MCQ

MCQ 11 Mark

Let $f$ and $g$ be real function be $f(x)=\sqrt{x+4}, x \geq-4$ and $g(x)=\sqrt{x-4}, x \geq 4$. Then function $f g$ is

✓
$\sqrt{x^2-16}$
B
$\sqrt{x^2-4}$
C
$\sqrt{x^2+1}$
D
$\sqrt{x^2-1}$

Answer

Correct option: A.

$\sqrt{x^2-16}$

(a) $\sqrt{x^2-16}$
Explanation:
$
\begin{aligned}
f g=f(x) g(x) & =\sqrt{x+4} \sqrt{x-4} \\
& =\sqrt{x^2-16}
\end{aligned}
$

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MCQ 21 Mark

The domain and range of the function $f$ given by $f(x)=2-|x-5|$ is

A
Domain $=R^{+}$, Range $=(-\infty, 1]$
✓
Domain $=R$, Range $=(-\infty, 2]$
C
Domain $=R$, Range $=(-\infty, 2)$
D
Domain $=R^{+}$, Range $=(-\infty, 2]$

Answer

Correct option: B.

Domain $=R$, Range $=(-\infty, 2]$

(B) Domain $=R$, Range $=(-\infty, 2]$
Explanation : We have, $f(x)=2-|x-5|$
Clearly, $f(x)$ is defined for all $x \in R$
$\therefore$ Domain of $f=R$
Now, $|x-5| \geq 0, \forall x \in R$
$\Rightarrow \quad-|x-5| \leq 0$
$\Rightarrow \quad 2-|x-5| \leq 2$
$\therefore f(x) \leq 2$
$\therefore$ Range of $f=(-\infty, 2]$.

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MCQ 31 Mark

If $[x]^2-5[x]+6=0$, where $[$.$] denote the greatest$ integer function, then

A
$x \in[3,4]$
B
$x \in(2,3]$
✓
$x \in[2,3]$
D
$x \in[2,4)$

Answer

Correct option: C.

$x \in[2,3]$

(C) $x \in[2,3]$
Explanation : Given,
$
\begin{aligned}
& & {[x]^2-5[x]+6 } & =0 \\
\Rightarrow & & {[x]^2-2[x]-3[x]+6 } & =0 \\
\Rightarrow & & {[x]([x]-2)-3([x]-2) } & =0 \\
\Rightarrow & & ([x]-3)([x]-2) & =0 \\
\Rightarrow & & {[x] } & =2,3
\end{aligned}
$
Hence, $x \in[2,3]$

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MCQ 41 Mark

Let $f(x)=\sqrt{1+x^2}$, then

A
$f(x y)=f(x) \cdot f(y)$
B
$f(x y) \geq f(x) \cdot f(y)$
✓
$f(x y) \leq f(x) \cdot f(y)$
D
None of these

Answer

Correct option: C.

$f(x y) \leq f(x) \cdot f(y)$

(C) $f(x y) \leq f(x) \cdot f(y)$
Explanation : Given, $f(x)=\sqrt{1+x^2}$
$
\begin{array}{l}
\Rightarrow \quad f(x y)=\sqrt{1+x^2 y^2} \\
\text { and } \quad f(x) \cdot f(y)=\sqrt{1+x^2} \cdot \sqrt{1+y^2} \\
=\sqrt{1+x^2+y^2+x^2 y^2} \\
\because \quad \sqrt{1+x^2 y^2} \leq \sqrt{1+x^2+y^2+x^2 y^2} \\
\text { Therefore, } f(x y) \leq f(x) \cdot f(y)
\end{array}
$

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MCQ 51 Mark

Domain of $\sqrt{a^2-x^2}(a>0)$ is

A
$(-a, a)$
✓
$[-a, a]$
C
$[0, a]$
D
$(-a, 0]$

Answer

Correct option: B.

$[-a, a]$

(B) $[-a, a]$
Explanation : Let $f(x)=\sqrt{a^2-x^2} ; f(x)$ is defined
$
\begin{array}{l}
\sqrt{a^2-x^2} \geq 0 \\
\Rightarrow \quad x^2-a^2 \leq 0 \\
\Rightarrow \quad x^2 \leq a^2 \\
\Rightarrow \quad x \leq \pm a \\
\Rightarrow \quad-a \leq x \leq a \\
\therefore \text { Domain of } f(x)=[-a, a] \text {. }
\end{array}
$

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MCQ 61 Mark

If $f(x)=x^3-\frac{1}{x^3}$, then $f(x)+f\left(\frac{1}{x}\right)$ is equal to

A
$2 x^3$
B
$\frac{2}{x^3}$
✓
$0$
D
1

Answer

Correct option: C.

$0$

(C) $0$
Explanation : $\quad f(x)=x^3-\frac{1}{x^3}$
$
f\left(\frac{1}{x}\right)=\frac{1}{x^3}-\frac{1}{\frac{1}{x^3}}=\frac{1}{x^3}-x^3
$
Hence, $\quad f(x)+f\left(\frac{1}{x}\right)=x^3-\frac{1}{x^3}+\frac{1}{x^3}-x^3=0$

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MCQ 71 Mark

If $f(x)=a x+b$, where $a$ and $b$ are integers, $f(-1)$ $=-5$ and $f(3)=3$, then $a$ and $b$ are equal to

A
$a=-3, b=-1$
✓
$a=2, b=-3$
C
$a=0, b=2$
D
$a=2, b=3$

Answer

Correct option: B.

$a=2, b=-3$

(B) $a=2, b=-3$
Explanation : We have, $f(x)=a x+b$
$\therefore \quad f(-1)=a(-1)+b$
$\Rightarrow \quad-5=-a+b\quad \ldots(i) $
Also, $\quad f(3)=a(3)+b$
$\Rightarrow \quad 3=3 a+b\quad \ldots(ii)$
On solving eqs. (i) and (ii), we get $a=2, b=-3$.

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