M.C.Q (1 Marks)

MCQ 11 Mark

The domain of definition of the function $\text{f(x)}=\sqrt{\text{x}-1}+\sqrt{3-\text{x}}$ is:

A
$[1,\infty)$
B
$\big(-\infty,3\big)$
C
$(1,3)$
✓
$\big[1,3\big]$

Answer

Correct option: D.

$\big[1,3\big]$

$\text{f(x)}=\sqrt{\text{x}-1}+\sqrt{3-\text{x}}$
For f(x) to be defined,
$(\text{x}-1)\geq0$
$\Rightarrow\text{x}\geq1\ ...(\text{i})$
and $(3-\text{x})\geq0$
From (i) and (ii),
$\text{x}\in[1,3]$

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MCQ 21 Mark

If $\text{x}\neq1$ and $\text{f(x)}=\frac{\text{x}+1}{\text{x}-1}$ is a real function, then $\text{f}(\text{f}(\text{f(2)}))$ is:

A
1
B
2
✓
3
D
4

Answer

Correct option: C.

$\text{f(x)}=\frac{\text{x}+1}{\text{x}-1}$
$\text{f}(\text{f}(\text{f(2)}))$
$=\text{f}\Big(\text{f}\Big(\frac{2+1}{2-1}\Big)\Big)$
$=\text{f}(\text{f}(3))$
$=\text{f}\Big(\frac{3+1}{3-1}\Big)$
$=\text{f}(2)=3$

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MCQ 31 Mark

The domain of definition of $\text{f(x)}=\sqrt{\frac{\text{x}+3}{(2-\text{x})(\text{x}-5)}}$ is:

✓
$(-\infty,-3]\cup(2,5)$
B
$(-\infty,-3]\cup(2,5)$
C
$(-\infty,-3]\cup[2,5]$
D
None of these.

Answer

Correct option: A.

$(-\infty,-3]\cup(2,5)$

$\text{f(x)}=\sqrt{\frac{\text{x}+3}{(2-\text{x})(\text{x}-5)}}$
For f(x) to be defined,
$(2-\text{x})(\text{x}-5)\neq0$
$\Rightarrow\text{x}\neq2,5\ ...(\text{i})$
Also, $\frac{(\text{x}+3)}{(2-\text{x})(\text{x}-5)}\geq0$
$\Rightarrow\frac{(\text{x}+3)(2-\text{x})(\text{x}-5)}{(2-\text{x})^2(\text{x}-5)^2}\geq0$
$\Rightarrow(\text{x}+3)(\text{x}-2)(\text{x}-5)\leq0$
$\Rightarrow\text{x}\in\big(-\infty,-3\big]\cap(2,5)\ ...(\text{ii})$
From (i) and (ii)
$\text{x}\in\big(-\infty,-3\big]\cup(2,5)$

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MCQ 41 Mark

The range of the function $\text{f(x)}=\frac{\text{x}^2-\text{x}}{\text{x}^2+2\text{x}}$ is:

A
$\text{R}$
B
$\text{R}-\{1\}$
✓
$\text{R}-\Big\{\frac{1}{2},1\Big\}$
D
None of these.

Answer

Correct option: C.

$\text{R}-\Big\{\frac{1}{2},1\Big\}$

$\text{f(x)}=\frac{\text{x}^2-\text{x}}{\text{x}^2+2\text{x}}$
Let, $\text{y}=\frac{\text{x}^2-\text{x}}{\text{x}^2+2\text{x}}$ $\big[\text{Also},\text{ x}\neq0\big]$
$\Rightarrow\text{y}=\frac{\text{x}(\text{x}-1)}{\text{x}(\text{x}+2)}$
$\Rightarrow\text{y}=\frac{(\text{x}-1)}{(\text{x}+2)}$
$\Rightarrow\text{xy}+2\text{y}=\text{x}-1$
$\Rightarrow\text{x}=\frac{2\text{y}+1}{1-\text{y}}$
Here, $1-\text{y}\neq0$
Or, $\text{y}\neq1$
Also, $\text{x}\neq0$
$\Rightarrow\frac{2\text{y}+1}{1-\text{y}}\neq0$
$\Rightarrow\text{y}\neq-\frac{1}{2}$
Thus, range $\text{(f)}=\text{R}-\Big\{-\frac{1}{2},1\Big\}$

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MCQ 51 Mark

If $f : R \rightarrow R$ and $g : R \rightarrow R$ are defined by $f(x) = 2x + 3$ and $g(x) = x^2 + 7,$ then the values of $x$ such that $g(f(x)) = 8$ are:

A
$1, 2$
B
$-1, 2$
✓
$-1, -2$
D
$1, -2$

Answer

Correct option: C.

$-1, -2$

$f(x) = 2x + 3$ and $g(x) = x^2 + 7$
$g(f(x)) = 8$
$\Rightarrow (f(x))^2 + 7 = 8$
$\Rightarrow (2x + 3)^2 + 7 = 8$
$\Rightarrow x^2 + 3x + 2 = 0$
$\Rightarrow (x + 2)(x + 1) = 0$
$\Rightarrow x = -1, -2$

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MCQ 61 Mark

If $\text{e}^{\text{f(x)}}=\frac{10+\text{x}}{10-\text{x}},\text{ x}\in(-10,10)$ and $\text{f(x)}=\text{kf}\Big(\frac{200\text{x}}{100+\text{x}^2}\Big),$ then k =

✓
0.5
B
0.6
C
0.7
D
0.8

Answer

Correct option: A.

0.5

$\text{e}^{\text{f(x)}}=\frac{10+\text{x}}{10-\text{x}}$
$\Rightarrow\text{ f(x)}=\log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)\ ...(\text{i})$
$\Rightarrow\ \text{f(x)}=\text{kf}\Big(\frac{200\text{x}}{100+\text{x}^2}\Big)$
$\Rightarrow\ \log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)=\text{k}\log_\text{e}\Bigg(\frac{10+\frac{200\text{x}}{100+\text{x}^2}}{10-\frac{200\text{x}}{100+\text{x}^2}}\Bigg)$ {from (1)}
$\Rightarrow\ \log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)=\text{k}\log_\text{e}\Big(\frac{1000+10\text{x}^2+200\text{x}}{1000+10\text{x}^2-200\text{x}}\Big)$
$\Rightarrow\ \log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)=\text{k}\log_\text{e}\bigg(\frac{(\text{x}+10)^2}{(\text{x}-10)^2}\bigg)$
$\Rightarrow\ \log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)=2\text{k}\log_\text{e}\frac{(\text{x}+10)}{(\text{x}+10)}$
$\Rightarrow\ 1=2\text{k}$
$\Rightarrow\ \text{k}=\frac{1}{2}=0.5$

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MCQ 71 Mark

Let f(x) = x, $\text{g(x)}=\frac{1}{\text{x}}$ and h(x) = f(x) g(x). Then, h(x) = 1

A
$\text{x}\in\text{R}$
B
$\text{x}\in\text{Q}$
C
$\text{x}\in\text{R}-\text{Q}$
✓
$\text{x}\in\text{R},\text{ x}\neq0$

Answer

Correct option: D.

$\text{x}\in\text{R},\text{ x}\neq0$

Given,
$\text{f(x)}=\text{x},\text{ g(x)}=\frac{1}{\text{x}}$ and $\text{h(x)}=\text{f(x)}\text{g(x)}$
Now,
$\text{h(x)}=\text{x}\times\frac{1}{\text{x}}=1$
We observe that the domain of f is R and the domain of g is R - {0}
$\therefore\ \text{Domain of h}=\text{Domain of f }\cap\text{ Domain of g}\\\ \ \ =\text{R }\cap\big[\text{R}-\{0\}\big]=\text{R}-\{0\}$
$\Rightarrow\text{x}\in\text{R},\text{ x}\neq0$

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MCQ 81 Mark

The range of the function f(x) = |x - 1| is:

A
$\big(-\infty,0\big)$
✓
$\big[0,\infty\big)$
C
$\big(0,\infty\big)$
D
$\text{R}$

Answer

Correct option: B.

$\big[0,\infty\big)$

$\text{f(x)}=|\text{x}-1|\geq0\ \forall\text{ x}\in\text{R}$
Thus, range $=\big[0,\infty\big)$

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MCQ 91 Mark

Let f : R → R be defined by f(x) = 2x + |x|. Then f(2x) + f(-x) - f(x) =

A
2x
✓
2|x|
C
-2x
D
-2|x|

Answer

Correct option: B.

2|x|

f(x) = 2x + |x|
Then, f(2x) + f(-x) - f(x)
= 2(2x) + 2|x| + (-2x) + |-x| - 2x + |x|
= 4x - 2x - 2x + 2|x| + |-x| - |x|
= 0 + 2|x| + |x| - |x| = 2|x|
= 2|x|

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MCQ 101 Mark

The domain of definition of the function $\text{f(x)}=\sqrt{\frac{\text{x}-2}{\text{x}+2}}+\sqrt{\frac{1-\text{x}}{1+\text{x}}}$ is:

A
$\big(-\infty,-2\big]\cap\big[2,-\infty\big)$
B
$\big[-1,1\big]$
✓
$\phi$
D
None of these.

Answer

Correct option: C.

$\phi$

$\text{f(x)}=\sqrt{\frac{\text{x}-2}{\text{x}+2}}+\sqrt{\frac{1-\text{x}}{1+\text{x}}}$
For f(x) to be defined,
$\text{x}+2\neq0$
$\Rightarrow\text{x}\neq-2\ ...(\text{i})$
And $1+\text{x}\neq0$
$\Rightarrow\text{x}\neq-1\ ...(\text{ii})$
Also, $\frac{\text{x}-2}{\text{x}+2}\geq0$
$\Rightarrow\frac{(\text{x}-2)(\text{x}-2)}{(\text{x}-2)^2}\ge0$
$\Rightarrow(\text{x}-2)(\text{x}+2)\geq0$
$\Rightarrow\text{x}\in(\infty,-2)\cup\big[2,\infty\big)\ ...(\text{iii})$
And $\frac{1-\text{x}}{1+\text{x}}\geq0$
$\Rightarrow\frac{(1-\text{x})(1+\text{x})}{(1+\text{x})^2}\geq0$
$\Rightarrow(1-\text{x})(1+\text{x})\geq0$
$\Rightarrow\text{ x}\in\big(-\infty,-1\big)\cup\big[1,\infty\big)\ ...(\text{iv})$
From (i), (ii), (iii) and (iv) we get
$\text{x }\in\phi$
Thus, domain $(\text{f(x)})=\phi$

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MCQ 111 Mark

If $\text{f(x)}=\cos(\log\text{x}),$ then value of $\text{f(x)}\text{f(4)}-\frac{1}{2}\Big\{\text{f}\Big(\frac{\text{x}}{4}\Big)+\text{f}(4\text{x})\Big\}$ is:

A
1
B
-1
✓
0
D
$\pm1$

Answer

Correct option: C.

Given, $\text{f(x)}=\cos(\log\text{x})$
Then, $\text{f(x)}\text{f(4)}-\frac{1}{2}\Big\{\text{f}\Big(\frac{\text{x}}{4}\Big)+\text{f}(4\text{x})\Big\}$
$=\cos(\log\text{x})\cos(\log4)+\frac{1}{2}\Big\{\cos\Big(\log\frac{\text{x}}{4}\Big)+\cos(\log4\text{x})\Big\}$
$=\frac{1}{2}\big[\cos(\log\text{x}+\log4\big)+\cos(\log\text{x}-\log4)\big]\\-\frac{1}{2}\Big\{\cos\Big(\log\frac{\text{x}}{4}\Big)+\cos(\log4\text{x})\Big\}$
$=\frac{1}{2}\Big\{\cos(\log4\text{x})+\cos\Big(\log\frac{\text{x}}{4}\Big)-\cos\Big(\log\frac{\text{x}}{4}\Big)-\cos(\log4\text{x})\Big\}$
$=\frac{1}{2}\times0=0$

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MCQ 121 Mark

If f : [-2, 2] → R is defined by $\text{f(x)}=\begin{cases}-1,&\text{for}-2\leq\text{x}\leq0\\\text{x}-1,&\text{for }0\leq\text{x}\leq2\end{cases}$ then $\{\text{x}\in[-2,2]:\text{x}\leq0\text{ and }\text{f}(\text{|x|})=\text{x}\}=$

A
$\{-1\}$
B
$\{0\}$
✓
$\Big\{-\frac{1}{2}\Big\}$
D
$\phi$

Answer

Correct option: C.

$\Big\{-\frac{1}{2}\Big\}$

Given,
$\text{f(x)}=\begin{cases}-1,&\text{for}-2\leq\text{x}\leq0\\\text{x}-1,&\text{for }0\leq\text{x}\leq2\end{cases}$
We know,
$|\text{x}|\geq0$
$\Rightarrow\text{f(|x|)}=|\text{x}|-1\ ...(\text{i})$
Also,
If $\text{x}\leq0,$ then $|\text{x}|=-\text{x}\ ...(\text{ii})$
$\{\text{x}\in[-2,2]:\text{x}\leq0\text{ and }\text{f}(\text{|x|})=\text{x}\}$
$=\{\text{x}:|\text{x}|-1=\text{x}\}$ [Using(i)]
$=\{\text{x}:-\text{x}-1=\text{x}\}$ [Using (ii)]
$=\Big\{\text{x}:2\text{x}=\frac{-1}{2}\Big\}$
$=\Big\{\text{x}:\text{x}=\frac{-1}{2}\Big\}$
$=\Big\{-\frac{1}{2}\Big\}$

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MCQ 131 Mark

The domain of definition of $\text{f(x)}=\sqrt{\text{x}-3-2\sqrt{\text{x}-4}}-\sqrt{\text{x}-3+2\sqrt{\text{x}-4}}$ is:

✓
$\big[4,\infty\big)$
B
$\big(-\infty,4\big]$
C
$(4,\infty)$
D
$(-\infty,4)$

Answer

Correct option: A.

$\big[4,\infty\big)$

$\text{f(x)}=\sqrt{\text{x}-3-2\sqrt{\text{x}-4}}-\sqrt{\text{x}-3+2\sqrt{\text{x}-4}}$
For f(x) to be defined, $\text{x}-4\geq0$
$\Rightarrow\text{x}-4\geq0$
$\Rightarrow\text{x}\geq4\ ...(\text{i})$
Also, $\text{x}-3-2\sqrt{\text{x}-4}\geq0$
$\Rightarrow\text{x}-3-2\sqrt{\text{x}-4}\geq0$
$\Rightarrow\text{x}-3\geq2\sqrt{\text{x}-4}$
$\Rightarrow(\text{x}-3)^2\geq\big(2\sqrt{\text{x}-4}\big)^2$
$\Rightarrow\text{x}^2+9-6\text{x}\geq4(\text{x}-4)$
$\Rightarrow\text{x}^2-10\text{x}+25\geq0$
$\Rightarrow\big(\text{x}-5\big)^2\geq0,$ which is always true.
Similarly, $\text{x}-3+2\sqrt{\text{x}-4}\geq0$ is always true.
Thus, domain $(\text{f(x)})=\big[4,\infty)$

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MCQ 141 Mark

Which one of the following is not a function?

A
$\{(\text{x, y}):\text{x},\text{ y}\in\text{R},\text{x}^2=\text{y}\}$
✓
$\{(\text{x, y}):\text{x},\text{ y}\in\text{R},\text{y}^2=\text{x}\}$
C
$\{(\text{x, y}):\text{x},\text{ y}\in\text{R},\text{x}^2=\text{y}^3\}$
D
$\{(\text{x, y}):\text{x},\text{y}\in\text{R},\text{y}=\text{x}^3\}$

Answer

Correct option: B.

$\{(\text{x, y}):\text{x},\text{ y}\in\text{R},\text{y}^2=\text{x}\}$

$y^2 = x$ gives two values of $y$ for a value of $x$
i.e. there are two images for a value of $x.$
For example: $(2)^2 = 4$ and $(-2)^2 =4 $
Thus, it is not a function.

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MCQ 151 Mark

If $\text{f(x)}=\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$ and $\text{g(x)}=\frac{3\text{x}+\text{x}^3}{1+3\text{x}^2},$ then $f(g(x))$ is equal to:

A
$f(3x)$
B
${f(x)}^3$
✓
$3f(x)$
D
$-f(x)$

Answer

Correct option: C.

$3f(x)$

$\text{f(x)}=\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$ and $\text{g(x)}=\frac{3\text{x}+\text{x}^3}{1+3\text{x}^2}$
Now,
$=\frac{1+\text{g(x)}}{1-\text{g(x)}}$
$=\frac{1+\frac{3\text{x}+\text{x}^3}{1+3\text{x}^2}}{1-\frac{3\text{x}+\text{x}^3}{1+3\text{x}^2}}$
$=\frac{1+3\text{x}^2+3\text{x}+\text{x}^3}{1+3\text{x}^2-3\text{x}-\text{x}^3}$
$=\frac{(1+\text{x})^3}{(1-\text{x})^3}$
Then, $\text{f}(\text{g(x)})=\log$
$=\log\Big(\frac{1+\text{g(x)}}{1-\text{g(x)}}\Big)$
$=\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)^3$
$=3\text{f}(\text{x})$

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MCQ 161 Mark

If $\text{f(x)}=\cos(\log\text{x}),$ then the value of $\text{f(x})\text{f}(\text{y})-\frac{1}{2}\Big\{\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f}\big(\text{x}\text{y}\big)\Big\}$ is:

A
-1
B
$\frac{1}{2}$
C
-2
✓
None of these.

Answer

Correct option: D.

None of these.

Given,
$\text{f(x)}=\cos(\log\text{x})$
$\therefore\ \text{f(y)}=\cos(\log\text{y})$
Now,
$\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)=\cos\Big(\cos\Big(\frac{\text{x}}{\text{y}}\Big)\Big)=\cos(\log\text{x}-\log\text{y})$
and
$\text{f(xy)}=\cos(\log\text{xy})=\cos(\log\text{x}+\log\text{y})$
$\Rightarrow\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}=\cos(\log\text{x}-\log\text{y})+\cos(\log\text{x}+\log\text{y})$
$\Rightarrow\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}=2\cos(\log\text{x})\cos(\log\text{y})$
$\Rightarrow\frac{1}{2}\Big[\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}\Big]=\cos(\log\text{x})\cos(\log\text{y})$
$\Rightarrow\text{f(x})\text{f}(\text{y})-\frac{1}{2}\Big\{\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f}\big(\text{x}\text{y}\big)\Big\}\\=\cos(\log\text{x})\cos(\log\text{y})-\cos(\log\text{x})\cos(\log\text{y})=0$

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MCQ 171 Mark

The range of the function $\text{f(x)}=\frac{\text{x}}{|\text{x}|}$ is:

A
R - {0}
B
R - {-1, 1}
✓
{-1, 1}
D
None of these.

Answer

Correct option: C.

{-1, 1}

$\text{f(x)}=\frac{\text{x}}{|\text{x}|}$
Let $\text{y}=\frac{\text{x}}{|\text{x}|}$
For x > 0, |x| = x
$\Rightarrow\text{y}=\frac{\text{x}}{\text{x}}=1$
For x < 0, = -x
$\Rightarrow\text{y}=\frac{\text{x}}{-\text{x}}=-1$
Thus, range of f(x) is {-1, 1}

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MCQ 181 Mark

The range of $\text{f(x)}=\cos[\text{x}],$ for $-\frac{\pi}{2}<\text{x}<\frac{\pi}{2}$ is:

A
$\{-1,1,0\}$
✓
$\{\cos1,\cos2,1\}$
C
$\{\cos1,-\cos1,1\}$
D
$[-1,1]$

Answer

Correct option: B.

$\{\cos1,\cos2,1\}$

Since, $\text{f(x)}=\cos[\text{x}],$ where $\frac{-\pi}{2}<\text{x}<\frac{\pi}{2}$
$-\frac{\pi}{2}<\text{x}<\frac{\pi}{2}$
$\Rightarrow-1.57<\text{x}<1.57$
$\Rightarrow[\text{x}]\ \in\ \{-1,0,1,2\}$
Thus, $\cos[\text{x}]=\{\cos(-1),\cos0,\cos1,\cos2\}$
Range of $\text{f(x)}=\{\cos1,1,\cos2\}$

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MCQ 191 Mark

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
{(1, 3), (2, 4)}
✓
{(1, 3), (2, 2), (3, 3)}
D
{(1, 2), (2, 3), (3, 2), (3, 4)}

Answer

Correct option: C.

{(1, 3), (2, 2), (3, 3)}

We have,
R = {(1, 3), (2, 2), (3, 3)}
We observe that each element of the given set has appeared as first component in one and only one ordered pair of R.
So, R = {(1, 3), (2, 2), (3, 3)} is a function.

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MCQ 201 Mark

If $2\text{f(x)}-3\text{f}\Big(\frac{1}{\text{x}}\Big)=\text{x}^2(\text{x}\neq0),$ then f(2) is equal to:

✓
$-\frac{7}{4}$
B
$\frac{5}{2}$
C
$-1$
D
None of these.

Answer

Correct option: A.

$-\frac{7}{4}$

$2\text{f(x)}-3\text{f}\Big(\frac{1}{\text{x}}\Big)=\text{x}^2\ ...(\text{i})$ $(\text{x}\neq0)$
Replacing x by $\frac{1}{\text{x}}$
$2\text{f}\Big(\frac{1}{\text{x}}\Big)-3\text{f(x)}=\frac{1}{\text{x}^2}\ ...(\text{ii})$
Solving equations (i) & (ii)
$-5\text{f(x)}=\frac{3}{\text{x}^2}+2\text{x}^2$
$\Rightarrow\text{f(x)}=\frac{-1}{5}\Big(\frac{3}{\text{x}^2}+2\text{x}^2\Big)$
Thus, $\text{f(2)}=\frac{-1}{5}\Big(\frac{3}{4}+2\times4\Big)$
$=\frac{-1}{5}\Big(\frac{3+32}{4}\Big)$
$=-\frac{7}{4}$

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MCQ 211 Mark

If $\text{f(x)}=\frac{2^{\text{x}}+2^{-\text{x}}}{2},$ then f(x + y)f(x - y) is equal to:

✓
$\frac{1}{2}\big[\text{f(2}\text{x})+\text{f}(2\text{y})\big]$
B
$\frac{1}{2}\big[\text{f(2}\text{x})-\text{f}(2\text{y})\big]$
C
$\frac{1}{4}\big[\text{f(2}\text{x})+\text{f}(2\text{y})\big]$
D
$\frac{1}{4}\big[\text{f(2}\text{x})-\text{f}(2\text{y})\big]$

Answer

Correct option: A.

$\frac{1}{2}\big[\text{f(2}\text{x})+\text{f}(2\text{y})\big]$

Given,
$\text{f(x)}=\frac{2^{\text{x}}+2^{-\text{x}}}{2}$
Now,
$\text{f}(\text{x}+\text{y})\text{f}(\text{x}-\text{y})=\Big(\frac{2^{\text{x}+\text{y}}+2^{-\text{x}-\text{y}}}{2}\Big)\Big(\frac{2^{\text{x}-\text{y}}+2^{-\text{x}+\text{y}}}{2}\Big)$
$\Rightarrow\ \text{f}(\text{x}+\text{y})\text{f}(\text{x}-\text{y})=\frac{1}{4}\big(2^{2\text{x}}+2^{-2\text{y}}+2^{2\text{y}}+2^{-2\text{x}}\big)$
$\Rightarrow\ \text{f}(\text{x}+\text{y})\text{f}(\text{x}-\text{y})=\frac{1}{2}\Big(\frac{2^{2\text{x}}+2^{-2\text{x}}}{2}+\frac{2^{2\text{y}}+2^{-2\text{y}}}{2}\Big)$
$\Rightarrow\text{f}(\text{x}+\text{y})\text{f}(\text{x}-\text{y})=\frac{1}{2}\big[\text{f}(2\text{x})+\text{f}(2\text{y})\big]$

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MCQ 221 Mark

If $[\text{x}^2]-5[\text{x}]+6=0,$ where [.] denotes the greatest integer function, then:

A
$\text{x}\in[3,4]$
B
$\text{x}\in\big(2,3\big]$
C
$\text{x}\in\big[2,3\big]$
✓
$\text{x}\in\big[2,4\big)$

Answer

Correct option: D.

$\text{x}\in\big[2,4\big)$

The given equation is $[\text{x}^2]-5[\text{x}]+6=0$
$[\text{x}^2]-5[\text{x}]+6=0$
$\Rightarrow[\text{x}^2\big]-3\big[\text{x}\big]-2\big[\text{x}\big]+6=0$
$\Rightarrow\big[\text{x}\big]\big([\text{x}]-3\big)-2\big([\text{x]}-3\big)=0$
$\Rightarrow\big([\text{x}]-2)\big([\text{x}]-3)=0$
$\Rightarrow\big[\text{x}\big]-2=0$ or $[\text{x}]-3=0$
$\Rightarrow[\text{x}]=2$ or $[\text{x}]=3$
$\Rightarrow\text{x}\in\big[2,3\big)$ or $\big[3,4\big)$
$\Rightarrow\text{x}\in\big[2,4\big)$

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MCQ 231 Mark

If $\text{f(x)}=\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big),$ then $\text{f}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ is equal to:

A
$\{f(x)\}^2$
B
$\{f(x)\}^3$
✓
$2f(x)$
D
$3f(x)$

Answer

Correct option: C.

$2f(x)$

$\text{f(x)}=\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$
Then, $\text{f}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
$=\log\Bigg(\frac{1+\frac{2\text{x}}{1+\text{x}^2}}{1-\frac{2\text{x}}{1+\text{x}^2}}\Bigg)$
$=\log\Bigg(\frac{\frac{1+\text{x}^2+2\text{x}}{1+\text{x}^2}}{1-\frac{2\text{x}}{1+\text{x}^2}}\Bigg)$
$=\log\bigg(\frac{(1+\text{x})^2}{(1-\text{x})^2}\bigg)$
$=2\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$
$=2(\text{f(x)})$

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MCQ 241 Mark

The range of the function $\text{f(x)}=\frac{\text{x}+2}{|\text{x}+2|},\text{ x}\neq-2$ is:

✓
{-1, 1}
B
{-1, 0, 1}
C
{1}
D
$(0,\infty)$

Answer

Correct option: A.

{-1, 1}

$\text{f(x)}=\frac{\text{x}+2}{|\text{x}+2|},\text{ x}\neq-2$
Let $\text{y}=\frac{\text{x}+2}{|\text{x}+2|}$
For |x + 2| > 0
Or x > -2
$\text{y}=\frac{\text{x}+2}{\text{x}+2}=1$
For |x + 2| < 0
Or x < -2
$\text{y}=\frac{\text{x}+2}{-(\text{x}+2)}=-1$
Thus, y = {-1, 1}
Or range f(x) = {-1, 1}

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MCQ 251 Mark

If $A = \{1, 2, 3\}$ and $B = \{x, y\},$ then the number of functions that can be defined from $A$ into $B$ is:

A
$12$
✓
$8$
C
$6$
D
$3$

Answer

Correct option: B.

$8$

Given,
Number of elements in set $A = 3$
Number of elements in set $B = 2$
Therefore, the number of functions that can be defined from $A$ into $B$ is $= 2^3 = 8$

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MCQ 261 Mark

If $3\text{f(x)}+5\text{f}\Big(\frac{1}{\text{x}}\Big)=\frac{1}{\text{x}}-3$ for all non-zero x, then f(x) =

A
$\frac{1}{14}\Big(\frac{3}{\text{x}}+5\text{x}-6\Big)$
B
$\frac{1}{14}\Big(-\frac{3}{\text{x}}+5\text{x}-6\Big)$
C
$\frac{1}{14}\Big(-\frac{3}{\text{x}}+5\text{x}+6\Big)$
✓
None os these.

Answer

Correct option: D.

None os these.

$3\text{f(x)}+5\text{f}\Big(\frac{1}{\text{x}}\Big)=\frac{1}{\text{x}}-3\ ...(\text{i})$
Multiplying (1) by 3,
$15\text{f}\Big(\frac{1}{\text{x}}\Big)+9\text{f(x)}=\frac{3}{\text{x}}-9\ ...(\text{ii})$
Replacing x by $\frac{1}{\text{x}}$ in (i)
$3\text{f}\Big(\frac{1}{\text{x}}\Big)+5\text{f(x)}=\text{x}-3$
Multiplying by 5
$15\text{f}\Big(\frac{1}{\text{x}}\Big)+25\text{f(x)}=5\text{x}-15\ ...(\text{iii})$
Solving (ii) and (iii),
$-16\text{f(x)}=\frac{3}{\text{x}}-5\text{x}+6$
$\Rightarrow\text{f(x)}=\frac{1}{16}\Big(-\frac{3}{\text{x}}+5\text{x}-6\Big)$
Disclaimer: The question in the book has some error, so, none of the options are matching with the solution. The solution is created according to the question given in the book.

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MCQ 271 Mark

The function f : R → R is defined by $\text{f(x)}=\cos^2\text{x}+\sin^4\text{x}.$ Then, f(R) =

A
$\Big[\frac{3}{4},1\Big]$
B
$\Big(\frac{3}{4},1\Big]$
✓
$\Big[\frac{3}{4},1\Big]$
D
$\Big(\frac{3}{4},1\Big)$

Answer

Correct option: C.

$\Big[\frac{3}{4},1\Big]$

Given,
$\text{f(x)}=\cos^2\text{x}+\sin^4\text{x}$
$\Rightarrow\text{f(x)}=1-\sin^2\text{x}+\sin^4\text{x}$
$\Rightarrow\text{f(x)}=\Big(\sin^2\text{x}-\frac{1}{2}\Big)^2+\frac{3}{4}$
The minimum value of $\text{f(x)}$ is $\frac{3}{4}$
Also,
$\sin^2\text{x}\leq1$
$\Rightarrow\ \sin^2\text{x}-\frac{1}{2}\leq\frac{1}{2}$
$\Rightarrow\ \Big(\sin^2\text{x}-\frac{1}{2}\Big)^2\leq\frac{1}{4}$
$\Rightarrow\ \Big(\sin^2\text{x}-\frac{1}{2}\Big)^2+\frac{3}{4}\leq\frac{1}{4}+\frac{3}{4}$
$\Rightarrow\ \text{f(x)}\leq1$
The maximum value of f(x) is 1
$\therefore\ \text{f(R)}=\Big(\frac{3}{4},1\Big)$

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MCQ 281 Mark

The domain of definition of $\text{f(x)}=\sqrt{4\text{x}-\text{x}^2}$ is:

A
$\text{R}-[0,4]$
B
$\text{R}-(0,4)$
C
$(0,4)$
✓
$[0,4]$

Answer

Correct option: D.

$[0,4]$

$\text{f(x)}=\sqrt{4\text{x}-\text{x}^2}$
Clearly, f(x) assumes real values if
$4\text{x}-\text{x}^2\geq0$
$\Rightarrow\text{x}(4-\text{x})\geq0$
$\Rightarrow-\text{x}(\text{x}-4)\geq0$
$\Rightarrow\text{x}(\text{x}-4)\leq0$
$\Rightarrow\text{x}\in[0,4]$
Hence, domain $(\text{f})=[0,4]$

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MCQ 291 Mark

If $\text{f(x)}=64\text{x}^3+\frac{1}{\text{x}^3}$ and $\alpha,\beta$ are the roots of $4\text{x}+\frac{1}{\text{x}}=3.$ Then,

✓
$\text{f}(\alpha)=\text{f}(\beta)=-9$
B
$\text{f}(\alpha)=\text{f}(\beta)=63$
C
$\text{f}(\alpha)\neq\text{f}(\beta)$
D
None of these.

Answer

Correct option: A.

$\text{f}(\alpha)=\text{f}(\beta)=-9$

$\text{f(x)}=64\text{x}^3+\frac{1}{\text{x}^3}$
$\Rightarrow\text{f(x)}=\Big(4\text{x}+\frac{1}{\text{x}}\Big)\Big(16\text{x}^2+\frac{1}{\text{x}^2}-4\Big)$
$\Rightarrow\text{f(x)}=\Big(4\text{x}+\frac{1}{\text{x}}\Big)\Big(\Big(4\text{x}+\frac{1}{\text{x}}\Big)^2-12\Big)$
$\Rightarrow\text{f}(\text{a})=\Big(4\alpha+\frac{1}{\alpha}\Big)\Big(\Big(4\alpha+\frac{1}{\alpha}\Big)^2-12\Big)$ and $\text{f}(\beta)=\Big(4\beta+\frac{1}{\beta}\Big)\Big(\Big(4\beta+\frac{1}{\beta}\Big)^2-12\Big)$
Since $\alpha$ and $\beta$ are the roots of $4\text{x}+\frac{1}{\text{x}}=3,$
$4\alpha+\frac{1}{\alpha}=3$ and $4\beta+\frac{1}{\beta}=3$
$\Rightarrow\text{f}(\alpha)=3\big((3)^2-12\big)=-9$ and $\text{f}(\beta)=3\big((3)^2-12\big)=-9$
$\Rightarrow\text{f}(\alpha)=\text{f}(\beta)=-9$

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MCQ 301 Mark

The domain of the function $\text{f(x)}=\sqrt{5|\text{x}|-\text{x}^2-16}$ is:

A
$(-3,-2)\cup(2,3)$
B
$\big[-3,-2\big]\cup\big[2,3\big) $
✓
$\big[-3,-2\big]\cup\big[2,3\big] $
D
None os these.

Answer

Correct option: C.

$\big[-3,-2\big]\cup\big[2,3\big] $

$\text{f(x)}=\sqrt{5|\text{x}|-\text{x}^2-16}$
For f(x) to be defined, $5|\text{x}|-\text{x}^2-6\geq0$
$\Rightarrow5|\text{x}|-\text{x}^2-6\geq0$
$\Rightarrow\text{x}^2-5|\text{x}|+6\leq0$
For x > 0, |x| = x
$\Rightarrow\text{x}^2-5\text{x}+6\leq0$
$\Rightarrow(\text{x}-2)(\text{x}-3)\leq0$
$\Rightarrow\text{x}\in[2,3]\ ...(\text{i})$
For x < 0, |x| = -x
$\Rightarrow\text{x}^2+5\text{x}+6\leq0$
$\Rightarrow(\text{x}+2)(\text{x}+3)\leq0$
$\Rightarrow\text{x}\in\big[-3,-2\big]\ ...(\text{ii})$
From (i) and (ii),
$\text{x}\in\big[-3,-2\big]\cup\big[2,3\big]$
Or, $\text{domain(f)}=\big[-3,-2\big]\cup\big[2,3\big]$

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MCQ 311 Mark

f is a real valued function given by $\text{f(x)}=27\text{x}^3+\frac{1}{\text{x}^3}$ and $\alpha,\beta$ are roots of $3\text{x}+\frac{1}{\text{x}}=12$ Then,

A
$\text{f}(\alpha)\neq\text{f}(\beta)$
B
$\text{f}(\alpha)=10$
C
$\text{f}(\beta)=-10$
✓
None of these.

Answer

Correct option: D.

None of these.

$\text{f(x)}=27\text{x}^3+\frac{1}{\text{x}^3}$
$\Rightarrow\text{f(x)}=\Big(3\text{x}+\frac{1}{\text{x}}\Big)\Big(9\text{x}^2+\frac{1}{\text{x}^2}-3\Big)$
$\Rightarrow\text{f(x)}=\Big(3\text{x}+\frac{1}{\text{x}}\Big)\Big(\Big(3\text{x}+\frac{1}{\text{x}}\Big)^2-9\Big)$
$\Rightarrow\text{f}(\alpha)=\Big(3\alpha+\frac{1}{\alpha}\Big)\Big(\Big(3\alpha+\frac{1}{\alpha}\Big)^2-9\Big)$
Since $\alpha$ and $\beta$ are the roots of $3\text{x}+\frac{1}{\text{x}}=12,$
$3\alpha+\frac{1}{\alpha}=12$ and $3\beta+\frac{1}{\beta}=12$
$\Rightarrow\text{f}(\alpha)=12\big((12)^2-9\big)$ and $\text{f}(\beta)=12\big((12)^2-9\big)$
$\Rightarrow\text{f}(\alpha)=\text{f}(\beta)=\big((12)^2-9\big)$

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MCQ 321 Mark

The domain of the function $\text{f(x)}=\sqrt{\frac{(\text{x}+1)(\text{x}-3)}{\text{x}-2}}$ is:

✓
$\big[-1,2\big)\cap\big[3,\infty\big)$
B
$(-1,2)\cap[3,\infty)$
C
$[-1,2]\cap[3,\infty]$
D
None of these.

Answer

Correct option: A.

$\big[-1,2\big)\cap\big[3,\infty\big)$

$\text{f(x)}=\sqrt{\frac{(\text{x}+1)(\text{x}-3)}{\text{x}-2}}$
For f(x) to be defined,
$(\text{x}-2)\neq0$
$\Rightarrow\text{x}\neq2\ ...(\text{i})$
Also,
$\frac{(\text{x}+1)(\text{x}-3)}{\text{x}-2}\geq0$
$\Rightarrow\frac{(\text{x}+1)(\text{x}-3)(\text{x}-2)}{(\text{x}-2)^2}\geq0$
$\Rightarrow(\text{x}+1)(\text{x}-3)(\text{x}-2)\geq0$
$\Rightarrow\text{x}\in\big[-1,2\big)\cup\big[3,\infty\big)\ ...(\text{ii})$
From (i) and (ii),
$\text{x}\in\big[-1,2\big)\cap\big[3,\infty\big)$

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MCQ 331 Mark

Let $\text{f(x)}=\sqrt{\text{x}^2+1}$ Then which of the following is correct?

A
$\text{f(xy)}=\text{f(x)}\text{f(y)}$
B
$\text{f(xy)}\geq\text{f(x)}\text{f(y)}$
✓
$\text{f(xy)}\leq\text{f(x)}\text{f(y)}$
D
None os these.

Answer

Correct option: C.

$\text{f(xy)}\leq\text{f(x)}\text{f(y)}$

Given, $\text{f(x)}=\sqrt{\text{x}^2+1}\ ...(\text{i})$
Replacing x by y in (i), we get
$\text{f(y)}=\sqrt{\text{y}^2+1}$
$\therefore\ \text{f(x)}\text{f(y)}=\sqrt{\text{x}^2+1}\sqrt{\text{y}^2+1}$
$=\sqrt{(\text{x}^2+1)(\text{y}^2+1)}$
$=\sqrt{\text{x}^2\text{y}^2+\text{x}^2+\text{y}^2+1}$
Also, replacing x by xy in (i), we get
$\text{f(xy)}=\sqrt{\text{x}^2\text{y}^2+1}$
Now,
$\text{x}^2\text{y}^2+1\leq\text{x}^2\text{y}^2+\text{x}^2+\text{y}^2+1$
$\Rightarrow\sqrt{\text{x}^2\text{y}^2+1}\leq\sqrt{\text{x}^2\text{y}^2+\text{x}^2+\text{y}^2+1}$
$\Rightarrow\text{f}(\text{xy})\leq\text{f(x)}\text{f(y)}$

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MCQ 341 Mark

If $\text{f(x)}=\cos(\log\text{x}),$ then the value of $\text{f(x}^2)\text{f}(\text{y}^2)-\frac{1}{2}\Big\{\text{f}\Big(\frac{\text{x}^2}{\text{y}^2}\Big)+\text{f}\big(\text{x}^2\text{y}^2\big)\Big\}$ is:

A
-2
B
-1
C
$\frac{1}{2}$
✓
None of these.

Answer

Correct option: D.

None of these.

Given,
$\text{f(x)}=\cos(\log\text{x})$
$\Rightarrow\ \text{f(x}^2)=\cos(\log(\text{x}^2))$
$\Rightarrow\ \text{f(x}^2)=\cos(2\log(\text{x}))$
Similarly,
$\text{f}(\text{y}^2)=\cos(2\log(\text{y}))$
Now,
$\text{f}\Big(\frac{\text{x}^2}{\text{y}^2}\Big)=\cos\Big(\log\Big(\frac{\text{x}^2}{\text{y}^2}\Big)\Big)=\cos\big(\log\text{x}^2-\log\text{y}^2\big)$
and $\text{f}(\text{x}^2\text{y}^2)=\cos(\log\text{x}^2\text{y}^2)=\cos\big(\log\text{x}^2+\log\text{y}^2\big)$
$\Rightarrow\ \text{f}\Big(\frac{\text{x}^2}{\text{y}^2}\Big)+\text{f}(\text{x}^2\text{y}^2)=\cos\big((2\log\text{x}-2\log\text{y})\big)+\cos\big((2\log\text{x}-2\log\text{y})\big)$
$\Rightarrow\ \text{f}\Big(\frac{\text{x}^2}{\text{y}^2}\Big)+\text{f}(\text{x}^2\text{y}^2)=2\cos(2\log\text{x})\cos(2\log\text{y})$
$\Rightarrow\frac{1}{2}\bigg[\Big(\text{f}\Big(\frac{\text{x}^2}{\text{y}^2}\Big)+\text{f}(\text{x}^2\text{y}^2)\bigg]=\cos(2\log)\cos(2\log\text{y})$
$\Rightarrow\ \text{f(x}^2)\text{f}(\text{y}^2)-\frac{1}{2}\Big\{\text{f}\Big(\frac{\text{x}^2}{\text{y}^2}\Big)+\text{f}\big(\text{x}^2\text{y}^2\big)\Big\}\\\ =\cos(2\log)\cos(2\log\text{y})-\cos(2\log)\cos(2\log\text{y})=0$

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MCQ 351 Mark

If $f : Q \rightarrow Q$ is defined as $f(x) = x^2,$ then $f^{-1}(9)$ is equal to:

A
$3$
B
$-3$
✓
$\{-3, 3\}$
D
$\phi$

Answer

Correct option: C.

$\{-3, 3\}$

If $f : A \rightarrow B,$ such that $\text{y}\in\text{B},$ then $\text{f}^{-1}(\text{y})=\{\text{x}\in\text{A}:\text{f(x)}=\text{y}\}$
In other words, $f^{-1}\{y\}$ is the set of pre$-$images of $y.$
Let $\text{f}^{-1}\{9\}=\text{x}$
Then, $\text{f(x)}=9$
$\Rightarrow\text{x}^2=9$
$\Rightarrow\text{x}=\pm3$
$\therefore\ \text{f}^{-1}\{9\}=\{-3,3\}$

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MCQ 361 Mark

The domain of definition of the function $\text{f(x)}=\log|\text{x}|$ is:

A
$\text{R}$
B
$\big(-\infty,0\big)$
C
$(0,\infty)$
✓
$\text{R}-\{0\}$

Answer

Correct option: D.

$\text{R}-\{0\}$

$\text{f(x)}=\log|\text{x}|$
For f(x) to be defined,
$|\text{x}|>0,$ which is always true.
But $|\text{x}|\neq0$
$\Rightarrow\text{x}\neq0$
Thus, $\text{domain(f)}=\text{R}-\{0\}$

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MCQ 371 Mark

Let $\text{A}=\{\text{x}\in\text{R}:\text{x}\neq0-4\leq\text{x}\leq4\}$ and $\text{f}:\text{A}\in\text{R}$ be defined by $\text{f(x)}=\frac{|\text{x}|}{\text{x}}$ for $\text{x}\in\text{A}$ Then A:

✓
$[1,-1]$
B
$\big[\text{x}:0\leq\text{x}\leq4\big]$
C
$\{1\}$
D
$\{\text{x}:-4\leq\text{x}\leq0\}$

Answer

Correct option: A.

$[1,-1]$

As, $\text{|x|}=\begin{cases}\text{x},\ \text{x}\geq0\\-\text{x}<0\end{cases}$
So, $\text{f(x)}=\frac{\text{x}}{|\text{x}|}$
When, $\text{x}<0\text{ i.e.,}\text{ x}\in\big[-4,0\big)$
$\text{f(x)}=\frac{\text{x}}{-\text{x}}=-1$
and when, $\text{x}>0\text{ i.e., x}\in\big(0,4\big]$
$\text{f(x)}=\frac{\text{x}}{\text{x}}=1$
So, range $\text{f}=\{-1,1\}$

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MCQ 381 Mark

Let f(x) = |x - 1|. Then:

A
$(\text{x}^2) = [\text{f(x)}]^2$
B
$\text{f(x + y) = f(x)f(y)}$
C
$\text{f(|x|) = |f(x)|}$
✓
None of these.

Answer

Correct option: D.

None of these.

$\text{f(x)}=|\text{x}-1|$
Since, $|\text{x}^2-1|\neq|\text{x}-1|^2$
$\text{f(x)}^2\neq(\text{f(x)})^2$
Thus, (i) is wrong.
Since, $|\text{x}+\text{y}-1|\neq|\text{x}-1||\text{y}-1|$
$\text{f}(\text{x}+\text{y})\neq\text{f(x)}\text{f(y)}$
Thus, (ii) is wrong.
Since, $|\text{|x|}-1\neq||\text{x}-1||=|\text{x}-1|$
$\text{f(|x|)}\neq|\text{f(x)}|$
Thus, (iii) is wrong.
Hence, none of the given options is the answer.

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MCQ 391 Mark

If $\text{f(x)}=\cos(\log_\text{e}),$ then $\text{f}\Big(\frac{1}{\text{x}}\Big)\text{f}\Big(\frac{1}{\text{y}}\Big)-\frac{1}{2}\Big\{\text{f(xy)}+\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)\Big\}$ is equal to:

A
$\cos(\text{x}-\text{y})$
✓
$\log(\cos(\text{x}-\text{y}))$
C
$1$
D
$\cos(\text{x}+\text{y})$

Answer

Correct option: B.

$\log(\cos(\text{x}-\text{y}))$

$\text{f(x)}=\cos(\log_\text{e}\text{x})$
$\Rightarrow\text{f}\Big(\frac{1}{\text{x}}\Big)=\cos\Big(\log_\text{e}\Big(\frac{1}{\text{x}}\Big)\Big)$
$\Rightarrow\text{f}\Big(\frac{1}{\text{x}}\Big)=\cos\Big(-\log_\text{e}(\text{x})\Big)$
$\Rightarrow\text{f}\Big(\frac{1}{\text{x}}\Big)=\cos\Big(\log_\text{e}(\text{x})\Big)$
Similarly,
$\text{f}\Big(\frac{1}{\text{y}}\Big)=\cos(\log_\text{e}\text{y})$
Now,
$\text{f(xy)}=\cos(\log_\text{e}\text{xy})=\cos\big(\log_\text{e}\text{x}+\log_\text{e}\text{y}\big)$
and
$\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)=\cos\Big(\log_\text{e}\frac{\text{x}}{\text{y}}\Big)=\cos\big(\log_\text{e}\text{x}-\log_\text{e}\text{y}\big)$
$\Rightarrow\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}=\cos\big(\log_\text{e}\text{x}+\log_\text{e}\text{y}\big)+\cos\big(\log_\text{e}\text{x}-\log_\text{e}\text{y}\big)$
$\Rightarrow\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}=2\cos\big(\log_\text{e}\text{x}\big)\cos(\log_\text{e}\text{y})$
$\Rightarrow\frac{1}{2}\Big[\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}\Big]=\cos\big(\log_\text{e}\text{x}\big)\cos\big(\log_\text{e}\text{y}\big)$
$\Rightarrow\text{f}\Big(\frac{1}{\text{x}}\Big)\text{f}\Big(\frac{1}{\text{y}}\Big)-\frac{1}{2}\Big\{\text{f(xy)}+\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)\Big\}=\cos\big(\log_\text{e}\text{x}\big)\cos\big(\log_\text{e}\text{y}\big)\\\ \ -\cos\big(\log_\text{e}\text{x}\big)\cos\big(\log_\text{e}\text{y}\big)=0$
Disclaimer: The question in the book has some error, so none of the options are matching with the solution. The solution is created according to the question given in the book.

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MCQ 401 Mark

If $\text{f(x)}=\sin[\pi^2]\text{x}+\sin[-\pi^2]\text{x},$ where [x] denotes the greatest integer less than or equal to x, then:

✓
$\text{f}\Big(\frac{\pi}{2}\Big)=1$
B
$\text{f}(\pi)=2$
C
$\text{f}\Big(\frac{\pi}{4}\Big)=-1$
D
None of these.

Answer

Correct option: A.

$\text{f}\Big(\frac{\pi}{2}\Big)=1$

$\text{f(x)}=\sin[\pi^2]\text{x}+\sin[-\pi^2]\text{x}$
$\Rightarrow\text{f(x)}=\sin\big[9.8\big]\text{x}+\sin\big[-9.8\big]\text{x}$
$\Rightarrow\text{f(x)}=\sin9\text{x}-\sin10\text{x}$
$\Rightarrow\text{f}\Big(\frac{\pi}{2}\Big)=\sin9\times\frac{\pi}{2}-\sin10\times\frac{\pi}{2}$
$\Rightarrow\text{f}\Big(\frac{\pi}{2}\Big)=1-0=1$

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MCQ 411 Mark

If $\text{f(x)}=\frac{\sin^{4}\text{x}+\cos^2\text{x}}{\sin^2\text{x}+\cos^4\text{x}}$ for $\text{x}\in\text{R},$ then f(2002) =

✓
1
B
2
C
3
D
4

Answer

Correct option: A.

Given,
$\text{f(x)}=\frac{\sin^{4}\text{x}+\cos^2\text{x}}{\sin^2\text{x}+\cos^4\text{x}}$
On dividing the numerator and denominator by $\cos^4\text{x},$ we get
$\text{f(x)}=\frac{\tan^4\text{x}+\sec^2\text{x}}{1+\tan^2\text{x}\sec^2\text{x}}$
$=\frac{1+\tan^4\text{x}+\tan^2\text{x}}{1+\tan^2\text{x}(1+\tan^2\text{x})}$
$=\frac{1+\tan^{4}\text{x}+\tan^{2}\text{x}}{1+\tan^{4}\text{x}+\tan^{2}\text{x}}=1$ $(\text{For every x}\in\text{R})$
$\text{For x}=2002,$
We have,
$\text{f}(2002)=1$

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MCQ 421 Mark

If f : R → R be given by for all $\text{f(x)}=\frac{4^{\text{x}}}{4^{\text{x}}+2}\text{ x }\in\text{R},$ then:

A
f(x) = f(1 - x)
B
f(x) + f(1 - x) = 0
✓
f(x) + f(1 - x) = 1
D
f(x) + f(x - 1) = 1

Answer

Correct option: C.

f(x) + f(1 - x) = 1

$\text{f(x)}=\frac{4^{\text{x}}}{4^{\text{x}}+2}\text{ x }\in\text{R},$
$\text{f}\big(\text{1}-\text{x}\big)=\frac{4^{1-\text{x}}}{4^{1-\text{x}}+2}$
$=\frac{4}{2\times4^{\text{x}}+4}$
$=\frac{2}{4^{\text{x}}+2}$
$\text{f(x)}+\text{f}(1-\text{x})=\frac{4^{\text{x}}}{4^{\text{x}}+2}+\frac{2}{4^{\text{x}}+2}$
$=\frac{4^{\text{x}}+2}{4^{\text{x}}+2}=1$

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MCQ 431 Mark

The range of $\text{f(x)}=\frac{1}{1-2\cos\text{x}}$ is:

A
$\Big[\frac{1}{3},1\Big]$
✓
$\Big[-1,\frac{1}{3}\Big]$
C
$\big(-\infty,-1\big)\cup\Big[\frac{1}{3},\infty\Big)$
D
$\Big[-\frac{1}{3},1\Big]$

Answer

Correct option: B.

$\Big[-1,\frac{1}{3}\Big]$

We know that $-1\leq\cos\text{x}\leq1$ for all $\text{x}\in\text{R}$
Now,
$-1\leq\cos\text{x}\leq1$
$\Rightarrow-1\leq\cos\text{x}\leq1$
$\Rightarrow-2\leq-2\cos\text{x}\leq2$
$\Rightarrow-1\leq1-2\cos\text{x}\leq3$ (Adding 1 ro each term)
But,
$\cos\text{x}\neq\frac{1}{2}$
$\Rightarrow1-2\cos\text{x}\in\big[-1,3\big]-\{0\}$
$\Rightarrow\frac{1}{1-2\cos\text{x}}\in\big(-\infty,-1\big]\cap\Big[\frac{1}{3},\infty\Big)$
$\therefore\ \text{Range of }\text{f(x)}=\big(-\infty,-1\big]\cap\Big[\frac{1}{3},\infty\Big)$
Disclaimer: The range of the function does not matches with either of the given options. The range matches with option (c) if it is given as $\big(-\infty,-1\big]\cap\Big[\frac{1}{3},\infty\Big)$

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MCQ 441 Mark

The domain of the function $\text{f(x)}=\sqrt{2-2\text{x}-\text{x}^2}$ is:

A
$\big[-\sqrt{3},\sqrt{3}\big]$
✓
$\big[-1,-\sqrt{3},-1+\sqrt{3}\big]$
C
$\big[-2,2\big]$
D
$\big[-2-\sqrt{3},-2+\sqrt{3}\big]$

Answer

Correct option: B.

$\big[-1,-\sqrt{3},-1+\sqrt{3}\big]$

$\text{f(x)}=\sqrt{2-2\text{x}-\text{x}^2}$
Since, $2-2\text{x}-\text{x}^2\geq0$
$\text{x}^2+2\text{x}-2\leq0$
$\Rightarrow\text{x}^2-2\text{x}-2+1-1\leq0$
$\Rightarrow(\text{x}-1)^2-\big(\sqrt{3}\big)^2\leq0$
$\Rightarrow\big[\text{x}-\big(1-\sqrt{3}\big)\big]\big[\text{x}-\big(1+\sqrt{3}\big)\big]\leq0$
$\Rightarrow\big(-1-\sqrt{3}\big)\leq\text{x}\leq(-1+\sqrt{3})$
Thus, domain $(\text{f})=\big[-1-\sqrt{3},-1+\sqrt{3}\big]$

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MCQ 451 Mark

Which of the following are functions?

A
$\{(\text{x, y}):\text{y}^2=\text{x, x, y}\in\text{R}\}$
✓
$\{(\text{x, y}):\text{y}=\text{|x|},\text{x, y}\in\text{R}\}$
C
$\{(\text{x, y}):\text{x}^2+\text{y}^2=1,\text{x, y}\in\text{R}\}$
D
$\{(\text{x, y}):\text{x}^2-\text{y}^2=1\text{x, y}\in\text{R}\}$

Answer

Correct option: B.

$\{(\text{x, y}):\text{y}=\text{|x|},\text{x, y}\in\text{R}\}$

For every value of $\text{x}\in\text{R},$ there is a unique value $\text{y} \in\text{R}$
i.e., there is a unique image for all values of $\text{x}\in\text{R},$
Also, values of x occur only once in the ordered pairs.
Thus, it is a function.

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