow R$ be two functions defined as $f (x)=|x|+x$ and $g (x)=|x|-x \forall x \in R$. Then $(fog)(x)$ for $x<0$ is
  • A
    $0$
  • B
    $4 x$
  • ✓
    $-4 x$
  • D
    $2 x$
Answer
Correct option: C.
$-4 x$
(C)
$|x|=-x$, $\quad$if $x<0$
$=x$, $\quad$if $x \geq 0$
Now, $(fog)(x)= f [ g (x)]$
$=|g(x)|+g(x)$
$=| | x|-x|+|x|-x$
When, $x<0$
$(fog)(x)=|-x-x|+(-x)-x$
$=-2 x-2 x$
$=-4 x$
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MCQ 222 Marks
Let $f (x)=x^2$ and $g (x)=\sin x$ for all $x \in R$. Then the set of all $x$ satisfying
(fogogof) $(x)=($ gogof $)(x)$, where
$( fog )(x)= f ( g (x))$ is
  • ✓
    $\pm \sqrt{n \pi}, n \in\{0,1,2, \ldots\}$
  • B
    $\pm \sqrt{n \pi}, n \in\{1,2, \ldots\}$
  • C
    $\frac{\pi}{2}+2 n \pi, n \in\{\ldots,-2,-1,0,1,2, \ldots\}$
  • D
    $2 n \pi, n \in\{\ldots,-2,-1,0,1,2, \ldots\}$
Answer
Correct option: A.
$\pm \sqrt{n \pi}, n \in\{0,1,2, \ldots\}$
(A)
$($gof$)(x)=\sin x^2 \Rightarrow\left(\right.$ gogof $(x)=\sin \left(\sin x^2\right)$
$\Rightarrow($ fogogof $)(x)=\left(\sin \left(\sin x^2\right)\right)^2=\sin ^2\left(\sin x^2\right)$
Now, $\sin ^2\left(\sin x^2\right)=\sin \left(\sin x^2\right)$
$\Rightarrow \sin \left(\sin x^2\right)=0,1$
$\Rightarrow \sin x^2= n \pi,(4 n +1) \frac{\pi}{2} n \in I$
$\Rightarrow \sin x^2=0 \Rightarrow x^2= n \pi$
$\Rightarrow x= \pm \sqrt{ n \pi} n \in W$
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MCQ 232 Marks
Two functions $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined as follows:
$f(x)=\left\{\begin{array}{l}0 ;(x \text { rational }) \\ 1 ;(x \text { irrational })\end{array}\right.$
$g(x)=\left\{\begin{array}{l}-1 ;(x \text { rational }) \\ 0 ;(x \text { irrational }),\end{array}\right.$
then $(gof)(e)+(fog)(\pi)=$
  • ✓
    -1
  • B
    $0$
  • C
    1
  • D
    2
Answer
Correct option: A.
-1
(A)
$( gof )( e )+( fog )(\pi)= g ( f ( e ))+ f ( g (\pi))$
$=g(1)+f(0)$
$=-1+0=-1$
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MCQ 242 Marks
Let $g(x)=1+x-[x]$ and
$f(x)=\left\{\begin{array}{l}-1, x<0 \\ 0, x=0, \\ 1, x>0\end{array}\right.$ then for all $x, f(g(x))$ is equal to
  • A
    $x$
  • ✓
    1
  • C
    $f(x)$
  • D
    $g (x)$
Answer
Correct option: B.
1
(B)
As $x-[x] \in[0,1), \forall x \in R$
$\therefore \quad 0 \leq x-[x]<1, \forall x \in R$
$\Rightarrow 1 \leq 1+x-[x]<2, \forall x \in R$
$\Rightarrow l \leq g (x)<2, \forall x \in R$
Hence, $f ( g (x))=1 \forall x \in R$
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MCQ 252 Marks
If $f(x)=\sin ^2 x+\sin ^2\left(x+\frac{\pi}{3}\right)+\cos x \cos \left(x+\frac{\pi}{3}\right)$ and $g\left(\frac{5}{4}\right)=1$, then $(gof)(x)$ is equal to:
  • A
    $\frac{1}{2}$
  • B
    $0$
  • C
    $\sin x$
  • ✓
    none of these
Answer
Correct option: D.
none of these
(D)
$f(x)=\sin ^2 x+\sin ^2\left(x+\frac{\pi}{3}\right)+\cos x \cos \left(x+\frac{\pi}{3}\right)$
$=\sin ^2 x+\left[\sin \left(x+\frac{\pi}{3}\right)\right]^2$ $+\cos x\left[\cos x \cos \frac{\pi}{3}-\sin x \sin \frac{\pi}{3}\right]$
$=\sin ^2 x+\left[\sin x \cos \frac{\pi}{3}+\cos x \sin \frac{\pi}{3}\right]^2$ $+\cos x\left[\frac{1}{2} \cos x-\frac{\sqrt{3}}{2} \sin x\right]$
$=\sin ^2 x+\left[\frac{\sin x}{2}+\frac{\sqrt{3}}{2} \cos x\right]^2$ $+\frac{\cos ^2 x}{2}-\frac{\sqrt{3}}{2} \sin x \cos x$
$=\sin ^2 x+\frac{\sin ^2 x}{4}+\frac{3}{4} \cos ^2 x+\frac{\cos ^2 x}{2}$ $+\frac{\sqrt{3}}{2} \sin x \cos x-\frac{\sqrt{3}}{2} \sin x \cos x$
$=\frac{5}{4}\left(\sin ^2 x+\cos ^2 x\right)$
$=\frac{5}{4}$
$( gof )(x)= g [ f (x)]= g \left(\frac{5}{4}\right)=1$
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MCQ 262 Marks
If $f(x)=2 x+1$ and $g(x)=\frac{x-1}{2}$ for all real $x$, then $(fog)^{-1}\left(\frac{1}{x}\right)$ is equal to
  • A
    $x$
  • ✓
    $\frac{1}{x}$
  • C
    $-x$
  • D
    $-\frac{1}{x}$
Answer
Correct option: B.
$\frac{1}{x}$
(B)
(fog) $(x)= f ( g (x))$
$= f \left(\frac{x-1}{2}\right)$
$=2\left(\frac{x-1}{2}\right)+1=x$
$\Rightarrow( fog )(x)=x$
$\Rightarrow x=( fog )^{-1}(x)$
Hence, $(f o g)^{-1}\left(\frac{1}{x}\right)=\frac{1}{x}$
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MCQ 272 Marks
If $f(x)=a x+b$ and $g(x)=c x+d$, then $f(g(x))=g(f(x))$ is equivalent to
  • A
    $f(c)=g(a)$
  • ✓
    $f(d)=g(b)$
  • C
    $f(a)=g(c)$
  • D
    $f(b)=g(b)$
Answer
Correct option: B.
$f(d)=g(b)$
(B)
Given, $f (x)= ax + b , g (x)= cx + d$
and $f ( g (x))= g ( f (x))$
$\Rightarrow f ( c x+ d )= g ( a x+ b )$
$\Rightarrow a ( c x+ d )+ b = c ( a x+ b )+ d$
$\Rightarrow ad + b = cb + d$
$\Rightarrow f ( d )= g ( b )$
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MCQ 282 Marks
If $f (x)=\frac{\alpha x}{x+1}, x \neq-1$, then for what value of $\alpha$ is $f ( f (x))=x$
  • A
    $\sqrt{2}$
  • B
    $-\sqrt{2}$
  • C
    1
  • ✓
    -1
Answer
Correct option: D.
-1
(D)
$f (x)=\frac{\alpha x}{x+1}$;
$f ( f (x))= f \left(\frac{\alpha x}{x+1}\right)=\frac{\alpha\left(\frac{\alpha x}{x+1}\right)}{\frac{\alpha x}{x+1}+1}$
But $f ( f (x))=x$
$\therefore \quad \frac{\alpha^2 x}{\alpha x+x+1}=x$
In L.H.S., Put $\alpha=-1$,
$\therefore \quad \frac{(-1)^2 x}{(-1) x+x+1}=\frac{x}{-x+x+1}=x ;$
$\therefore \quad \alpha=-1$
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MCQ 292 Marks
If $f(x)=\left(25-x^4\right)^{1 / 4}$ for $0 < x<\sqrt{5}$, then $f\left(f\left(\frac{1}{2}\right)\right)=$
  • A
    $2^{-4}$
  • B
    $2^{-3}$
  • C
    $2^{-2}$
  • ✓
    $2^{-1}$
Answer
Correct option: D.
$2^{-1}$
(D)
Here, $f \left(\frac{1}{2}\right)=\left(25-\frac{1}{16}\right)^{\frac{1}{4}}=\left(\frac{399}{16}\right)^{\frac{1}{4}}$
$\Rightarrow f \left[ f \left(\frac{1}{2}\right)\right]= f \left(\left(\frac{399}{16}\right)^{\frac{1}{4}}\right)$
$=\left(25-\frac{399}{16}\right)^{\frac{1}{4}}$
$=\left(\frac{1}{16}\right)^{\frac{1}{4}}$
$=\frac{1}{2}$
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MCQ 302 Marks
If $f (x)=\frac{1-x}{1+x}$, then $f [ f (\cos 2 \theta)]=$
  • A
    $\tan 2 \theta$
  • B
    $\sec 2 \theta$
  • ✓
    $\cos 2 \theta$
  • D
    $\cot 2 \theta$
Answer
Correct option: C.
$\cos 2 \theta$
(C)
$f [ f (\cos 2 \theta)]= f \left[\frac{1-\cos 2 \theta}{1+\cos 2 \theta}\right]$
$=f\left(\tan ^2 \theta\right)$
$=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}$
$=\cos 2 \theta$
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MCQ 312 Marks
If $x \neq 1$ and $f (x)=\frac{x+1}{x-1}$ is a real function, then $f(f(f(2)))$ is
  • A
    1
  • B
    2
  • ✓
    3
  • D
    4
Answer
Correct option: C.
3
(C)
Here, $f(2)=\frac{2+1}{2-1}=3$
$\therefore \quad f(f(2))=f(3)=\frac{3+1}{3-1}=\frac{4}{2}=2$
$\therefore \quad f(f(f(2)))=f(2)=\frac{2+1}{2-1}=3$
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MCQ 322 Marks
If $g(x)=x^2+x-2$ and $\frac{1}{2}(gof)(x)=2 x^2-5 x+2$ then $f (x)$ is equal to
  • A
    $2 x+3$
  • ✓
    $2 x-3$
  • C
    $2 x^2+3 x+1$
  • D
    $2 x^2-3 x-1$
Answer
Correct option: B.
$2 x-3$
(B)
Given, $g (x)=x^2+x-2$
and $\frac{1}{2}(\operatorname{gof})(x)=2 x^2-5 x+2$
$\Rightarrow g ( f (x))=4 x^2-10 x+4$
$\Rightarrow( f (x))^2+ f (x)-2=4 x^2-10 x+4$
$\Rightarrow( f (x))^2+ f (x)-\left(4 x^2-10 x+6\right)=0$
$\Rightarrow f (x)=\frac{-1 \pm \sqrt{1+16 x^2-40 x+24}}{2}$
$=\frac{-1 \pm(4 x-5)}{2}$
$=2 x-3,-2 x+2$
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MCQ 332 Marks
If $f(x)=\sin ^2 x+\sin ^2\left(x+\frac{\pi}{3}\right)+\cos \left(x+\frac{\pi}{3}\right) \cos x$ and $g\left(\frac{5}{4}\right)=1$, then $gof(x)$ is
  • A
    a polynomial of first degree in $\sin x$ and $\cos x$
  • ✓
    a constant function
  • C
    a polynomial of second degree in $\sin x$ and $\cos x$
  • D
    none of these
Answer
Correct option: B.
a constant function
(B)
Given,
$f(x)=\sin ^2 x+\sin ^2\left(x+\frac{\pi}{3}\right)+\cos \left(x+\frac{\pi}{3}\right) \cos x$
$\begin{array}{l}=\frac{1}{2}\left\{1-\cos 2 x+1-\cos \left(2 x+\frac{2 \pi}{3}\right)\right. \left.+\cos \left(2 x+\frac{2 \pi}{3}\right)+\cos \frac{\pi}{3}\right\}\end{array}$
$=\frac{1}{2}\left[\frac{5}{2}-\left\{\cos 2 x+\cos \left(2 x+\frac{2 \pi}{3}\right)\right\}+\cos \left(2 x+\frac{\pi}{3}\right)\right]$
$=\frac{1}{2}\left[\frac{5}{2}-2 \cos \left(2 x+\frac{\pi}{3}\right) \cos \frac{\pi}{3}+\cos \left(2 x+\frac{\pi}{3}\right)\right]$
$=\frac{5}{4}$
$\therefore \operatorname{gof}(x)=g[f(x)]=g\left(\frac{5}{4}\right)$
$=1 \quad \ldots\left[\because g\left(\frac{5}{4}\right)=1\right]$
Hence, $\operatorname{gof}(x)$ is a constant function.
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MCQ 342 Marks
If $f$ be the greatest integer function and $g$ be the modulus function, then (gof) $\left(-\frac{5}{3}\right)-( fog )\left(-\frac{5}{3}\right)$=
  • ✓
    1
  • B
    -1
  • C
    2
  • D
    4
Answer
Correct option: A.
1
(A)
Given, (gof) $\left(-\frac{5}{3}\right)-( fog )\left(-\frac{5}{3}\right)$
$=g\left\{f\left(\frac{-5}{3}\right)\right\}-f\left\{g\left(\frac{-5}{3}\right)\right\}$
$=g(-2)-f\left(\frac{5}{3}\right)=2-1=1$
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MCQ 352 Marks
If the functions $f, g, h$ are defined from the sets of real numbers $R$ to $R$ such that
$f(x)=x^2-1, g(x)=\sqrt{x^2+1}, h(x)=\left\{\begin{array}{l}0, \text { if }{ }_{x=0} \\x, \text { if }{ }_{x>0} ,\end{array}\right.$
then the composite function $(hofog)(x)=$
  • A
    $\left\{\begin{array}{ll}0, & x=0 \\ x^2, & x>0 \\ -x^2, & x<0\end{array}\right.$
  • ✓
    $\left\{\begin{array}{l}0, x=0 \\ x^2, x \neq 0\end{array}\right.$
  • C
    $\left\{\begin{array}{l}0, x \leq 0 \\ x^2, x>0\end{array}\right.$
  • D
    none of these
Answer
Correct option: B.
$\left\{\begin{array}{l}0, x=0 \\ x^2, x \neq 0\end{array}\right.$
(B)
$(\operatorname{hofog})(x)=(\operatorname{hof})( g (x))$
$=(\operatorname{hof})\left(\sqrt{x^2+1}\right)$
$= h \left( f \left(\sqrt{x^2+1}\right)\right)$
$= h \left[\left(\sqrt{x^2+1}\right)^2-1\right]$
$= h \left(x^2+1-1\right)$
$= h \left(x^2\right)=\left\{\begin{array}{l}0, \text { if } x=0 \\ x^2, \text { if } x \neq 0\end{array}\right.$
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MCQ 362 Marks
$f: R \rightarrow R$ and $g:[0, \infty) \rightarrow R$ is defined by $f ( x )=x^2$ and $g (x)=\sqrt{x}$. Which one of the following is not true?
  • ✓
    $fog(-4)=4$
  • B
    $\operatorname{gof}(-2)=2$
  • C
    $\operatorname{gof}(4)=4$
  • D
    $fog(2)=2$
Answer
Correct option: A.
$fog(-4)=4$
(A)
Consider, fog $(-4)= f [ g (-4)]$
But $g(-4)$ is not defined.
$\therefore \quad$ fog $(-4)=4$ is not true
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MCQ 372 Marks
The inverse of the function $y=\frac{10^x-10^{-x}}{10^x+10^{-x}}$ is
  • A
    $\log _{10}(2-x)$
  • ✓
    $\frac{1}{2} \log _{10} \frac{1+x}{1-x}$
  • C
    $\frac{1}{2} \log _{10}(2 x-1)$
  • D
    $\frac{1}{4} \log _{10} \frac{2 x}{2-x}$
Answer
Correct option: B.
$\frac{1}{2} \log _{10} \frac{1+x}{1-x}$
(B)
Let $y= f (x)=\frac{10^x-10^{-x}}{10^x+10^{-x}}$
$\therefore \quad y=\frac{10^{2 x}-1}{10^{2 x}+1}$
$\Rightarrow 10^{2 x}=\frac{1+y}{1-y}$
$\Rightarrow 2 x=\log _{10} \frac{1+y}{1-y}$
$\Rightarrow x=\frac{1}{2} \log _{10} \frac{1+y}{1-y}$
$\Rightarrow f ^{-1}(y)=\frac{1}{2} \log _{10} \frac{1+y}{1-y}$
$\Rightarrow f ^{-1}(x)=\frac{1}{2} \log _{10} \frac{1+x}{1-x}$
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MCQ 382 Marks
If $f (x)=\frac{2 x-1}{x+5}(x \neq 5)$, then $f ^{-1}(x)$ is equal to
  • A
    $\frac{x+5}{2 x-1}, x \neq \frac{1}{2}$
  • ✓
    $\frac{5 x+1}{2-x}, x \neq 2$
  • C
    $\frac{5 x-1}{2-x}, x \neq 2$
  • D
    $\frac{x-5}{2 x+1}, x \neq \frac{1}{2}$
Answer
Correct option: B.
$\frac{5 x+1}{2-x}, x \neq 2$
(B)
Let $f (x)=y \Rightarrow x= f ^{-1}(y)$.
Now, $y=\frac{2 x-1}{x+5},(x \neq-5)$
$x y+5 y=2 x-1 \Rightarrow 5 y+1=2 x-x y$
$\Rightarrow x(2-y)=5 y+1$
$\Rightarrow x=\frac{5 y+1}{2-y}$
$\Rightarrow f ^{-1}(y)=\frac{5 y+1}{2-y}$
$\therefore f ^{-1}(x)=\frac{5 x+1}{2-x}, x \neq 2$
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MCQ 392 Marks
If $f : R \rightarrow R$ be a mapping defined by $f ( x )=x^3+5$, then $f ^{-1}(x)$ is equal to
  • A
    $\frac{1}{x^3+5}$
  • B
    $(x+5)^{\frac{1}{3}}$
  • C
    $(5-x)^{\frac{1}{3}}$
  • ✓
    $(x-5)^{\frac{1}{3}}$
Answer
Correct option: D.
$(x-5)^{\frac{1}{3}}$
(D)
Let $f (x)=y \Rightarrow x= f ^{-1}(y)$
Now, $y=x^3+5$
$\Rightarrow y-5=x^3$
$\Rightarrow x=(y-5)^{\frac{1}{3}}$
$\Rightarrow f ^{-1}(y)=(y-5)^{\frac{1}{3}}$
$\Rightarrow f ^{-1}(x)=(x-5)^{\frac{1}{3}}$
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MCQ 402 Marks
If $f: R \rightarrow R$ is defined by $f(x)=|x|$, then
  • A
    $f ^{-1}(x)=-x$
  • B
    $f ^{-1}(x)=\frac{1}{|x|}$
  • ✓
    The function $f ^{-1}(x)$ does not exist
  • D
    $f ^{-1}(x)=\frac{1}{x}$
Answer
Correct option: C.
The function $f ^{-1}(x)$ does not exist
(C)
$f (x)=|x|$
$f (x)=\left\{\begin{array}{lll}x, & \text { if } & x \geq 0 \\ -x, & \text { if } & x<0\end{array}\right.$
Therefore, the function $f ^{-1}(x)$ does not exist.
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MCQ 412 Marks
The inverse of the function $f (x)=\frac{ e ^x- e ^{-x}}{ e ^x+ e ^{-x}}+2$ is
  • A
    $\log _e\left(\frac{x-2}{x-1}\right)^{\frac{1}{2}}$
  • ✓
    $\log _e\left(\frac{x-1}{3-x}\right)^{\frac{1}{2}}$
  • C
    $\log _e\left(\frac{x}{2-x}\right)^{\frac{1}{2}}$
  • D
    $\log _e\left(\frac{x-1}{x+1}\right)^{-2}$
Answer
Correct option: B.
$\log _e\left(\frac{x-1}{3-x}\right)^{\frac{1}{2}}$
(B)
Let $y= f (x)=\frac{ e ^x- e ^{-x}}{ e ^x+ e ^{-x}}+2$
$\therefore \quad y-2=\frac{ e ^{2 x}-1}{ e ^{2 x}+1}$
$\Rightarrow(y-2) e ^{2 x}+y-2= e ^{2 x}-1$
$\Rightarrow e ^{2 x}=\frac{1-y}{y-3}=\frac{y-1}{3-y}$
$\Rightarrow 2 x=\log _{ e }\left(\frac{y-1}{3-y}\right)$
$\Rightarrow x=\frac{1}{2} \log _{ e }\left(\frac{y-1}{3-y}\right)$
$\Rightarrow f ^{-1}(y)=\frac{1}{2} \log _e\left(\frac{y-1}{3-y}\right)$
$\Rightarrow f ^{-1}(x)=\log _{ e }\left(\frac{x-1}{3-x}\right)^{\frac{1}{2}}$
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MCQ 422 Marks
If the function $f:[1, \infty) \rightarrow[1, \infty)$ is defined by $f(x)=2^{x(x-1)}$, then $f^{-1}(x)$ is
  • A
    $\left(\frac{1}{2}\right)^{x(x-1)}$
  • ✓
    $\frac{1}{2}\left(1+\sqrt{1+4 \log _2 x}\right)$
  • C
    $\frac{1}{2}\left(1-\sqrt{1+4 \log _2 x}\right)$
  • D
    Not defined
Answer
Correct option: B.
$\frac{1}{2}\left(1+\sqrt{1+4 \log _2 x}\right)$
(B)
Given, $f (x)=2^{x(x-1)}$
$\Rightarrow x(x-1)=\log _2 f (x)$
$\Rightarrow x^2-x-\log _2 f (x)=0$
$\Rightarrow x=\frac{1 \pm \sqrt{1+4 \log _2 f (x)}}{2}$
Only $x=\frac{1+\sqrt{1+4 \log _2 f (x)}}{2}$ lies in the domain
$\therefore \quad f ^{-1}(x)=\frac{1}{2}\left[1+\sqrt{1+4 \log _2 x}\right]$
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MCQ 432 Marks
If $f:[0, \pi / 2) \rightarrow R$ is defined as
$f(\theta)=\left|\begin{array}{ccc}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{array}\right|$. Then the range of f is
  • A
    $(2, \infty)$
  • B
    $(-\infty,-2]$
  • ✓
    $[2, \infty)$
  • D
    $(-\infty, 2]$
Answer
Correct option: C.
$[2, \infty)$
(C)
$f(\theta)=\left|\begin{array}{ccc}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{array}\right|$
$\therefore f(\theta)=2 \sec ^2 \theta \geq 2$
∴ range of $f$ is $[2, \infty)$.
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MCQ 442 Marks
The range of function $f (x)=\log _{ e } \sqrt{4-x^2}$ is given by
  • A
    $(0, \infty)$
  • B
    $(-\infty, \infty)$
  • ✓
    $\left(-\infty, \log _e 2\right]$
  • D
    $\left(\log _e 2, \infty\right)$
Answer
Correct option: C.
$\left(-\infty, \log _e 2\right]$
(C)
Let $y=\log _e \sqrt{4-x^2} \Rightarrow e ^y=\sqrt{4-x^2}$
$\Rightarrow e ^{2 y}=4-x^2 \Rightarrow x^2=4- e ^{2 y} \Rightarrow x=\sqrt{4- e ^{2 y}}$
$\therefore 4-e^{2 y} \geq 0$
$\Rightarrow e ^{2 y} \leq 4 \Rightarrow 2 y \leq \log _{ e } 4$
$\Rightarrow y \leq \frac{1}{2} \log _{ e } 4 \Rightarrow y \leq \log _{ e } 2$
$\therefore y \in\left(-\infty, \log _e 2\right]$
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MCQ 452 Marks
The range of the function $f(x)=\log _e\left(3 x^2+4\right)$ is
  • A
    $\left[\log _e 2, \infty\right]$
  • B
    $\left[\log _e 3, \infty\right)$
  • C
    $\left[2 \log _e 3, \infty\right)$
  • ✓
    $\left[2 \log _e 2, \infty\right)$
Answer
Correct option: D.
$\left[2 \log _e 2, \infty\right)$
(D)
Let $y=\log _e\left(3 x^2+4\right)$
$\Rightarrow 3 x^2+4= e ^y$
$\Rightarrow x^2=\frac{ e ^y- 4}{3}$
Since, $x^2 \geq 0$
$\therefore \frac{ e ^y-4}{3} \geq 0 \Rightarrow e ^y-4 \geq 0 \Rightarrow y \geq \log _{ e } 4$
$\Rightarrow y \geq 2 \log _c 2$
So, range $=\left[2 \log _{ e } 2, \infty\right)$
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MCQ 462 Marks
The range of $f(x)=\cos x-\sin x$ is
  • A
    (-1, 1)
  • B
    [-1, 1)
  • C
    $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
  • ✓
    $[-\sqrt{2}, \sqrt{2}]$
Answer
Correct option: D.
$[-\sqrt{2}, \sqrt{2}]$
(D)
Since maximum and minimum values of $\cos -\sin x$ are $\sqrt{2}$ and $-\sqrt{2}$ respectively, therefore range of $f (x)$ is $[-\sqrt{2}, \sqrt{2}]$.
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MCQ 472 Marks
Range of the function $f (x)=\sqrt{x^2+x+1}$ is equal to
  • A
    $[0, \infty]$
  • ✓
    $\left[\frac{\sqrt{3}}{2}, \infty\right)$
  • C
    $\left(\frac{\sqrt{3}}{2}, \frac{-\sqrt{3}}{2}\right)$
  • D
    (0,0)
Answer
Correct option: B.
$\left[\frac{\sqrt{3}}{2}, \infty\right)$
(B)
Here, $f (x)=\sqrt{x^2+x+1}$
$\Rightarrow y^2=x^2+x+1$
$\Rightarrow x^2+x+\left(1-y^2\right)=0$
$\Rightarrow x=\frac{-1+\pm{1-4\left(1-y^2\right)}}{2}$
$\Rightarrow x=\frac{-1 \pm \sqrt{4 y^2-3}}{2}$
For $x$ real, $4 y^2-3 \geq 0$
$\therefore y \geq \pm \frac{\sqrt{3}}{2}$
$\therefore R _{ f }=\left[\frac{\sqrt{3}}{2}, \infty\right)$
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MCQ 482 Marks
Let f: R → R be defined as $f(x)=\frac{x^2-x+4}{x^2+x+4}$
Then the range of the function $f (x)$ is
  • ✓
    $\left[\frac{3}{5}, \frac{5}{3}\right]$
  • B
    $\left(\frac{3}{5}, \frac{5}{3}\right)$
  • C
    $\left(-\infty, \frac{3}{5}\right) \cup\left(\frac{5}{3}, \infty\right)$
  • D
    $\left[-\frac{5}{3},-\frac{3}{5}\right]$
Answer
Correct option: A.
$\left[\frac{3}{5}, \frac{5}{3}\right]$
(A)
Let $y=\frac{x^2-x+4}{x^2+x+4}$
$\Rightarrow(y-1) x^2+(y+1) x+4 y-4=0$
For real value of $x, b^2-4 ac \geq 0$
$\Rightarrow(y+1)^2-4(y-1)(4 y-4) \geq 0$
$\Rightarrow-15 y^2+34 y-15 \geq 0$
$\Rightarrow 15 y^2-34 y+15 \geq 0$
$\Rightarrow\left(y-\frac{3}{5}\right)\left(y-\frac{5}{3}\right) \leq 0$
$\Rightarrow \frac{3}{5} \leq y \leq \frac{5}{3}$
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MCQ 492 Marks
The range of the function $f(x)=\frac{x^2-3 x+2}{x^2+x-6}$ is
  • A
    $R -\left[\frac{1}{5}, 1\right]$
  • B
    R
  • ✓
    $R -\{1\}$
  • D
    $R-\{-3,2\}$
Answer
Correct option: C.
$R -\{1\}$
(C)
$f (x)$ is defined for $x^2+x-6 \neq 0$, i.e., $x \neq-3,2$
$\therefore \quad \operatorname{Dom}(f)=R-\{-3,2\}$
Let $y=\frac{x^2-3 x+2}{x^2+x-6}=\frac{x-1}{x+3}$
$\Rightarrow x-\frac{3 y+1}{y-1}$
$x$ is real for $y-1 \neq 0$, i.e., $y \neq 1$
Hence, $\operatorname{range}(f)=R-\{1\}$
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MCQ 502 Marks
The range of the function $f(x)=\frac{1+x^2}{x^2}$ is
  • A
    (0,1)
  • B
    [0, 1]
  • ✓
    $(1, \infty)$
  • D
    $[1, \infty)$
Answer
Correct option: C.
$(1, \infty)$
(C)
$f (x)$ is defined for all $x \in R -\{0\}$.
So, $\operatorname{dom}(f)=R-\{0\}$
Let $y=\frac{1+x^2}{x^2}$
$\Rightarrow x= \pm \sqrt{\frac{1}{y-1}}$
For $x$ to be real, $y-1>0 \Rightarrow y \in(1, \infty)$
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MCQ 512 Marks
Range of the function $f(x)=\frac{1}{3 x+2}$ is
  • A
    R
  • ✓
    $R -\{0\}$
  • C
    $(0, \infty)$
  • D
    $R-\left\{-\frac{2}{3}\right\}$
Answer
Correct option: B.
$R -\{0\}$
(B)
$\operatorname{Dom}(f)=R-\left\{-\frac{2}{3}\right\}$
For Range(f), let $y= f (x)=\frac{1}{3 x+2}$
$\therefore \quad 3 x+2=\frac{1}{y} \Rightarrow x=\frac{1}{3}\left(\frac{1}{y}-2\right)$
$x$ is real if $y \neq 0$.
Hence, $R _{ f }= R -\{0\}$
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MCQ 522 Marks
The domain of the function $\cos ^{-1}\left(\log _2\left(x^2+5 x+8\right)\right)$ is
  • A
    [2,3]
  • B
    [-2,2]
  • C
    [3,1]
  • ✓
    [-3,-2]
Answer
Correct option: D.
[-3,-2]
(D)
$-1 \leq \log _2\left(x^2+5 x+8\right)<1$
$\Rightarrow \frac{1}{2} \leq\left(x^2+5 x+8\right) \leq 2$
$\Rightarrow x^2+5 x+\frac{15}{2} \geq 0$
$\Rightarrow x^2+2\left(\frac{5}{2}\right) x+\left(\frac{5}{2}\right)^2-\left(\frac{5}{2}\right)^2+\frac{15}{2} \geq 0$
$\Rightarrow\left(x+\frac{5}{2}\right)^2+\frac{5}{4} \geq 0$ and $x^2+5 x+6 \leq 0$
$\Rightarrow(x+3)(x+2) \leq 0$
$\Rightarrow x \in[-3,-2]$
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MCQ 532 Marks
The domain of the function $f (x)=\sqrt{\cos ^{-1}\left(\frac{1-|x|}{2}\right)}$ is
  • A
    (-3,3)
  • ✓
    [-3,3]
  • C
    $(-\infty,-3) \cup(3, \infty)$
  • D
    $(-\infty,-3] \cup[3, \infty)$
Answer
Correct option: B.
[-3,3]
(B)
$f (x)=\sqrt{\cos ^{-1}\left(\frac{1-|x|}{2}\right)}$
$\therefore -1 \leq \frac{1-|x|}{2} \leq 1$
$\Rightarrow-2-1 \leq-|x| \leq 2-1$
$\Rightarrow-3 \leq-|x| \leq 1$
$\Rightarrow-1 \leq|x| \leq 3$
$\Rightarrow x \in[-3,3]$
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MCQ 542 Marks
Domain of the function $f(x)=\sin ^{-1}\left(1+3 x+2 x^2\right)$ is
  • A
    $(-\infty, \infty)$
  • B
    $(-1,1)$
  • ✓
    $\left[-\frac{3}{2}, 0\right]$
  • D
    $\left(-\infty, \frac{-1}{2}\right) \cup(2, \infty)$
Answer
Correct option: C.
$\left[-\frac{3}{2}, 0\right]$
(C)
$-1 \leq 1 + 3 x + 2 x^2 \leq 1$
Case I : $2 x^2+3 x+1 \geq-1 ; 2 x^2+3 x+2 \geq 0$
$x=\frac{-3 \pm \sqrt{9-16}}{6}=\frac{-3 \pm i \sqrt{7}}{6}$ (imaginary).
Case II :$2 x^2+3 x+1 \leq 1$
$\Rightarrow 2 x^2+3 x \leq 0 \Rightarrow 2 x\left(x+\frac{3}{2}\right)<0$
$\Rightarrow \frac{-3}{2} \leq x \leq 0 \Rightarrow x \in\left[-\frac{3}{2}, 0\right]$
In case I, we get imaginary value hence, rejected
∴ Domain of function $=\left[\frac{-3}{2}, 0\right]$.
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MCQ 552 Marks
The domain of the function $f (x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^2}}$ is
  • A
    [1,2)
  • ✓
    ( 2, 3)
  • C
    [1,2]
  • D
    [2, 3]
Answer
Correct option: B.
( 2, 3)
(B)
To define $f (x), 9-x^2>0 \Rightarrow|x|<3$
$\Rightarrow-3< x< 3,$ ...(i)
and $-1 \leq(x-3) \leq 1$
$\Rightarrow 2 \leq x \leq 4$ ...(ii)
From (i) and (ii), $2 \leq x<3$ i.e., $[2,3)$.
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MCQ 562 Marks
The domain of the function $f (x)=\exp \left(\sqrt{5 x-3-2 x^2}\right)$ is
  • A
    $\left[1, \frac{-3}{2}\right]$
  • B
    $\left[\frac{3}{2}, \infty\right]$
  • C
    $(-\infty, 1]$
  • ✓
    $\left[1, \frac{3}{2}\right]$
Answer
Correct option: D.
$\left[1, \frac{3}{2}\right]$
(D)
$f (x)= e ^{\sqrt{5 x-3-2 x^2}}$
$\Rightarrow 5 x-3-2 x^2 \geq 0$
$\Rightarrow(x-1)\left(x-\frac{3}{2}\right) \leq 0$
$\therefore \quad D _{ f }=\left[1, \frac{3}{2}\right]$
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MCQ 572 Marks
The domain of the function $f (x)=\log _{3+x}\left(x^2-1\right)$ is
  • A
    $(-3,-1) \cup(1, \infty)$
  • B
    $[-3,-1) \cup[1, \infty)$
  • ✓
    $(-3,-2) \cup(-2,-1) \cup(1, \infty)$
  • D
    $[-3,-2) \cup(-2,-1) \cup[1, \infty)$
Answer
Correct option: C.
$(-3,-2) \cup(-2,-1) \cup(1, \infty)$
(C)
$f (x)$ is to be defined when $x^2-1>0$
$\Rightarrow x^2>1, \Rightarrow x<-1$ or $x>1$ and $3+x>0$
$\therefore \quad x>-3$ and $x \neq-2$
$\therefore \quad D_f=(-3,-2) \cup(-2,-1) \cup(1, \infty)$
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MCQ 582 Marks
The domain of the function $f(x)=\sqrt{\log \frac{1}{|\sin x|}}$ is
  • A
    $R-\{2 n \pi, n \in I\}$
  • ✓
    $R -\{ n \pi, n \in I )$
  • C
    $R -(-\pi, \pi)$
  • D
    $(-\infty, \infty)$
Answer
Correct option: B.
$R -\{ n \pi, n \in I )$
(B)
$f (x)=\sqrt{\log \frac{1}{|\sin x|}}$
$\Rightarrow \sin x \neq 0 \Rightarrow x \neq n \pi+(-1)^{ n } 0 \Rightarrow x \neq n \pi$
Domain of $f (x)= R -\{ n \pi, n \in I \}$.
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MCQ 592 Marks
The domain of definition of $f(x)=\sqrt{\frac{1-|x|}{2-|x|}}$ is
  • A
    $(-\infty,-1) \cup(2, \infty)$
  • ✓
    $[-1,1] \cup(2, \infty) \cup(-\infty,-2)$
  • C
    $(-\infty, 1) \cup(2, \infty)$
  • D
    $[-1,1] \cup(2, \infty)$
Answer
Correct option: B.
$[-1,1] \cup(2, \infty) \cup(-\infty,-2)$
(B)
$\frac{1-|x|}{2-|x|} \geq 0$
$\Rightarrow \frac{|x|-1}{|x|-2} \geq 0$
$\Rightarrow|x| \leq 1$ as $|x|>2$
$\Rightarrow x \in(-\infty,-2) \cup(2, \infty) \cup[-1,1]$
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MCQ 602 Marks
Domain of $f(x)=\log |\log x|$ is
  • A
    $(0, \infty)$
  • B
    $(1, \infty)$
  • ✓
    $(0,1) \cup(1, \infty)$
  • D
    $(-\infty, 1)$
Answer
Correct option: C.
$(0,1) \cup(1, \infty)$
(C)
$f (x)=\log |\log x|, f (x)$ is defined if $|\log x|>0$ and $x>0$, i.e., if $x>0$ and $x \neq 1$
$\ldots[\because |\log x|>0$ if $x \neq 1]$
$\Rightarrow x \in(0,1) \cup(1, \infty)$.
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MCQ 612 Marks
The domain of the function $y= f (x)=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}$ is
  • ✓
    $[-2,1)$, excluding 0
  • B
    $[-3,-2]$, excluding -2.5
  • C
    $[0,1]$, excluding 0
  • D
    none of these
Answer
Correct option: A.
$[-2,1)$, excluding 0
(A)
$D_f=D_g \cap D_h$
where $g(x)=\frac{1}{\log _{10}(1-x)}$ and $h(x)=\sqrt{2+x}$
Now, $D _{ g }=\left\{x \in R : 1-x>0, \log _{10}(1-x) \neq 0\right\}$
$-\{x \in R : x<1,1-x \neq 1\}$
$=\{x \in R : x<1, x \neq 0\}$
and $D _{ h }=\{x \in R : x+2 \geq 0\}$
$=\{x \in R : x \geq-2\}$
$\therefore \quad D_f=[(-\infty, 1)-\{0\}] \cap[-2, \infty)$
$=[-2,1)-\{0\}$
View full question & answer→
MCQ 622 Marks
The domain of $f(x)=\frac{\log _2(x+3)}{x^2+3 x+2}$ is
  • A
    $R-\{-2\}$
  • B
    $(-2,+\infty)$
  • C
    $R -\{-1,-2,-3\}$
  • ✓
    $(-3, \infty)-\{-1,-2\}$
Answer
Correct option: D.
$(-3, \infty)-\{-1,-2\}$
(D)
Here, $x+3>0$ and $x^2+3 x+2 \neq 0$
$\therefore \quad x>-3$ and $(x+1)(x+2) \neq 0$, i.e., $x \neq-1,-2$.
$\therefore \quad$ Domain $=(-3, \infty)-\{-1,-2\}$.
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MCQ 632 Marks
The domain of the fuction $\sqrt{\log \left(x^2-6 x+6\right)}$ is
  • A
    $(-\infty, \infty)$
  • B
    $(-\infty, 3-\sqrt{3}) \cup(3+\sqrt{3}, \infty)$
  • ✓
    $(-\infty, 1] \cup[5, \infty)$
  • D
    $[0, \infty)$
Answer
Correct option: C.
$(-\infty, 1] \cup[5, \infty)$
(C)
The function $f (x)=\sqrt{\log \left(x^2-6 x+6\right)}$ is defined, when $\log \left(x^2-6 x+6\right) \geq 0$
$\Rightarrow x^2-6 x+6 \geq 1 \Rightarrow(x-5)(x-1) \geq 0$
This inequality holds, if $x \leq 1$ or $x \geq 5$.
Hence, the domain of the function will be $(-\infty, 1] \cup[5, \infty)$.
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MCQ 642 Marks
The domain of the function
$f(x)=\sin ^{-1}\left(\frac{8.3^{x-2}}{1-3^{2 (x-1)}}\right)$ is
  • A
    $(-\infty, 0]$
  • B
    $[2, \infty)$
  • ✓
    $(-\infty, 0) \cup[2, \infty)$
  • D
    $(-\infty,-1) \cup[1, \infty)$
Answer
Correct option: C.
$(-\infty, 0) \cup[2, \infty)$
(C)
$f (x)$ is defined for
$-1 \leq \frac{8.3^{x-2}}{1-3^{2(x-1)}} \leq 1$
$\Rightarrow-1 \leq \frac{\left(3^2-1\right)\left(3^{x-2}\right)}{1-3^{2 x-2}} \leq 1$
$\Rightarrow-1 \leq \frac{3^x-3^{x-2}}{1-3^{2 x-2}} \leq 1$
$\Rightarrow \frac{3^x-3^{x-2}}{1-3^{2 x-2}}+1 \geq 0$ and $\frac{3^x-3^{x-2}}{1-3^{2 x-2}}-1 \leq 0$
$\Rightarrow \frac{1+3^x-3^{x-2}-3^{2 x-2}}{1-3^{2 x-2}} \geq 0$ and
$\frac{3^x-3^{x-2}-1+3^{2 x-2}}{1-3^{2 x-2}} \leq 0$
$\Rightarrow \frac{\left(3^x+1\right)\left(3^{x-2}-1\right)}{\left(3^x \cdot 3^{x-2}-1\right)} \geq 0$ and $\frac{\left(3^x-1\right)\left(3^{x-2}+1\right)}{\left(3^{2 x-2}-1\right)} \geq 0$
$\Rightarrow \frac{\left(3^{x-2}-1\right)}{\left(3^x \cdot 3^{x-2}-1\right)} \geq 0$ and $\frac{\left(3^x-1\right)}{\left(3^{2 x-2}-1\right)} \geq 0$
$\Rightarrow \frac{\left(3^x-3^2\right)}{\left(3^{2 x}-3^2\right)} \geq 0$ and $\frac{\left(3^x-1\right)}{\left(3^{2 x}-3^2\right)} \geq 0$
$\Rightarrow \frac{\left(3^x-3^2\right)}{\left(3^x-3\right)} \geq 0$ and $\frac{\left(3^x-1\right)}{\left(3^x-3\right)} \geq 0$
$\Rightarrow x \in(-\infty, 1] \cup[2, \infty)$ and $x \in(-\infty, 0] \cup(1, \infty)$
$\Rightarrow x \in(-\infty, 0] \cup[2, \infty)$
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MCQ 652 Marks
The domain of the function
$f(x)=\sqrt{x^2-5 x+6}+\sqrt{2 x+8-x^2}$, is
  • A
    [2,3]
  • B
    [-2,4]
  • ✓
    $[-2,2] \cup[3,4]$
  • D
    $[-2,1] \cup[2,4]$
Answer
Correct option: C.
$[-2,2] \cup[3,4]$
(C)
$f (x)$ is defined, if
$x^2-5 x+6 \geq 0$ and $2 x+8-x^2 \geq 0$
$\Rightarrow(x-2)(x-3) \geq 0$ and $(x-4)(x+2) \leq 0$
$\therefore \quad x \in(-\infty, 2] \cup[3, \infty)$ and $x \in[-2,4]$
$\therefore \quad x \in[-2,2] \cup[3,4]$
View full question & answer→
MCQ 662 Marks
The domain of $\sin ^{-1}\left[\log \left(\frac{x}{3}\right)\right]$ is
  • ✓
    [1,9]
  • B
    (-1,9)
  • C
    [-9,1]
  • D
    [-9,-1]
Answer
Correct option: A.
[1,9]
(A)
$y=\sin ^{-1}\left[\log _3\left(\frac{x}{3}\right)\right]$
$\therefore \quad-1 \leq \log _3\left(\frac{x}{3}\right) \leq 1$
$\therefore \quad \frac{1}{3} \leq \frac{x}{3} \leq 3$
$\therefore \quad 1 \leq x \leq 9$
$\therefore \quad x \in[1,9]$
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MCQ 672 Marks
Domain of the function $\frac{\sqrt{1+x}-\sqrt{1-x}}{x}$ is
  • A
    (-1,1)
  • B
    (- 1, 1) - {0}
  • C
    [-1,1]
  • ✓
    [- 1, 1] - {0}
Answer
Correct option: D.
[- 1, 1] - {0}
(D)
$1+x \geq 0$
$\Rightarrow x \geq-1 ; 1-x \geq 0$
$\Rightarrow x \leq 1, x \neq 0$
Hence, domain is $[-1,1]-\{0\}$.
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MCQ 682 Marks
Domain of the function $f(x)=\sqrt{\frac{x}{1+x}}$ is
  • ✓
    $(-\infty,-1) \cup(0, \infty)$
  • B
    R
  • C
    $[0, \infty)$
  • D
    $(-\infty,-1)$
Answer
Correct option: A.
$(-\infty,-1) \cup(0, \infty)$
(A)
For domain, take $\frac{x}{1+x} \geq 0$
$\therefore D _{ f }=(-\infty,-1) \cup[0, \infty)$
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MCQ 692 Marks
If If f: R → S defined by $f(x)=\sin x-\sqrt{3} \cos \ x+1$ is onto, them the interval of S is
  • ✓
    [-1, 3]
  • B
    [1, 1]
  • C
    [0, 1]
  • D
    [0, -1]
Answer
Correct option: A.
[-1, 3]
(A)
$-\sqrt{1+(-\sqrt{3})^2} \leq(\sin x-\sqrt{3} \cos x) \leq \sqrt{1+(-\sqrt{3})^2}$
$\therefore -2 \leq(\sin x-\sqrt{3} \cos x) \leq 2$
$\therefore-2+1 \leq(\sin x-\sqrt{3} \cos x+1) \leq 2+1$
$\therefore-1 \leq(\sin x-\sqrt{3} \cos x+1) \leq 3$
i.e., range $=[-1,3]$
$\therefore $ For $f$ to be onto $S=[-1,3]$.
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MCQ 702 Marks
If the number of elements in the sets G and A are 3 and 4 respectively, then match the items of List I with those of List II.
 List I List II
(a) The number of non bijective functions from GG to GI.24
(b) The number of bijective functions from A to AII.0
(c) The number of functions from G to GAIII.1728
(d) The number of surjective functions from A to AAIV.12
  V.19683
  • ✓
    abcd
    VIIIIII
  • B
    abcd
    VIIIIVII
  • C
    abcd
    IIIIVVII
  • D
    abcd
    IIIIIIIV
Answer
Correct option: A.
abcd
VIIIIII
(A)
(a) $n( G \times G )=9$
The number of functions from a finite set A into a finite set $B =[ n ( B )]^{ n ( A )}$
we get, The number of bijections from a finite set A onto a finite set B is
$n ( A )!; \quad$ if $n ( A )= n ( B )$
0 ; $\quad$otherwise
The number of non-biective functions from $G \times G$ to G is $3^9-0=19,683$
(b) The number of bijections from a finite set A onto a finite set B is
$n ( A )!; \quad$ if $n ( A )= n ( B )$
0 ; $\quad$otherwise
we get, The Number of bijective functions from A to $A =4!=24$
(c) $n ( G \times A )=12$
The number of functions from a finite set A into a finite set $B =[ n ( B )]^{ n ( A )}$
we get, The number of functions G to $G \times A$
$=12^3=1728$
(d) $n ( A \times A )=16$
∴ The number of surjective functions from
A to $A \times A$ is 0 .
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MCQ 712 Marks
The function f : R $\rightarrow$ R defined by $f (x)= e ^x$ is
  • A
    Onto
  • B
    Many-one
  • ✓
    One-one and into
  • D
    Many one and onto
Answer
Correct option: C.
One-one and into
(C)
Let $f \left(x_1\right)= f \left(x_2\right)$ for $x_1, x_2 \in R$
$\Rightarrow e ^{x_1}= e ^{x_2}$
$\Rightarrow x_1=x_2$
⇒ f is one one
Let $y= e ^x$
$\Rightarrow \log y=x$
Note that the negative numbers and zero has no pre-image in the domain of the function.
⇒ function f is into.
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MCQ 722 Marks
A function f from the set of natural numbers to integers defined by $f(n)=\left\{\begin{array}{l}\frac{n-1}{2}, \text { when } n \text { is odd } \\ -\frac{n}{2}, \text { when } n \text { is even }\end{array}\right.$, is
  • A
    One-one but not onto
  • B
    Onto but not one-one
  • C
    One-one and onto both
  • D
    Neither one-one nor onto
Answer
(C)
$f : N \rightarrow I$
$\begin{array}{l}f(1)-0, f(2)=-1, f(3)=1, f(4)=-2, f(5)=2 \\ \text {and } f(6)=-3 \text { so on. }\end{array}$
Image
In this type of function every element of set A has unique image in set B and there is no element left in set B. Hence f is one-one and onto function.
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MCQ 732 Marks
Let f : N ⟶ N defined by
$f(n)=\left\{\begin{array}{l}\frac{n+1}{2} \text { if } n \text { is odd } \\ \frac{n}{2} \text { if } n \text { is even }\end{array}\right.$ then $f$ is even
  • ✓
    onto but not one-one
  • B
    one-one and onto
  • C
    neither one-one nor onto
  • D
    Neither one-one nor onto
Answer
Correct option: A.
onto but not one-one
(A)
$f : N \rightarrow N$
$f(n)=\left\{\begin{array}{ll}\frac{n+1}{2} & \text { if } n \text { isodd } \\ \frac{n}{2} & \text { if niseven }\end{array}\right.$
Now for $n=1, f(1)=\frac{1+1}{2}=1$
and if $n=2, f(2)=\frac{2}{2}=1$
$\therefore f(1)=f(2)$, But $1+2$.
$\therefore f (x)$ is not one-one.
$f (x)=\frac{ n +1}{2}$ if n is odd
if $y=\frac{ n +1}{2}$ then $n =2 y-1, \forall y$
Also, $f (x)=\frac{ n }{2}$ if n is even i.e., $y=\frac{ n }{2}$
or $n =2 y \forall y$
$\therefore f (x)$ is onto.
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MCQ 742 Marks
$f: R \rightarrow R, f(x)=x^2+3 x+4$ is __________
  • A
    one-one and onto
  • B
    one-one and not onto
  • ✓
    many-one and not onto
  • D
    many-one and not onto
Answer
Correct option: C.
many-one and not onto
(C)
$f (x)=x^2+3 x+4$
$=x^2+3 x+\frac{9}{4}-\frac{9}{4}+4$
$=\left(x^2+3 x+\frac{9}{4}\right)+\frac{7}{4}$
$=\left(x+\frac{3}{2}\right)^2+\frac{7}{4}$
Hence, $f$ is many-one and not onto
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MCQ 752 Marks
Let f : R $\rightarrow$ R be defined by $f (x)=x^4$, then
  • A
    f is one-one and onto
  • B
    f may be one-one and onto
  • C
    f is one-one but not onto
  • ✓
    f is neither one-one nor onto
Answer
Correct option: D.
f is neither one-one nor onto
(D)
Let $x_1, x_2 \in R$ such that $f \left(x_1\right)= f \left(x_2\right)$
$\Rightarrow x_1= \pm x_2$
$\therefore f \left(x_1\right)= f \left(x_2\right)$ does not impty that $x_1=x_2$
$\therefore f$ is not one-one.
Consider an element 2 in the co-domain R .
There doesnot exist any $x$ is domain R such that $f (x)=2$.
$\therefore f$ is not onto.
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MCQ 762 Marks
Which one of the following is a bijective function on the set of real numbers?
  • ✓
    $2 x-5$
  • B
    $|x|$
  • C
    $x^2$
  • D
    $x^2+1$
Answer
Correct option: A.
$2 x-5$
(A)
$|x|$ is not one-one; $x^2$ is not one-one;
$x^2+1$ is not one-one.
But $2 x-5$ is one-one because
$f (x)= f (y) \Rightarrow 2 x-5=2 y-5 \Rightarrow x=y$
Now, $f (x)=2 x-5$ is onto.
$\therefore f (x)=2 x-5$ is bijective.
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MCQ 772 Marks
Set A has 3 elements and set B has 4 elements. The number of injection that can be defined from A to B is
  • A
    144
  • B
    12
  • ✓
    24
  • D
    64
Answer
Correct option: C.
24
(C)
The number of one-one functions that can be defined from a set A into a finite set B is
${ }^{ n ( B )} P _{ n ( A )} \quad ;$ if $n ( B ) \geq n ( A )$
0 ; otherwise
the total number of injective functions from a set A containing 3 elements to a set B containing 4 elements $={ }^4 P_3=24$.
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MCQ 782 Marks
Number of bijective function from a set of 10 elements to itself is
  • A
    5!
  • ✓
    10!
  • C
    15!
  • D
    8!
Answer
Correct option: B.
10!
(B)
The number of bijections from a finite set A onto a finite set B is
$n ( A )!; \quad$ if $n ( A )= n ( B )$
0 ;$\quad$ otherwise
the number of bijective function from a set of 10 elements to itself $=10$ !
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MCQ 792 Marks
If $A =\{x \mid x \in N, x \leq 5\}$,$B=\left\{x \mid x \in Z, x^2-5 x+6=0\right\},$then the number of onto functions from $A$ to $B$ is
  • A
    23
  • B
    32
  • C
    2
  • ✓
    30
Answer
Correct option: D.
30
(D)
$A =\{1,2,3,4,5\}$
$B=\{2,3\}$
The number of onto functions, that can be defined from a finite set A , containing n elements onto a finite set B , containing 2 elements $=2^n-2$
the number of onto functions from A to B
$=2^5-2=32-2=30$
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MCQ 802 Marks
A is a set having 6 distinct elements. The number of distinct functions from A to A which are not bijection is
  • A
    $6!-6$
  • B
    $6^6-6$
  • ✓
    $6^6-6!$
  • D
    $6!$
Answer
Correct option: C.
$6^6-6!$
(C)
The number of functions from a finite set A into a finite set $B =[ n ( B )]^{ n ( A )}$
The number of bijections from a finite set A onto a finite set B is
$n ( A )!; \quad$ if $n ( A )= n ( B )$
0 ; $\quad$otherwise
total number of distinct functions from
$A \rightarrow A = n ^{ n }=6^6$, and
number of bijections $=n!=6!$
∴ Number of functions which are not bijections
$=6^6-6!$
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MCQ 812 Marks
Mapping f : R $\rightarrow$ R which is defined as f(x) = cos x, x $\in$ R will be
  • ✓
    Neither one-one nor onto
  • B
    One-one
  • C
    Onto
  • D
    One-one onto
Answer
Correct option: A.
Neither one-one nor onto
(A)
Let $x_1, x_2 \in R$, then $f\left(x_1\right)-\cos x_1$, &
$f \left(x_2\right)=\cos x_2$, Now $f \left(x_1\right)= f \left(x_2\right)$
$\Rightarrow \cos x_1=\cos x_2 \Rightarrow x_1=2 n \pi \pm x_2$
$\Rightarrow x_1 \neq x_2$,
$\therefore \quad$ it is not one-one.
Again the value of f-image of $x$ lies in between - 1 to 1
$\Rightarrow f [ R ]=\{ f (x):-1 \leq f (x) \leq 1\}$
So other numbers of co-domain (besides -1 and 1) is not f-image. $f[R] \in R$, so it is also not onto. So this mapping is neither one-one nor onto.
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MCQ 822 Marks
If R denotes the set of all real numbers, then the function f : R $\rightarrow$ R defined by f(x) = [x] is
  • A
    One-one only
  • B
    Onto only
  • C
    Both one-one and onto
  • ✓
    Neither one-one nor onto
Answer
Correct option: D.
Neither one-one nor onto
(D)
Let $f \left(x_1\right)= f \left(x_2\right) \Rightarrow\left[x_1\right]=\left[x_2\right] \nRightarrow x_1=x_2$
{For example, if $x_1=1.4, x_2=1.5$, then
$[1.4]=[1.5]=1\}$
$\therefore f$ is not one-one.
Also, $f$ is not onto as its range $I$ (set of integers) is a proper subset of its co-domain R.
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MCQ 832 Marks
If $f : R -\{3\} \rightarrow R -\{1\}$ be defined by $f(x)=\frac{x-2}{x-3}$, then $f$ is
  • A
    one-one into
  • ✓
    one-one onto
  • C
    many-one into
  • D
    many-one onto
Answer
Correct option: B.
one-one onto
(B)
$f(x)-\frac{x-2}{x-3}, x\neq3$
Let $y= f (x) \Rightarrow y=\frac{x-2}{x-3} \Rightarrow x=\frac{2-3 y}{1-y}$
$\Rightarrow y \neq 1 \Rightarrow$ Range of $f (x)$ is $R -\{1\}$
So, f is onto
For one-one, let $f \left(x_1\right)= f \left(x_2\right)$
$\Rightarrow \frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3} \Rightarrow x_1=x_2$
Hence, $f$ is one-one.
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MCQ 842 Marks
If $f:[0, \infty) \rightarrow[0, \infty)$ and $f(x)=\frac{x}{1+x}$, then $f$ is
  • A
    One-one and onto
  • ✓
    One-one but not onto
  • C
    Onto but not one-one
  • D
    Neither one-one nor onto
Answer
Correct option: B.
One-one but not onto
(B)
For $0 \in[0, \infty)$ in co-domain we cannot find
any $x \in[0, \infty)$ in domain such that $f (x)=0$
⇒ function is one-one but not onto.
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MCQ 852 Marks
The function $f(x)=\sin \left(\log \left(x+\sqrt{\left.x^2+1\right)}\right)\right.$ is
  • A
    Even function
  • ✓
    Odd function
  • C
    Neither even nor odd
  • D
    peridic function
Answer
Correct option: B.
Odd function
(B)
$f (x)=\sin \left(\log \left(x+\sqrt{1+x^2}\right)\right)$
$\rightarrow f (-x)=\sin \left[\log \left(-x+\sqrt{1+x^2}\right)\right]$
$\Rightarrow f (-x)=\sin \log \left(\left(\sqrt{1+x^2}-x\right) \frac{\left(\sqrt{1+x^2}+x\right)}{\left(\sqrt{1+x^2}+x\right)}\right)$
$\Rightarrow f (-x)=\sin \log \left[\frac{1}{x+\sqrt{1+x^2}}\right]$
$\Rightarrow f (-x)=\sin \left[-\log \left(x+\sqrt{1+x^2}\right)\right]$
$\Rightarrow f (-x)=-\sin \left[\log \left(x+\sqrt{1+x^2}\right)\right]$
$\Rightarrow f (-x)=- f (x)$
$\therefore f (x)$ is odd function.
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MCQ 862 Marks
The function $f (x)=\sec \left[\log \left(x+\sqrt{1+x^2}\right)\right]$ is
  • A
    Odd
  • ✓
    Even
  • C
    Neither odd not even
  • D
    Constant
Answer
Correct option: B.
Even
(B)
$f (-x)=\sec \left[\log \left(-x+\sqrt{1+(-x)^2}\right)\right]$
$=\sec \left[\log \left(-x+\sqrt{1+x^2}\right)\right]$
$=\sec \left[\log \left(\sqrt{1+x^2}-x\right)\right]$
$=\sec \left[\log \left(\frac{1+x^2-x^2}{\sqrt{1+x^2}+x}\right)\right]$
$=\sec \left[\log \left(\frac{1}{\sqrt{1+x^2}+x}\right)\right]$
$=\sec \left[-\log \left(\sqrt{1+x^2}+x\right)\right]$
$=\sec \left[\log \left(\sqrt{1+x^2}+x\right)\right]$
$\therefore f (x)$ is an even function.
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MCQ 872 Marks
If $f (x)=\log \frac{1+x}{1-x}$, then $f (x)$ is
  • A
    Even function
  • B
    $f \left(x_1\right) f \left(x_2\right)= f \left(x_1+x_2\right)$
  • C
    $\frac{ f \left(x_1\right)}{ f \left(x_2\right)}= f \left(x_1-x_2\right)$
  • ✓
    Odd function
Answer
Correct option: D.
Odd function
(D)
Here, $f (x)=\log \frac{1+x}{1-x}$
and $f(-x)=\log \left(\frac{1-x}{1+x}\right)=\log \left(\frac{1+x}{1-x}\right)^{-1}$
$=-\log \left(\frac{1+x}{1-x}\right)=-f(x)=f(-x)$
$\Rightarrow f (x)$ is an odd function.
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MCQ 882 Marks
If the real valued function $f(x)=\frac{a^x-1}{x^x\left(a^x+1\right)}$ is even, then n equals
  • A
    2
  • B
    $\frac{-2}{3}$
  • C
    $\frac{1}{4}$
  • ✓
    $-\frac{1}{3}$
Answer
Correct option: D.
$-\frac{1}{3}$
(D)
Since $f (x)$ is even, $f (-x)= f (x)$
$\therefore \quad \frac{ a ^{-x}-1}{(-x)^{ n }\left( a ^{-x}+1\right)}=\frac{ a ^x-1}{x^{ n }\left( a ^x+1\right)}$
$\Rightarrow \frac{1- a ^x}{(-1)^{ n } x^{ n }\left(1+ a ^x\right)}=\frac{ a ^x-1}{x^{ n }\left( a ^x+1\right)}$
$\Rightarrow \frac{-1}{(-1)^{ n }}=1 \Rightarrow-1=(-1)^{ n }$
$\therefore \quad n =-\frac{1}{3}$ can satisfy the equation.
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MCQ 892 Marks
Which of the following functions is an odd function?
  • ✓
    $f (x)=\sqrt{1+x+x^2}-\sqrt{1-x+x^2}$
  • B
    $f (x)=x\left(\frac{ a ^x+1}{ a ^x-1}\right)$
  • C
    $f (x)=\log _{10}\left(\frac{1-x^2}{1+x^2}\right)$
  • D
    $f (x)= k ($ constant $)$
Answer
Correct option: A.
$f (x)=\sqrt{1+x+x^2}-\sqrt{1-x+x^2}$
(A)
Let $f (x)=\sqrt{1+x+x^2}-\sqrt{1-x+x^2}$, then
$f (-x)=\sqrt{1-x+x^2}-\sqrt{1+x+x^2}$
Here, $f (-x)=- f (x)$
$\therefore f (x)$ is an odd function.
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MCQ 902 Marks
Which of the following is an even function?
  • A
    $\sqrt{x}$
  • ✓
    $x^2+\sin ^2 x$
  • C
    $\sin ^3 x$
  • D
    all of above
Answer
Correct option: B.
$x^2+\sin ^2 x$
(B)
Let $f(x)=x^2+\sin ^2 x$
Here, $f (-x)= f (x)$
$\therefore f (x)$ is an even function.
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MCQ 912 Marks
Consider the function $f (x)=\cos x^2$. Then
  • A
    f is of period $2 \pi$
  • B
    f is of period $\sqrt{2 \pi}$
  • ✓
    f is not periodic
  • D
    f is of periodic $\pi$
Answer
Correct option: C.
f is not periodic
(C)
Let $f (x)$ is periodic with period T.
Then, $\cos (x+ T )^2=\cos x^2$ for all $x \in R$
$\Rightarrow \cos (x+ T )^2-\cos x^2=0$
$\Rightarrow-2 \sin \left(\frac{(x+ T )^2+x^2}{2}\right) \sin \left(\frac{(x+ T )^2-x^2}{2}\right)=0$
$\forall x \in R$
$\Rightarrow(x+ T )^2-x^2= n \pi$ or $(x+ T )^2+x^2= n \pi$
$\forall x \in R$
Here $T$ is denendent on the value of $r$
$\Rightarrow f (x)$ is not periodic.
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MCQ 922 Marks
Let f be a real valued function, satisfying $f (x+y)= f (x) f (y)$ for all $x, y \in R$ Such that, $f (1)= a$. Then, $f (x)=$
  • ✓
    $a^x$
  • B
    $a x$
  • C
    $x^{ a }$
  • D
    $\log x$
Answer
Correct option: A.
$a^x$
(A)
The general expression for the function
satisfying $f (x+y)= f (x) f (y)$ for all $x, y \in R$ is
$f (x)=[ f (1)]^x= a ^x$ for all $x, y \in R$.
$\ldots[\because f(1)=a]$
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MCQ 932 Marks
If $[x]$ denotes the greatest integer $\leq x$, then $\left[\frac{2}{3}\right]+\left[\frac{2}{3}+\frac{1}{99}\right]+\left[\frac{2}{3}+\frac{2}{99}\right]+\ldots .+\left[\frac{2}{3}+\frac{98}{99}\right]=$
  • A
    99
  • B
    98
  • ✓
    66
  • D
    65
Answer
Correct option: C.
66
(C)
Given expression
$=\sum_{i=0}^{98}\left[\frac{2}{3}+\frac{i}{99}\right]$
$=\sum_{i=0}^{32}\left[\frac{2}{3}+\frac{i}{99}\right]+\sum_{i=33}^{98}\left[\frac{2}{3}+\frac{i}{99}\right]$
$=0+\sum_{i=33}^{98}\left[\frac{2}{3}+\frac{i}{99}\right]$
$\ldots\left[\because \frac{2}{3} \leq \frac{2}{3}+\frac{ i }{99}<1\right.$ for $\left.i =0,1,2, \ldots, 32\right]$
$=66$
$\ldots\left[\begin{array}{c}\because \text { each term in the summation is one or more } \\ \text { but less than } 2 \text { when } i=33,34,35, \ldots, 98\end{array}\right]$
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MCQ 942 Marks
If $f (x)+2 f \left(\frac{1}{x}\right)=3 x, x \neq 0$, and $S =\{x \in R : f (x)= f (-x)\} ;$ then S
  • A
    Contains exactly one element
  • ✓
    Contains exactly two elements
  • C
    Contains more than two elements
  • D
    Is an empty set
Answer
Correct option: B.
Contains exactly two elements
(B)
$f(x)+2 f\left(\frac{1}{x}\right)-3 x$....(i)
$\therefore f\left(\frac{1}{x}\right)+2 f(x)=\frac{3}{x}$....(ii)
From (i) and (ii), we get
$3 f (x)=\frac{6}{x}-3 x$
$\Rightarrow f (x)=\frac{2}{x}-x \Rightarrow f (-x)=-\frac{2}{x}+x$
Since, $f (x)= f (-x)$
$\therefore \quad \frac{2}{x}-x=-\frac{2}{x}+x$
$\Rightarrow \frac{4}{x}=2 x \Rightarrow x^2=2 \Rightarrow x= \pm \sqrt{2}$
$\therefore$ option (B) is the correct answer.
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MCQ 952 Marks
The graph of the function y = f(x) is symmetrical about the line x = 2, then
  • A
    $f(x)=-f(-x)$
  • ✓
    $f (2+x)= f (2-x)$
  • C
    $f(x)=f(-x)$
  • D
    $f (x+2)= f (x-2)$
Answer
Correct option: B.
$f (2+x)= f (2-x)$
(B)
$f (x)= f (-x) \Rightarrow f (0+x)= f (0-x)$ is
symmetrical about $x=0$.
$\therefore f (2+x)= f (2-x)$ is symmetrical about $x=2$.
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MCQ 962 Marks
If $f(x)=\cos \left[\pi^2\right] x+\cos \left[-\pi^2\right] x$, then
  • A
    $f\left(\frac{\pi}{4}\right)=2$
  • B
    $f(-\pi)=2$
  • C
    $f(\pi)=1$
  • ✓
    $f\left(\frac{\pi}{2}\right)=-1$
Answer
Correct option: D.
$f\left(\frac{\pi}{2}\right)=-1$
(D)
$f (x)=\cos \left[\pi^2\right] x+\cos \left[-\pi^2\right] x$
$f(x)=\cos (9 x)+\cos (-10 x)$
$\cdots\left[\begin{array}{c}\because \pi=3.14 \Rightarrow[9.85]=9 \\ \text { and }[-9.85]=-10\end{array}\right]$
$=\cos (9 x)+\cos (10 x)$
$=2 \cos \left(\frac{19 x}{2}\right) \cos \left(\frac{x}{2}\right)$
$\therefore \quad f\left(\frac{\pi}{2}\right)=2 \cos \left(\frac{19 \pi}{4}\right) \cos \left(\frac{\pi}{4}\right) ;$
$\therefore \quad f\left(\frac{\pi}{2}\right)=2 \times \frac{-1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=-1$
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MCQ 972 Marks
If $e ^{ f (x)}=\frac{10+x}{10-x}, x \in(-10,10)$ and $f ( x )= kf \left(\frac{200 x}{100+x^2}\right)$, then $k =$
  • ✓
    0.5
  • B
    0.6
  • C
    0.7
  • D
    0.8
Answer
Correct option: A.
0.5
(A)
$e ^{ f (x)}=\frac{10+x}{10-x}, x \in(-10,10)$
$\Rightarrow f (x)=\log \left(\frac{10+x}{10-x}\right)$
$\Rightarrow f \left(\frac{200 x}{100+x^2}\right)=\log \left[\frac{10+\frac{200 x}{100+x^2}}{10-\frac{200 x}{100+x^2}}\right]$
$=\log \left[\frac{10(10+x)}{10(10-x)}\right]^2$
$=2 \log \left(\frac{10+x}{10 x}\right)$
$=2 f (x)$
$\therefore f (x)=\frac{1}{2} f \left(\frac{200 x}{100+x^2}\right) \Rightarrow k =\frac{1}{2}=0.5$
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MCQ 982 Marks
If $f (x)=\log \left[\frac{1+x}{1-x}\right]$, then $f \left[\frac{2 x}{1+x^2}\right]$ is equal to
  • A
    $[ f (x)]^2$
  • B
    $[f(x)]^3$
  • ✓
    $2 f (x)$
  • D
    $3 f (x)$
Answer
Correct option: C.
$2 f (x)$
(C)
$f (x)=\log \left[\frac{1+x}{1-x}\right]$
$\therefore f\left(\frac{2 x}{1+x^2}\right)=\log \left[\frac{1+\frac{2 x}{1+x^2}}{1-\frac{2 x}{1+x^2}}\right]$
$=\log \left[\frac{x^2+1+2 x}{x^2+1-2 x}\right]$
$=\log \left[\frac{1+x}{1-x}\right]^2$
$=2 \log \left[\frac{1+x}{1-x}\right]$
$=2 f (x)$
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MCQ 992 Marks
Given the function $f (x)=\frac{ a + a }{2}, a > 2$, then
  • ✓
    $2 f (x) \cdot f (y)$
  • B
    $f (x) \cdot f (y)$
  • C
    $\frac{ f (x)}{ f (y)}$
  • D
    $\frac{1}{2} f (x) f (y)$
Answer
Correct option: A.
$2 f (x) \cdot f (y)$
(A)
$f (x+y)+ f (x-y)$
$=\frac{1}{2}\left[ a ^{x+y}+ a ^{-x-y}+ a ^{x-y}+ a ^{-x+y}\right]$
$=\frac{1}{2}\left[ a ^x\left( a ^y+ a ^{-y}\right)+ a ^{-x}\left( a ^y+ a ^{-y}\right)\right]$
$=\frac{1}{2}\left( a ^x+ a ^{-x}\right)\left( a ^y+ a ^{-y}\right)$
$=2 f (x) f (y)$
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MCQ 1002 Marks
If $f(x)=\cos (\log x)$, then$f(x) f(y)-\frac{1}{2}[f(x / y)+f(x y)]=$
  • A
    -1
  • B
    $\frac{1}{2}$
  • C
    -2
  • ✓
    $0$
Answer
Correct option: D.
$0$
(D)
Given, $f (x)=\cos (\log x) \Rightarrow f (y)=\cos (\log y)$
Then, $f (x) . f (y)-\frac{1}{2}\left[ f \left(\frac{x}{y}\right)+ f (x y)\right]$
$=\cos (\log x) \cos (\log y)$
$-\frac{1}{2}\left[\cos \left(\log \frac{x}{y}\right)+\cos (\log x y)\right]$
$=\cos (\log x) \cos (\log y)$
$-\frac{1}{2}[2 \cos (\log x) \cos (\log y)]=0$
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MCQ 1012 Marks
If for non-zero $x$, a.f $(x)+$ b.f $\left(\frac{1}{x}\right)=\frac{1}{x}-5$, where $a \neq b$, then $f(2)=$
  • A
    $\frac{3(2 b+3 a)}{2\left(a^2-b^2\right)}$
  • ✓
    $\frac{3(2 b-3 a)}{2\left(a^2-b^2\right)}$
  • C
    $\frac{3(3 a-2 b)}{2\left(a^2-b^2\right)}$
  • D
    $\frac{6}{a+b}$
Answer
Correct option: B.
$\frac{3(2 b-3 a)}{2\left(a^2-b^2\right)}$
(B)
a.f $(x)+$ b.f $\left(\frac{1}{x}\right)=\frac{1}{x}-5$
On replacing $x$ by $\frac{1}{x}$, b.f $(x)+ a . f \left(\frac{1}{x}\right)=x-5$
Solving two equations,
$f (x)=\frac{1}{ a ^2- b ^2}\left(\frac{ a }{x}- b x\right)-\frac{5}{ a + b }$
$\therefore f(2)=\frac{3(2 b-3 a)}{2\left(a^3-b^3\right)}$
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MCQ 1022 Marks
If $f (x)=x+\frac{1}{x}$, such that $[ f (x)]^3= f \left(x^3\right)+\lambda f \left(\frac{1}{x}\right)$, then $\lambda=$
  • A
    1
  • ✓
    3
  • C
    -3
  • D
    -1
Answer
Correct option: B.
3
(B)
$f (x)=x+\frac{1}{x} \Rightarrow f \left(x^3\right)=x^3+\frac{1}{x^3}$
$\therefore[ f (x)]^3=\left(x+\frac{1}{x}\right)^3=\left(x^3+\frac{1}{x^3}\right)+3\left(x+\frac{1}{x}\right)$
$\therefore[ f (x)]^3= f \left(x^3\right)+3 f (x)$
$\therefore[ f (x)]^3= f \left(x^5\right)+3 f \left(\frac{1}{x}\right) \Rightarrow \lambda=3$
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MCQ 1032 Marks
The value of b and c for which the identity $f (x+1)- f (x)=8 x+3$ is satisfied, where $f(x)=b x^2+c x+d$, are
  • A
    b = 2, c = 1
  • ✓
    b = 4, c = -1
  • C
    b = -1, c = 4
  • D
    b = -1, c = 1
Answer
Correct option: B.
b = 4, c = -1
(B)
$f (x+1)- f (x)=8 x+3$
$\begin{array}{r}\Rightarrow\left[ b (x+1)^2+ c (x+1)+ d \right]-\left( b x^2+ c x+ d \right) \\ =8 x+3\end{array}$
$\Rightarrow(2 b) x+( b + c )=8 x+3$
$\Rightarrow 2 b=8, b+c=3$
$\Rightarrow b=4, c=-1$
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MCQ 1042 Marks
If $f(x)=x^2-2 x+3$, then the value of $x$ for which $f ( x )= f ( x +1)$ is
  • ✓
    $1 / 2$
  • B
    $1 / 3$
  • C
    1
  • D
    3
Answer
Correct option: A.
$1 / 2$
(A)
$f (x)= f (x+1)$
$\therefore \quad x^2-2 x+3=(x+1)^2-2(x+1)+3$
$\therefore \quad x^2-2 x=x^2+2 x+1-2 x-2 \Rightarrow x=1 / 2$
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MCQ 1052 Marks
Let f : R → R satisfy f(x) f(y) = f(xy) for all real numbers x and y. If $f (2)=4$, then $f \left(\frac{1}{2}\right)=$
  • A
    $0$
  • ✓
    $\frac{1}{4}$
  • C
    $\frac{1}{2}$
  • D
    1
Answer
Correct option: B.
$\frac{1}{4}$
(B)
Since, $f(x) f(y)-f(x y)$
$\therefore f(1) \cdot f(2)=f(2)$
$\therefore f(1) \cdot 4=4$
$\therefore f(1)=1 \quad\ldots(i)$
Also, $f(2) \cdot f\left(\frac{1}{2}\right)=f(1)$
$\therefore 4 \times f\left(\frac{1}{2}\right)=1 \quad\ldots[From \ i]$
$\therefore f \left(\frac{1}{2}\right)=\frac{1}{4}$
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MCQ 1062 Marks
If for two functions g and f, gof is both injective and surjective, then which of the following is true?
  • ✓
    g and f should be injective and surjective
  • B
    g should be injective and surjective
  • C
    f should be injective and surjective
  • D
    None of them may be surjective and injective
Answer
Correct option: A.
g and f should be injective and surjective
A
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MCQ 1072 Marks
If $f(x)=\frac{3 x+4}{5 x-7}, g(x)=\frac{7 x+4}{5 x-3}$ then $f[g(x)]=$
  • A
    -41
  • ✓
    x
  • C
    -x
  • D
    41
Answer
Correct option: B.
x
(B)
$f [ g (x)]=\frac{3[g(x)]+4}{5[g(x)]-7}=\frac{3\left[\frac{7 x+4}{5 x-3}\right]+4}{5\left[\frac{7 x+4}{5 x-3}\right]-7}=x$
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MCQ 1082 Marks
If $f(x)=\frac{x-1}{x+1}$, then $f\left(\frac{1}{f(x)}\right)$ equals
  • A
    $0$
  • B
    1
  • C
    x
  • ✓
    $\frac{1}{2}$
Answer
Correct option: D.
$\frac{1}{2}$
(D)
$f (x)=\frac{x-1}{x+1}$
$\Rightarrow f \left(\frac{1}{ f (x)}\right)= f \left(\frac{x+1}{x-1}\right)=\frac{\frac{x+1}{x-1}-1}{\frac{x+1}{x-1}+1}=\frac{1}{x}$
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MCQ 1092 Marks
If $f (x)=1-\frac{1}{x}$, then $f \left( f \left(\frac{1}{x}\right)\right)$ is
  • A
    $\frac{1}{x}$
  • B
    $\frac{1}{1+x}$
  • ✓
    $\frac{x}{x-1}$
  • D
    $\frac{1}{x-1}$
Answer
Correct option: C.
$\frac{x}{x-1}$
(C)
$f \left( f \left(\frac{1}{x}\right)\right)= f \left(1-\frac{1}{1 / x}\right)$
$= f (1-x)=\frac{x}{x-1}$
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MCQ 1102 Marks
Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be given by $f(x)=x^2$ and $g(x)=x^3+1$, then $(fog)(x)$
  • A
    $x^6+1$
  • B
    $x^6-1$
  • C
    $\left(x^3-1\right)^2$
  • ✓
    $\left(x^3+1\right)^2$
Answer
Correct option: D.
$\left(x^3+1\right)^2$
(D)
$(fog)(x)=f[g(x)]=f\left(x^3+1\right)=\left(x^3+1\right)^2$
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MCQ 1112 Marks
If $f(x)=x^2+1$, then the value of (fof) (x) is equal to
  • A
    $x^4+1$
  • ✓
    $x^4+2 x^2+2$
  • C
    $x^4+x^2+1$
  • D
    none of these
Answer
Correct option: B.
$x^4+2 x^2+2$
(B)
$f ( f (x))= f \left(x^2+1\right)=\left(x^2+1\right)^2+1=x^4+2 x^2+2$
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MCQ 1122 Marks
If $f (x)=3 x-1, g(x)=x^2+1$ then $f [ g (x)]=$
  • ✓
    $3 x^2+2$
  • B
    $9 x^2-6 x+2$
  • C
    $3 x^2-2$
  • D
    $9 x^2+6 x-2$s
Answer
Correct option: A.
$3 x^2+2$
A
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MCQ 1132 Marks
If $f(x)=x^2, g(x)=5 x-6$, then $g[f(x)]=$
  • A
    $25 x^2-60 x+36$
  • B
    $5 x^2+6$
  • C
    $25 x^2+60 x-36$
  • ✓
    $5 x^2-6$
Answer
Correct option: D.
$5 x^2-6$
(D)
$g [ f (x)]=5[ f (x)]-6=5 x^2-6$
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MCQ 1142 Marks
If f = {(1, 4), (2, 5), (3, 6)} and g = {(4, 8), (5, 7), (6, 9)} then gof is
  • A
    {}
  • ✓
    ((1, 8), (2, 7), (3, 9)}
  • C
    ((1,7), (2, 8), (3, 9)}
  • D
    {(1, 8), (2, 5), (3, 9))
Answer
Correct option: B.
((1, 8), (2, 7), (3, 9)}
(B)
$(gof)(1)=g(f(1))=g(4)=8$,
$(gof)(2)=g(f(2))=g(5)=7$
and $(go f)(3)=g(f(3))=g(6)=9$
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MCQ 1152 Marks
If : R $\rightarrow$ R is defined as $f(x)=x^2-3 x+4$ for all $x \in R$, then $f ^{-1}(2)$ is equal to
  • ✓
    {1,2}
  • B
    (1,2)
  • C
    [1, 2]
  • D
    none of these
Answer
Correct option: A.
{1,2}
(A)
$f ^{-1}(y)=\{x \in R : y= f (x)\}$
$\Rightarrow f ^{-1}(2)=\{x \in R : 2= f (x)\}$
$=\left\{x \in R : x^2-3 x+4=2\right\}$
$=\left\{x \in R : x^2-3 x+2=0\right\}=\{1,2\}$
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MCQ 1162 Marks
If f : N $\rightarrow$ N f(x) = x + 3, then $f ^{-1}(x)=$ __________
  • A
    x + 3
  • ✓
    does not exists
  • C
    x - 3
  • D
    3 - x
Answer
Correct option: B.
does not exists
(B)
For $l \in N$ in co-domain we cannot find any
$x \in N$ in domain such that $f (x)=1$
∴ function is into $\Rightarrow f ^{-1}(x)$ does not exist.
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MCQ 1172 Marks
Inverse of the function $y=2 x-3$ is
  • ✓
    $\frac{x+3}{2}$
  • B
    $\frac{x-3}{2}$
  • C
    $\frac{1}{2 x-3}$
  • D
    $\frac{1}{x+3}$
Answer
Correct option: A.
$\frac{x+3}{2}$
(A)
$y=2 x-3 \Rightarrow x=\frac{y+3}{2}$
$\Rightarrow f ^{-1}(y)=\frac{y+3}{2} \Rightarrow f ^{-1}(x)=\frac{x+3}{2}$
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MCQ 1182 Marks
The range of the function $f (x)=\frac{x+2}{|x+2|}$ is
  • A
    {0,1}
  • ✓
    {-1, 1}
  • C
    R
  • D
    R - {- 2}
Answer
Correct option: B.
{-1, 1}
(B)
$f(x)=\frac{x+2}{|x+2|}$
$f(x)=\left\{\begin{array}{cc}-1, & x<-2 \\ 1, & x>-2\end{array}\right.$
$\therefore \quad$ Range of $f (x)$ is $\{-1,1\}$.
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MCQ 1192 Marks
The range of the function $f (x)=\sqrt{9-x^2}$ is
  • A
    (0,3)
  • ✓
    [0, 3]
  • C
    (0, 3]
  • D
    [0, 3)
Answer
Correct option: B.
[0, 3]
(B)
$f(r)=\sqrt{9-x^2}$
$f (0)=3, f (3)=0$
$\therefore \quad 0 \leq f (x) \leq 3$
$\therefore \quad x \in[0,3]$
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MCQ 1202 Marks
If the domain of function $f ( x )=x^2-6 x+7$ is $(-\infty, \infty)$, then the range of function is
  • A
    $(-\infty, \infty)$
  • ✓
    $[-2, \infty)$
  • C
    (-2, 3)
  • D
    $(-\infty,-2)$
Answer
Correct option: B.
$[-2, \infty)$
(B)
$x^2-6 x+7=(x-3)^2-2$
Here, minimum value is -2 and maximum $\infty$.
Hence, range of function is $[-2, \infty)$.
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MCQ 1212 Marks
The domain of the function of f: R $\rightarrow$ R defined by $f (x)=\sqrt{x^2-7 x+12}$ is
  • A
    $(-\infty, 3] \cup(4, \infty)$
  • B
    (3,4)
  • ✓
    $(-\infty, 3] \cup[4, \infty)$
  • D
    $(-\infty, 3] \cap[4, \infty)$
Answer
Correct option: C.
$(-\infty, 3] \cup[4, \infty)$
(C)
$x^2-7 x+12 \geq 0$
$\Rightarrow(x-4)(x-3) \geq 0$
$\Rightarrow x \in(-\infty, 3] \cup[4, \infty)$
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MCQ 1222 Marks
Domain of the function $\sqrt{\log \left\{\left(5 x-x^2\right) / 6\right\}}$ is
  • A
    (2, 3)
  • ✓
    [2, 3]
  • C
    [1, 2]
  • D
    [1, 3]
Answer
Correct option: B.
[2, 3]
(B)
$\log \left\{\frac{5 x-x^2}{6}\right\} \geq 0 \Rightarrow \frac{5 x-x^2}{6} \geq 1$
$\Rightarrow x^2-5 x+6 \leq 0$ or $(x-2)(x-3) \leq 0$.
Hence, $2 \leq x \leq 3$.
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MCQ 1232 Marks
Domain of the function $\log \left|x^2-9\right|$ is
  • A
    R
  • B
    $R -[-3,3]$
  • ✓
    $R -\{-3,3\}$
  • D
    $\{-3,3\}$
Answer
Correct option: C.
$R -\{-3,3\}$
(C)
For $x=-3,3,\left|x^2-9\right|=0$
Therefore, $\log \left|x^2-9\right|$ does not exist at
$x=-3,3$.
Hence, domain of function is $R -\{-3,3\}$
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MCQ 1242 Marks
Domain of function $f(x)=\sin ^{-1} 5 x$ is
  • A
    $\left(-\frac{1}{5}, \frac{1}{5}\right)$
  • ✓
    $\left[-\frac{1}{5}, \frac{1}{5}\right]$
  • C
    R
  • D
    $\left(0, \frac{1}{5}\right)$
Answer
Correct option: B.
$\left[-\frac{1}{5}, \frac{1}{5}\right]$
(B)
$-1 \leq 5 x \leq 1 \Rightarrow \frac{-1}{5} \leq x \leq \frac{1}{5}$
Hence, domain is $\left[\frac{-1}{5}, \frac{1}{5}\right]$.
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MCQ 1252 Marks
If $f (x)=\frac{1}{\sqrt{5 x-7}}$, then $\operatorname{dom}( f )=$
  • A
    $R-\left\{\frac{7}{5}\right\}$
  • B
    $\left[\frac{7}{5}, \infty\right)$
  • C
    $\left[\frac{5}{7}, \infty\right)$
  • ✓
    $\left(\frac{7}{5}, \infty\right)$
Answer
Correct option: D.
$\left(\frac{7}{5}, \infty\right)$
(D)
For $\operatorname{Dom}(f), 5 x-7>0 \Rightarrow x>\frac{7}{5}$
Hence, $D_f=\left(\frac{7}{5}, \infty\right)$
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MCQ 1262 Marks
If f : R $\rightarrow$ R then f(x) = |x| is
  • A
    One-one but not onto
  • B
    Onto but not one-one
  • C
    One-one and onto
  • ✓
    Many-one
Answer
Correct option: D.
Many-one
(D)
$f(-1)=f(1)=1$
∴ function is many-one function.
$\therefore f$ is neither one-one nor onto.
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MCQ 1302 Marks
If in greatest integer function, the domain is a set of real numbers, then range will be set of
  • A
    Real numbers
  • B
    Rational numbers
  • C
    Imaginary numbers
  • ✓
    Integers
Answer
Correct option: D.
Integers
(D)
$[x]= I$ (Integers only)
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MCQ 1312 Marks
Let f = {(1, 1) , (2, 4), (0, -2), (-1, -5)} be a linear function from Z into Z. Then, f(x) is
  • ✓
    $f(x)=3 x-2$
  • B
    $f(x)=6 x-8$
  • C
    $f(x)=5 x-2$
  • D
    $f(x)=7 x+2$
Answer
Correct option: A.
$f(x)=3 x-2$
(A)
$f=\{(1,1),(2,4),(0,-2),(-1,-5)\}$ be a linear
function from $Z$ to $Z$. The function satisfies the above points, if $f (x)=3 x-2$
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MCQ 1322 Marks
If f(x) is periodic function with period T, then the function f(ax + b) where a > 0, is periodic with period
  • A
    Т/b
  • B
    aT
  • C
    ЬТ
  • ✓
    T/a
Answer
Correct option: D.
T/a
(D)
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MCQ 1332 Marks
If $f(x)=4 x-x^2$, then $f(a+1)-f(a-1)=$
  • ✓
    4(2 - a)
  • B
    2(4 - a)
  • C
    4(2 + a)
  • D
    2(4 + a)
Answer
Correct option: A.
4(2 - a)
(A)
$f(a+1)-f(a-1)$
$=4(a+1)-(a+1)^2-\left[4(a-1)-(a-1)^2\right]$
$=4(2-a)$
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MCQ 1342 Marks
If f(x) = ax + 6 and f(1) = 11 then a =
  • A
    6
  • B
    17
  • C
    11
  • ✓
    5
Answer
Correct option: D.
5
(D)
$f (x)= a x+6 \Rightarrow f (1)= a (1)+6= a +6$
$f(1)=11 \Rightarrow 11=a+6 \Rightarrow a=5$
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MCQ 1372 Marks
If $f(x)=x^2-6 x+9,0 \leq x \leq 4$, then $f(3)=$
  • A
    4
  • B
    1
  • ✓
    $0$
  • D
    does not exist
Answer
Correct option: C.
$0$
(C)
$f (x)=x^2-6 x+9,0 \leq x \leq 4$
$f(3)=(3)^2-6(3)+9=0$
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MCQ 1382 Marks
If $f (x)=x^2+\frac{1}{x}, x \neq 0$ then $f \left(\frac{1}{x}\right)=$
  • ✓
    $\frac{1}{x^2}+x$
  • B
    $\frac{1}{x}+x^2$
  • C
    $\frac{1}{x^2}-x$
  • D
    $\frac{1}{x}-x^2$
Answer
Correct option: A.
$\frac{1}{x^2}+x$
(A)
$f (x)=x^2+\frac{1}{x}$
$f \left(\frac{1}{x}\right)=\left(\frac{1}{x}\right)^2+\frac{1}{\left(\frac{1}{x}\right)}=\frac{1}{x^2}+x$
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MCQ 1392 Marks
If a function $f (x)$ is given as $f (x)=x^2-3 x+2$ for all $x \in R$, then $f (a+h)=$
  • A
    $a^2+(2 a+3) h-3 a+2+h^2$
  • B
    $a^2+(2 a-3) h+3 a+2+h^2$
  • ✓
    $a^2+(2 a-3) h-3 a+2+h^2$
  • D
    $a^2+(2 a+3) h+3 a+2+h^2$
Answer
Correct option: C.
$a^2+(2 a-3) h-3 a+2+h^2$
(C)
$f(x)=x^2-3 x+2$
$f(a+h)=(a+h)^2-3(a+h)+2$
$=a^2+(2 a-3) h-3 a+2+h^2$
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MCQ 1402 Marks
Let $f : R \rightarrow R$ be defined by
$f (x)=\left\{\begin{array}{lc}2 x ; & x > 3 \\ x^2 ; & 1 < x \leq 3 \\ 3 x ; & x \leq 1\end{array}\right.$
Then $f(-1)+f(2)+f(4)$ is
  • ✓
    9
  • B
    14
  • C
    5
  • D
    10
Answer
Correct option: A.
9
(A)
$f (x)=\left\{\begin{array}{rc}2 x ; & x >3 \\ x^2 ; & 1< x \leq 3 . \\ 3 x ; & x \leq 1\end{array}\right.$
$\therefore \quad$ as $x=-1, f(x)=f(-1)=3(-1)=-3$
as $x=2, f (x)= f (2)=(2)^2=4$
as $x=4, f (x)= f (4)=2(4)=8$
$\therefore \quad f ( - 1 ) + f ( 2 ) + f ( 4 ) = 9$
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MCQ 1412 Marks
If a function f(x) is given as $f(x)=x^2-3 x+2$ for all $x \in R$, then f(- 1) =
  • ✓
    6
  • B
    $0$
  • C
    2
  • D
    8
Answer
Correct option: A.
6
(A)
$f (x)=x^2-3 x+2$
$\Rightarrow f(-1)=(-1)^2-3(-1)+2$
$=6$
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MCQ 1422 Marks
If the function $f : N \rightarrow N$ is defined by $f(x)=\sqrt{x}$, then $\frac{f(25)}{f(16)+f(1)}$ is equal to
  • A
    $\frac{5}{6}$
  • B
    $\frac{5}{7}$
  • C
    $\frac{5}{3}$
  • ✓
    1
Answer
Correct option: D.
1
(D)
$f (x)=\sqrt{x}$
$\Rightarrow \frac{ f (25)}{ f (16)+ f (1)}=\frac{\sqrt{25}}{\sqrt{16}+\sqrt{1}}=\frac{5}{5}=1$
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MCQ 1432 Marks
Let A = {1, 2, 3} and B = {2, 3, 4} then which of the following is a function from A to B?
  • A
    $\{(1,2),(1,3),(2,3),(3,3)\}$
  • B
    (0, 3), (2, 4))
  • ✓
    {(1, 3), (2, 3), (3, 3)}
  • D
    $\{(1,2),(2,3),(3,4),(3,2)\}$
Answer
Correct option: C.
{(1, 3), (2, 3), (3, 3)}
C
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