M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip

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M.C.Q (1 Marks)

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35 questions · 2 auto-graded MCQ + 33 self-marked written.

Question 11 Mark

$\sin\Big\{2\cos^{-1}\Big(\frac{-3}{5}\Big)\Big\}$ is equal to:

$\frac{6}{25}$
$\frac{24}{25}$
$\frac{4}{5}$
$-\frac{24}{25}$

Answer

$-\frac{24}{25}$

Solution:
Let $\cos^{-1}\Big(-\frac{3}{5}\Big)=\text{x},0\leq\text{x}\leq\pi$
Then, $\cos\text{x}=-\frac{3}{5}$
$\therefore\ \sin\text{x}=\sqrt{1-\cos^2\text{x}}=\sqrt{1-\Big(-\frac{3}{5}\Big)^2}=\sqrt{\frac{16}{25}}=\frac{4}{5}$
Now,
$\sin\Big\{2\cos^{-1}\Big(\frac{-3}{5}\Big)\Big\}=\sin(2\text{x})$
$=2\sin\text{x}\cos\text{x}$
$=2\times\frac{4}{5}\times\frac{-3}{5}$
$=-\frac{24}{25}$

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Question 21 Mark

If $\tan^{-1}\frac{\text{x}+1}{\text{x}-1}+\tan^{-1}\frac{\text{x}-1}{\text{x}}=\tan^{-1}(-7),$ then the value of x is:

0
-2
1
2

Answer

2

Solution:
$\tan^{-1}\frac{\text{x}+1}{\text{x}-1}+\tan^{-1}\frac{\text{x}-1}{\text{x}}=\tan^{-1}(-7),$
$\Rightarrow\tan^{-1}\Bigg(\frac{\frac{\text{x}+1}{\text{x}-1}+\frac{\text{x}-1}{\text{x}}}{1-\frac{\text{x}+1}{\text{x}-1}\times\frac{\text{x}-1}{\text{x}}}\Bigg)=\tan^{-1}(-7)$
$\Rightarrow\tan^{-1}\Big(\frac{\text{x}^2+\text{x}+\text{x}^2-2\text{x}+1}{\text{x}^2-\text{x}-(\text{x}^2-1)}\Big)\tan^{-1}(-7)$
$\Rightarrow\frac{2\text{x}^2-\text{x}+1}{-\text{x}+1}=-7$
$\Rightarrow2\text{x}^2-\text{x}+1=7\text{x}-7$
$\Rightarrow2\text{x}^2-8\text{x}+8=0$
$\Rightarrow\text{x}^2-4\text{x}+4=0$
$\Rightarrow(\text{x}-2)^2=0$
$\Rightarrow\text{x}=2$

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Question 31 Mark

The number of solutions of the equation
$\tan^{-1}2\text{x}+\tan^{-1}3\text{x}=\frac{\pi}{4}$ is:

2
3
1
none of these

Answer

2

Solution:
We know that $\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big).$
$\therefore\ \tan^{-1}2\text{x}+\tan^{-1}3\text{x}=\frac{\pi}{4}$
$\Rightarrow\tan^{-1}\Big(\frac{2\text{x}+3\text{x}}{1-2\text{x}\times3\text{x}}\Big)=\frac{\pi}{4}$
$\Rightarrow\frac{2\text{x}+3\text{x}}{1-2\text{x}\times3\text{x}}=\tan\frac{\pi}{4}$
$\Rightarrow\frac{5\text{x}}{1-6\text{x}^2}=1$
$\Rightarrow5\text{x}=1-6\text{x}^2$
$\Rightarrow6\text{x}^2+5\text{x}-1=0$

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Question 41 Mark

If $\sin^{-1}\Big(\frac{2\text{a}}{1-\text{a}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{a}^2}{1+\text{a}^2}\Big)=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big),$ where $\text{a},\text{x}\in(0,1),$ then the value of x is:

$0$
$\frac{\text{a}}{2}$
$\text{a}$
$\frac{2\text{a}}{1-\text{a}^2}$

Answer

$\frac{2\text{a}}{1-\text{a}^2}$

Solution:
$\sin^{-1}\Big(\frac{2\text{a}}{1-\text{a}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{a}^2}{1+\text{a}^2}\Big)=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)$
Let, $\text{a}=\tan\theta\Rightarrow\theta=\tan^{-1}\text{a}$
$\sin^{-1}(\sin2\theta)+\cos^{-1}(\cos2\theta)=2\tan^{-1}(\text{x})$
$2\theta+2\theta=2\tan^{-1}(\text{x})$
$4\theta=2\tan^{-1}(\text{x})$
$2\tan^{-1}\text{a}=\tan^{-1}(\text{x})$
$\tan^{-1}\Big(\frac{2\text{a}}{1-\text{a}^2}\Big)=\tan^{-1}(\text{x})$
$\text{x}=\frac{2\text{a}}{1-\text{a}^2}$

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Question 51 Mark

$2\tan^{-1}\big\{\text{cosec}\big(\tan^{-1}\text{x}\big)-\tan\big(\cot^{-1}\text{x}\big)\big\}$ is equal to:

$\cot^{-1}\text{x}$
$\cot^{-1}\text{x}$
$\tan^{-1}\text{x}$
$\text{none of these}$

Answer

$\tan^{-1}\text{x}$

Solution:
$\therefore\ 2\tan^{-1}\big\{\text{cosec}\big(\tan^{-1}\text{x}\big)-\tan\big(\cot^{-1}\text{x}\big)\big\}$
$=2\tan^{-1}\Big\{\text{cosec}\big(\tan^{-1}\text{x}\big)-\tan\Big(\tan^{-1}\frac{1}{\text{x}}\Big)\Big\}$
$=\frac{3}{29}=2\tan^{-1}\Big\{\text{cosec}\big(\tan^{-1}\text{x}\big)-\frac{1}{\text{x}}\Big\}$
$=2\tan^{-1}\Big\{\text{cosec y}-\frac{1}{\tan\text{y}}\Big\}$
$=2\tan^{-1}\Big\{\frac{1-\cos\text{y}}{\sin\text{y}}\Big\}$
$=2\tan^{-1}\bigg\{\frac{2\sin^2\frac{\text{y}}{2}}{\sin\text{y}}\bigg\}$
$=2\tan^{-1}\bigg\{\frac{2\sin^2\frac{\text{y}}{2}}{2\sin\frac{\text{y}}{2}\cos\frac{\text{y}}{2}}\bigg\}$
$=2\tan^{-1}\Big\{\tan\frac{\text{y}}{2}\Big\}$
$=\text{y}$
$=\tan^{-1}\text{x}$

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Question 61 Mark

The value of $\sin^{-1}\Big(\cos\frac{33\pi}{5}\Big)$ is:

$\frac{33}{5}$
$-\frac{\pi}{10}$
$\frac{\pi}{10}$
$\frac{7\pi}{5}$

Answer

$-\frac{\pi}{10}$

Solution:
$\sin^{-1}\Big(\cos\frac{33\pi}{5}\Big)$
$=\sin^{-1}\Big(\cos\Big(6\pi+\frac{3\pi}{5}\Big)\Big)$
$=\sin^{-1}\Big(\cos\Big(\frac{3\pi}{5}\Big)\Big)$
$=\sin^{-1}\Big(\sin\Big(\frac{\pi}{2}-\frac{3\pi}{5}\Big)\Big)$
$=\frac{\pi}{2}-\frac{3\pi}{5}$
$=\frac{-\pi}{10}$

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Question 71 Mark

If $\alpha=\tan^{-1}\Big(\frac{\sqrt3\text{x}}{2\text{y}-\text{x}}\Big),\beta=\tan^{-1}\Big(\frac{2\text{x}-\text{y}}{\sqrt3\text{y}}\Big),$ then $\alpha-\beta=$

$\frac{\pi}{6}$
$\frac{\pi}{3}$
$\frac{\pi}{2}$
$-\frac{\pi}{3}$

Answer

$\frac{\pi}{6}$

Solution:
We have
$\alpha=\tan^{-1}\Big(\frac{\sqrt3\text{x}}{2\text{y}-\text{x}}\Big),\beta=\tan^{-1}\Big(\frac{2\text{x}-\text{y}}{\sqrt3\text{y}}\Big)$
Now, $\alpha-\beta=\tan^{-1}\Big(\frac{\sqrt3\text{x}}{2\text{y}-1}\Big)-\tan^{-1}\Big(\frac{2\text{x}-\text{y}}{\sqrt3\text{y}}\Big)$
$=\tan^{-1}\begin{pmatrix}\frac{\frac{\sqrt3\text{x}}{2\text{y}-\text{x}}-\frac{2\text{x}-\text{y}}{\sqrt3\text{y}}}{1+{\frac{\sqrt3\text{x}}{2\text{y}-\text{x}}\times\frac{2\text{x}-\text{y}}{\sqrt3\text{y}}}}\end{pmatrix}$
$=\tan^{-1}\begin{pmatrix}\frac{\frac{3\text{xy}-4\text{xy}+2\text{y}^2+2\text{x}^2-\text{xy}}{\sqrt{3}\text{y}(2\text{y}-\text{x})+\sqrt{3}\text{x}(2\text{z}-\text{y})}}{\sqrt{3}\text{y}(2\text{y}-\text{x})}\end{pmatrix}$
$=\tan^{-1}\Big(\frac{3\text{xy}-4\text{xy}+2\text{y}^2+2\text{x}^2-\text{xy}}{2\sqrt3\text{y}^2-\sqrt3\text{xy}+2\sqrt3\text{x}^2-\sqrt3\text{xy}}\Big)$
$=\tan^{-1}\Big(\frac{2\text{y}^2+2\text{x}^2-2\text{xy}}{2\sqrt3\text{y}^2+2\sqrt3\text{x}^2-2\sqrt3\text{xy}}\Big)$
$=\tan^{-1}\Big(\frac{1}{\sqrt3}\Big)=\frac{\pi}{6}$

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MCQ 81 Mark

If $\tan^{-1}\Big\{\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}\Big\}=\alpha,$ then $x^2 =$

✓
$\sin2\alpha$
B
$\sin\alpha$
C
$\cos2\alpha$
D
$\cos\alpha$

Answer

Correct option: A.

$\sin2\alpha$

$\tan^{-1}\Big\{\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}\Big\}=\alpha$
$\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}=\tan\alpha$
$\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}=\tan\alpha$
$\frac{1+\text{x}^2-2\sqrt{1-\text{x}^2}\sqrt{1+\text{x}^2}+1-\text{x}^2}{1+\text{x}^2-1+\text{x}^2}=\tan\alpha$
$\frac{1-\sqrt{1-\text{x}^4}}{\text{x}^2}=\tan\alpha$
$1-\sqrt{1-\text{x}^4}=\text{x}^2\tan\alpha$
$\big(1-\text{x}^2\tan\alpha\big)^2=1-\text{x}^4$
$1-2\text{x}^2\tan\alpha+\text{x}^4\tan^2\alpha=1-\text{x}^4$
$\text{x}^4-2\text{x}^2\tan\alpha+\text{x}^4\tan^2\alpha=0$
$\text{x}^2\big(\text{x}^2-2\tan\alpha+\text{x}^2\tan^2\alpha\big)=0$
$\text{x}^2=\frac{2\tan\alpha}{1+\tan^2\alpha}$
$\text{x}^2=\frac{2\tan\alpha}{\sec^2\alpha}$
$\text{x}^2=2\tan\alpha\cos^2\alpha$
$\text{x}^2=2\sin\alpha\cos\alpha=2\sin\alpha$

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Question 91 Mark

If x < 0, y < 0 such that xy = 1, then $\tan^{-1}\text{x}+\tan^{-1}\text{y}$ equals:

$\frac{\pi}{2}$
$-\frac{\pi}{2}$
$-\pi$
$\text{none of these}$

Answer

$-\frac{\pi}{2}$

Solution:
We know that $\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$
x < 0, y < 0 such that
xy = 1
Let x = -a and y = -b, where a and b both are positive.
$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$
$=\tan^{-1}\Big(\frac{-\text{a}-\text{a}}{1-1}\Big)$
$=\tan^{-1}(-\infty)$
$=\tan^{-1}\Big\{\tan\Big(-\frac{\pi}{2}\Big)\Big\}$
$=-\frac{\pi}{2}$

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Question 101 Mark

If $\cos^{-1}\frac{\text{x}}{3}+\cos^{-1}\frac{\text{y}}{2}=\frac{\theta}{2},$ then, $4\text{x}^2-12\text{xy}\cos^2\frac{\theta}{2}+9\text{y}^2=$

$36$
$36-36\cos\theta$
$18-18\cos\theta$
$18+18\cos\theta$

Answer

$18-18\cos\theta$

Solution:
$\cos^{-1}\text{x}+\cos^{-1}\text{y}=\cos^{-1}\Big(\text{xy}\sqrt{1-\text{x}^2}\sqrt{1-\text{y}^2}\Big)$
$\Rightarrow\cos^{-1}\frac{\text{x}}{3}+\cos^{-1}\frac{\text{y}}{2}=\frac{\theta}{2}$
$\Rightarrow\cos^{-1}\Bigg(\frac{\text{x}}{3}\times\frac{\text{y}}{2}-\sqrt{1-\Big(\frac{\text{x}}{3}\Big)^2}\sqrt{1-\Big(\frac{\text{y}}{2}\Big)^2}\Bigg)=\frac{\theta}{2}$
$\Rightarrow\frac{\text{xy}}{6}-\sqrt{1-\Big(\frac{\text{x}^2}{9}\Big)}\sqrt{1-\Big(\frac{\text{y}^2}{4}\Big)}=\cos\frac{\theta}{2}$
$\Rightarrow\frac{\text{xy}-6\cos\frac{\theta}{2}}{6}=\frac{\sqrt{9-\text{x}^2}\sqrt{4-\text{y}^2}}{6}$
$\Rightarrow\text{xy}-6\cos\frac{\theta}{2}=\sqrt{9-\text{x}^2}\sqrt{4-\text{y}^2}$
Taking square on both sides,
$\Rightarrow\text{x}^2\text{y}^2-12\text{xy}\cos\frac{\theta}{2}+36\cos^2\frac{\theta}{2}=\big(9-\text{x}^2\big)\big(4-\text{y}^2\big)$
$\Rightarrow\text{x}^2\text{y}^2-12\text{xy}\cos\frac{\theta}{2}+36\cos^2\frac{\theta}{2}=36-9\text{y}^2-4\text{x}^2+\text{x}^2\text{y}^2$
$\Rightarrow4\text{x}^2+9\text{y}^2-12\text{xy}\cos^2\frac{\theta}{2}=36\Big(1-\cos^2\frac{\theta}{2}\Big)$
$\Rightarrow4\text{x}^2+9\text{y}^2-12\text{xy}\cos^2\frac{\theta}{2}=36\Big(1-\frac{1+\cos\theta}{2}\Big)$
$\Rightarrow4\text{x}^2+9\text{y}^2-12\cos^2\frac{\theta}{2}=18-18\cos\theta$

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Question 111 Mark

If $\cos^{-1}\frac{\text{x}}{2}+\cos^{-1}\frac{\text{y}}{2}=\theta,$ then $\Rightarrow9\text{x}^2-12\text{xy}\cos\theta+4\text{y}^2$ is equal to:

$36$
$-36\sin^2\theta$
$36\sin^2\theta$
$-36\cos^2\theta$

Answer

$36\sin^{2}\theta$

Solution:
We know
$\cos^{-1}\text{x}+\cos^{-1}\text{y}=\cos^{-1}\Big[\text{xy}-\sqrt{1-\text{x}^2}\sqrt{1-\text{y}^2}\Big]$
Now,
$\cos^{-1}\frac{\text{x}}{2}+\cos^{-1}\frac{\text{y}}{2}=\theta$
$\Rightarrow\cos^{-1}\Big[\frac{\text{x}}{2}\frac{\text{y}}{3}-\sqrt{1-\frac{\text{x}^2}{4}}\sqrt{1-\frac{\text{y}^2}{3}}\Big]=\theta$
$\Rightarrow\frac{\text{x}}{2}\frac{\text{y}}{3}-\sqrt{1-\frac{\text{x}^2}{4}}\sqrt{1-\frac{\text{y}^2}{3}}=\cos\theta$
$\Rightarrow\text{xy}-\sqrt{4-\text{x}^2}\sqrt{9-\text{y}^2}=6\cos\theta$
$\Rightarrow\sqrt{4-\text{x}^2}\sqrt{9-\text{y}^2}=\text{xy}-6\cos\theta$
$\Rightarrow(4-\text{x}^2)(9-\text{y}^2)=\text{x}^2\text{y}^2+36\cos^2\theta-12\text{xy}\cos\theta$ (Squaring both the sides)
$\Rightarrow36-4\text{y}^2-9\text{x}^2+\text{x}^2\text{y}^2=\text{x}^2\text{y}^2+36\cos^{2}\theta-12\text{xy}\cos\theta$
$\Rightarrow36-4\text{y}^2-9\text{x}^2=36\cos^2\theta-12\text{xy}\cos\theta$
$\Rightarrow9\text{x}^2-12\text{xy}\cos\theta+4\text{y}^2=36-36\cos^2\theta$
$\Rightarrow9\text{x}^2-12\text{xy}\cos\theta+4\text{y}^2=36\sin^2\theta$

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Question 121 Mark

In a $\triangle\text{ABC},$ if C is a right angle, then $\tan^{-1}\Big(\frac{\text{a}}{\text{b}+\text{c}}\Big)+\tan^{-1}\Big(\frac{\text{b}}{\text{c}+\text{b}}\Big)=$

$\frac{\pi}{3}$
$\frac{\pi}{4}$
$\frac{5\pi}{2}$
$\frac{\pi}{6}$

Answer

$\frac{\pi}{4}$

Solution:
We know,
$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$
$\therefore\ \tan^{-1}\Big(\frac{\text{a}}{\text{b}+\text{c}}\Big)+\tan^{-1}\Big(\frac{\text{b}}{\text{c}+\text{a}}\Big)=\tan^{-1}=\begin{pmatrix}\frac{\frac{\text{a}}{\text{b}+\text{c}}+\frac{\text{b}}{\text{c}+\text{a}}}{1-\frac{\text{a}}{\text{b}+\text{c}}\times\frac{\text{b}}{\text{c}+\text{a}}}\end{pmatrix}$
$=\tan^{-1}=\begin{pmatrix}\frac{\frac{\text{ac}+\text{a}^2+\text{b}^2+\text{bc}}{(\text{b}+\text{c})(\text{c}+\text{a})}}{\frac{\text{ac}+\text{c}^2+\text{bc}}{(\text{b}+\text{c})(\text{c}+\text{a})}}\end{pmatrix}$
$=\tan^{-1}\Big(\frac{\text{ac}+\text{c}^2+\text{bc}}{\text{ac}+\text{c}^2+\text{bc}}\Big)$ $\big[\because\text{a}^2+\text{b}^2=\text{c}^2\big]$
$=\tan^{-1}(1)$
$=\tan^{-1}\Big(\tan\frac{\pi}{4}\Big)$
$=\frac{\pi}{4}$

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Question 131 Mark

The value of $\tan\Big\{\cos^{-1}\frac{1}{5\sqrt2}-\sin^{-1}\frac{4}{\sqrt{17}}\Big\}$ is:

$\frac{\sqrt{29}}{3}$
$\frac{29}{3}$
$\frac{\sqrt3}{29}$
$\frac{3}{29}$

Answer

$\frac{3}{29}$

Solution:
Let, $\cos^{-1}\frac{1}{5\sqrt2}=\text{y}$ and $\sin^{-1}\frac{4}{\sqrt{17}}=\text{z}$
$\therefore\ \cos\text{y}=\frac{1}{5\sqrt2}\Rightarrow\sin\text{y}=\frac{7}{5\sqrt2}\Rightarrow\tan\text{y}=7$
$\sin\text{z}=\frac{4}{\sqrt{17}}\Rightarrow\cos\text{z}=\frac{1}{\sqrt{17}}\Rightarrow\tan\text{z}=4$
$\therefore\ \tan\Big(\cos^{-1}\frac{1}{5\sqrt2}-\sin^{-1}\frac{1}{\sqrt{17}}\Big)=\tan(\text{y}-\text{z})$
$=\frac{\tan\text{y}-\tan\text{z}}{1+\tan\text{y}\tan\text{z}}$
$=\frac{7-4}{1+7\times4}$
$=\frac{3}{29}$

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Question 141 Mark

The number of real solution of the equation
$\sqrt{1+\cos2\text{x}}=\sqrt2\sin^{-1}(\sin\text{x}),-\pi\leq\text{x}\leq\pi$ is:

0
1
2
infinite

Answer

2

Solution:
$\sqrt{1+\cos2\text{x}}=\sqrt2\sin^{-1}(\sin\text{x}),-\pi\leq\text{x}\leq\pi$
$\Rightarrow\sqrt{2\cos^2\text{x}}=\sqrt2(-\pi-\text{x})$
$\Rightarrow|\cos\text{x}|=\text{x}$
If $\cos\text{x}$ is positive then $\cos\text{x}=-\pi-\text{x}$
It does not satisfy any value in the interval $\Big(-\pi,-\frac{\pi}{2}\Big)$
For the interval $\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]$
$\cos\text{x}=\text{x}$
It gives the value of x in the $\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]$
For the interval $\Big[-\frac{\pi}{2},\pi\Big]$
$-\cos\text{x}=\pi-\text{x}$
$\cos\text{x}=\text{x}-\pi$
It gives one value of x in the interval $\Big[\frac{\pi}{2},\pi\Big].$
Two real solution in the interval $[-\pi,\pi]$

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Question 151 Mark

$\tan^{-1}\frac{1}{11}+\tan^{-1}\frac{2}{11}$ is equal to:

0
$\frac{1}{2}$
-1
None of these

Answer

None of these

Solution:
$\tan^{-1}\frac{1}{11}+\tan^{-1}\frac{2}{11}$
$=\tan^{-1}\Bigg(\frac{\frac{1}{11}+\frac{2}{11}}{1-\frac{2}{11}\times\frac{1}{11}}\Bigg)$
$=\tan^{-1}\Bigg(\frac{\frac{3}{11}}{1-\frac{2}{121}}\Bigg)$
$=\tan^{-1}\Big(\frac{33}{119}\Big)$

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Question 161 Mark

$\sin\big[\cot^{-1}\big\{\tan\big(\cos^{-1}\text{x}\big)\big\}\big]$ is equal to:

$\text{x}$
$\sqrt{1-\text{x}^2}$
$\frac{1}{\text{x}}$
$\text{none of these}$

Answer

$\text{x}$

Solution:
Put $\cos^{-1}\text{x}=\text{u}$
$\sin\big[\cot^{-1}\big\{\tan\big(\cos^{-1}\text{x}\big)\big\}\big]$
$=\sin\big[\cot^{-1}\{\tan(\text{u})\}\big]$
$=\sin\Big[\cot^{-1}\Big\{\cot\Big(\frac{\pi}{2}-\text{u}\Big )\Big\}\Big]$
$=\sin\Big[\frac{\pi}{2}-\text{u}\Big]$
$=\cos\text{u}$
$=\text{x}$ $\big(\therefore\ \cos^{-1}\text{x}=\text{u}\Rightarrow\text{x}=\cos\text{u}\big)$

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Question 171 Mark

If $\tan^{-1}3+\tan^{-1}\text{x}=\tan^{-1}8,$ then x =

$5$
$\frac{1}{5}$
$\frac{5}{14}$
$\frac{14}{5}$

Answer

$\frac{1}{5}$

Solution:
We know that $\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\frac{\text{x}+\text{y}}{1-\text{xy}}$
Now,
$\tan^{-1}3+\tan^{-1}\text{x}=\tan^{-1}8$
$\Rightarrow\tan^{-1}\Big(\frac{3+\text{x}}{1-3\text{x}}\Big)=\tan^{-1}8$
$\Rightarrow\frac{3+\text{x}}{1-3\text{x}}=8$
$\Rightarrow3+\text{x}=8-24\text{x}$
$\Rightarrow3-8=-24\text{x}-\text{x}$
$\Rightarrow-5=-25\text{x}$
$\Rightarrow\text{x}=\frac{5}{25}=\frac{1}{5}$

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Question 181 Mark

If $3\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)-4\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)+2\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)=\frac{\pi}{3}$ is equal to:

$\frac{1}{\sqrt3}$
$-\frac{1}{\sqrt3}$
$\sqrt3$
$-\frac{\sqrt3}{4}$

Answer

$\frac{1}{\sqrt3}$

Solution:
Let $\text{x}=\tan\text{y}$
Then,
$3\sin^{-1}\Big(\frac{\tan2\text{y}}{1+\tan^2\text{y}}\Big)-4\cos^{-1}\Big(\frac{1-\tan^2\text{y}}{1+\tan^2\text{y}}\Big)+2\tan^{-1}\Big(\frac{2\tan\text{y}}{1-\tan^2\text{y}}\Big)=\frac{\pi}{3}$
$\Rightarrow3\sin^{-1}(\sin2\text{y})-4\cos^{-1}(\cos2\text{y})+2\tan^{-1}(\tan2\text{y})=\frac{\pi}{3}$
$\Big[\because\ \sin2\text{y}=\Big(\frac{2\tan\text{y}}{1+\tan^2\text{y}}\Big),\cos2\text{y}=\Big(\frac{1-\tan^2\text{y}}{1+\tan^2\text{y}}\Big)\text{ and }\tan2\text{y}\Big(\frac{2\tan\text{y}}{1-\tan^2\text{y}}\Big)\Big]$
$\Rightarrow3\times2\text{y}-4\times2\text{y}+2\times2\text{y}=\frac{\pi}{3}$
$\Rightarrow6\text{y}-8\text{y}+4\text{y}=\frac{\pi}{3}$
$\Rightarrow2\text{y}=\frac{\pi}{3}$
$\Rightarrow\text{y}=\frac{\pi}{6}$
$\Rightarrow\tan^{-1}\text{x}=\frac{\pi}{6}$ $\big[\because\ \tan^{-1}\text{x}=\text{y}\big]$
$\Rightarrow\text{x}=\tan\frac{\pi}{6}$
$\Rightarrow\text{x}=\frac{1}{\sqrt3}$

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Question 191 Mark

If $\tan^{-1}(\cot\theta)=2\theta,$ then $\theta=$

$\pm\frac{\pi}{3}$
$\pm\frac{\pi}{4}$
$\pm\frac{\pi}{6}$
$\text{none of these}$

Answer

$\pm\frac{\pi}{6}$

Solution:
We have, $\tan^{-1}(\cot\theta)=2\theta$
$\Rightarrow\tan2\theta=\cot\theta$
$\Rightarrow\frac{2\tan\theta}{1-\tan^2\theta}=\frac{1}{\tan\theta}$
$\Rightarrow2\tan^2\theta=1-\tan^2\theta$
$\Rightarrow3\tan^2\theta=1$
$\Rightarrow\tan^2\theta=\frac{1}{3}$
$\Rightarrow\tan\theta=\pm\frac{1}{\sqrt3}$

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Question 201 Mark

If $\text{f}(\text{x})=\text{e}^{\cos^{-1}\big\{\sin\big(\text{x}+\frac{\pi}{3}\big)\big\}}$ then $\text{f}\Big(\frac{8\pi}{9}\Big)=$

$\text{e}^{\frac{5\pi}{18}}$
$\text{e}^{\frac{13\pi}{18}}$
$\text{e}^{\frac{-2\pi}{18}}$
$\text{none of these}$

Answer

$\text{e}^{\frac{13\pi}{18}}$

Solution:
Given
$\text{f}(\text{x})=\text{e}^{\cos^{-1}\big\{\sin\big(\text{x}+\frac{\pi}{3}\big)\big\}}$
Then,
$\text{f}\Big(\frac{8\pi}{9}\Big)=\text{e}^{\cos^{-1}\big\{\sin\big(\frac{8\pi}{9}+\frac{\pi}{3}\big)\big\}}$
$=\text{e}^{\cos^{-1}\big\{\sin\big(\frac{11\pi}{9}\big)\big\}}=\text{e}^{\cos^{-1}\big\{\cos\frac{\pi}{2}+\frac{13\pi}{18}\big\}}$ $\Big[\because\ \cos\Big(\frac{\pi}{2}+\theta\Big)=\sin\theta\Big]$
$=\text{e}^{\cos^{-1}\big\{\cos\big(\frac{13\pi}{18}\big)\big\}}$
$=\text{e}^{\frac{13\pi}{18}}$

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Question 211 Mark

If $\theta=\sin^{-1}\{\sin(-600^\circ)\},$ then one of the possible values of $\theta$ is:

$\frac{\pi}{3}$
$\frac{\pi}{2}$
$\frac{2\pi}{3}$
$-\frac{2\pi}{3}$

Answer

$\frac{\pi}{3}$

Solution:
$\theta=\sin^{-1}\{\sin(-600^\circ)\}$
$\theta=\sin^{-1}[\sin(-600^\circ)]$
$\theta=\sin^{-1}[-\sin(180^\circ\times3+60)]$
$\theta=\sin^{-1}[-\{-\sin(60^\circ)\}]$
$\theta=\sin^{-1}(\sin(60^\circ))$
$\theta=\sin^{-1}\Big(\frac{\sqrt3}{2}\Big)$
$\theta=\frac{\pi}{3}$

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Question 221 Mark

If $\text{u}=\cot^{-1}\sqrt{\tan\theta}-\tan^{-1}\sqrt{\tan\theta}$ then, $\tan\Big(\frac{\pi}{4}-\frac{\text{u}}{2}\Big)=$

$\sqrt{\tan\theta}$
$\sqrt{\cot\theta}$
$\tan\theta$
$\cot\theta$

Answer

$\sqrt{\tan\theta}$

Solution:
Let $\text{y}=\sqrt{\tan\theta}$
Then,
$\Rightarrow\text{u}=\cot^{-1}\sqrt{\tan\theta}-\tan^{-1}\sqrt{\tan\theta}$
$\Rightarrow\text{u}=\cot^{-1}\text{y}-\tan^{-1}\text{y}$ $\Big[\because\ \tan^{-1}\text{x}+\cot^{-1}\text{x}=\frac{\pi}{2}\Big]$
$\Rightarrow2\tan^{-1}\text{y}=\frac{\pi}{2}-\text{u}$
$\Rightarrow\tan^{-1}\text{y}=\frac{\pi}{4}-\frac{\text{u}}{2}$
$\Rightarrow\text{y}=\tan\Big(\frac{\pi}{4}-\frac{\text{u}}{2}\Big)$
$\Rightarrow\sqrt{\tan\theta}=\tan\Big(\frac{\pi}{4}-\frac{\text{u}}{2}\Big)$ $\Big[\because\ \text{y}=\sqrt{\tan\theta}\Big]$

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Question 231 Mark

The value of $\cos^{-1}\Big(\cos\frac{5\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{5\pi}{3}\Big)$ is:

$\frac{\pi}{2}$
$\frac{5\pi}{3}$
$\frac{10\pi}{3}$
$0$

Answer

0

Solution:
We have
$\cos^{-1}\Big(\cos\frac{5\pi}{3}\Big)+\sin^{-1}\Big(\sin\frac{5\pi}{3}\Big)$
$=\cos^{-1}\Big\{\cos\Big(2\pi-\frac{\pi}{3}\Big)\Big\}+\sin^{-1}\Big\{\sin\Big(2\pi-\frac{\pi}{3}\Big)\Big\}$
$=\cos^{-1}\Big\{\cos\Big(\frac{\pi}{3}\Big)\Big\}+\sin^{-1}\Big\{-\sin\Big(\frac{\pi}{3}\Big)\Big\}$
$=\cos^{-1}\Big\{\cos\Big(\frac{\pi}{3}\Big)\Big\}-\sin^{-1}\Big\{\sin\Big(\frac{\pi}{3}\Big)\Big\}$
$=\frac{\pi}{3}-\frac{\pi}{3}$
$=0$

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MCQ 241 Mark

The value of $\sin\big(2\big(\tan^{-1}0.75\big)\big)$ is equal to:

A
$0.75$
B
$1.5$
✓
$0.96$
D
$\sin^{-1 }1.5$

Answer

Correct option: C.

$0.96$

$\sin\big(2\big(\tan^{-1}0.75\big)\big)=\sin\big(2\tan^{-1}0.75\big)$
$=\sin\Big(\sin^{-1}\frac{2\times0.75}{1+(0.75)^2}\Big)$
$=\sin\big(\sin^{-1}0.96\big)$
$=0 .96$
Hence, the correct answer is option $(c).$

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Question 251 Mark

$\cos^{-1}\frac{\text{x}}{\text{a}}+\cos^{-1}\frac{\text{y}}{\text{b}}=\alpha,$ then $\frac{\text{x}^2}{\text{a}^2}-\frac{2\text{xy}}{\text{ab}}\cos\alpha+\frac{\text{y}^2}{\text{b}^2}=$

$\sin^2\alpha$
$\cos^2\alpha$
$\tan^2\alpha$
$\cot^2\alpha$

Answer

$\sin^2\alpha$

Solution:
$\cos^{-1}\text{x}+\cos^{-1}\text{y}=\cos^{-1}\Big(\text{xy}-\sqrt{1-\text{x}^2}\sqrt{1-\text{y}^2}\Big)$
Consider, $\cos^{-1}\frac{\text{x}}{\text{a}}+\cos^{-1}\frac{\text{y}}{\text{b}}=\alpha$
$\Rightarrow\cos^{-1}\bigg(\frac{\text{x}}{\text{a}}\times\frac{\text{y}}{\text{b}}-\sqrt{1-\frac{\text{x}^2}{\text{a}^2}}\sqrt{1-\frac{\text{y}^2}{\text{a}^2}}\bigg)$
$\Rightarrow\frac{\text{x}}{\text{a}}\times\frac{\text{y}}{\text{b}}-\sqrt{1-\frac{\text{x}^2}{\text{a}^2}}\sqrt{1-\frac{\text{y}^2}{\text{a}^2}}=\cos\alpha$
$\Rightarrow\frac{\text{x}}{\text{a}}\times\frac{\text{y}}{\text{b}}-\cos\alpha=\sqrt{1-\frac{\text{x}^2}{\text{a}^2}}\sqrt{1-\frac{\text{y}^2}{\text{a}^2}}$
Squaring on both sides,
$\Rightarrow\frac{\text{x}^2\text{y}^2}{\text{a}^2\text{b}^2}+\cos^2\alpha-\frac{2\text{xy}}{\text{ab}}\cos\alpha=\Big(1-\frac{\text{x}^2}{\text{a}^2}\Big)\Big(1-\frac{\text{y}^2}{\text{a}^2}\Big)$
$\Rightarrow\frac{\text{x}^2\text{y}^2}{\text{a}^2\text{b}^2}+\cos^2\alpha-\frac{2\text{xy}}{\text{ab}}\cos\alpha=1-\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{a}^2}+\frac{\text{x}^2\text{y}^2}{\text{a}^2\text{b}^2}$
$\Rightarrow\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{a}^2}-\frac{2\text{xy}}{\text{ab}}\cos\alpha=1-\cos^2\alpha$
$\Rightarrow\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{a}^2}-\frac{2\text{xy}}{\text{ab}}\cos\alpha=\sin^2\alpha$

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Question 261 Mark

If $4\cos^{-1}\text{x}+\sin^{-1}\text{x}=\pi,$ then the value of x is:

$\frac{3}{2}$
$\frac{1}{\sqrt2}$
$\frac{\sqrt3}{2}$
$\frac{2}{\sqrt3}$

Answer

$\frac{\sqrt3}{2}$

Solution:
We know that $\sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}$
$4\cos^{-1}\text{x}+\sin^{-1}\text{x}=\pi$
$\Rightarrow4\cos^{-1}\text{x}+\frac{\pi}{2}-\cos^{-1}\text{x}=\pi $
$\Rightarrow3\cos^{-1}\text{x}=\pi-\frac{\pi}{2}$
$\Rightarrow3\cos^{-1}\text{x}=\frac{\pi}{2}$
$\Rightarrow\cos^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow\text{x}=\cos\frac{\pi}{6}$

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Question 271 Mark

The value of $\tan\Big(\cos^{-1}\frac{3}{5}+\tan^{-1}\frac{1}{4}\Big)=$

$\frac{19}{8}$
$\frac{8}{19}$
$\frac{19}{2}$
$\frac{3}{4}$

Answer

$\frac{19}{8}$

Solution:
$\tan\Big(\cos^{-1}\frac{3}{5}+\tan^{-1}\frac{1}{4}\Big)$
$=\tan\Bigg(\tan^{-1}\frac{\sqrt{1-\frac{9}{25}}}{\frac{3}{5}}+\tan^{-1}\frac{1}{4}\Bigg)$
$=\tan\Bigg(\tan^{-1}\frac{\frac{4}{5}}{\frac{3}{5}}+\tan^{-1}\frac{1}{4}\Bigg)$
$=\tan\Big(\tan^{-1}\frac{4}{3}+\tan^{-1}\frac{1}{4}\Big)$
$=\tan\bigg(\tan^{-1}\frac{\frac{4}{3}+\frac{1}{4}}{1-\frac{1}{3}}\bigg)$
$=\frac{\frac{16+3}{12}}{\frac{2}{3}}$
$=\frac{19}{8}$
Hence, the correct answer is option (a).

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Question 281 Mark

The domain of $\cos^{-1}\big(\text{x}^2-4\big)$ is:

$[3,5]$
$[-1,1]$
$\Big[-\sqrt5,-\sqrt3\Big]\cup\Big[\sqrt3,\sqrt5\Big]$
$\Big[-\sqrt5,-\sqrt3\Big]\cap\Big[\sqrt3,\sqrt5\Big]$

Answer

$\Big[-\sqrt5,-\sqrt3\Big]\cup\Big[\sqrt3,\sqrt5\Big]$

Solution:
Let, $\cos^{-1}\big(\text{x}^2-4\big)=\text{y}$
$\Rightarrow\cos\text{y}=\text{x}^2-4$
$\Rightarrow-1\leq\text{x}^2-4\leq1$
$\Rightarrow3\leq\text{x}^2\leq5$
$\Rightarrow\pm\sqrt3\leq\text{x}\pm\sqrt5$
$\text{x}\in\Big[-\sqrt5,-\sqrt3\Big]\cup\Big[\sqrt3,\sqrt5\Big]$

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Question 291 Mark

If $\alpha=\tan^{-1}\Big(\tan\frac{5\pi}{4}\Big)$ and $\beta=\tan^{-1}\Big(-\tan\frac{2\pi}{3}\Big),$ then:

$4\alpha=3\beta$
$3\alpha=4\beta$
$\alpha-\beta=\frac{7\pi}{12}$
$\text{none of these}$

Answer

$4\alpha=3\beta$

Solution:
We know that $\tan^{-1}(\tan\text{x})=\text{x}$
$\therefore\ \alpha=\tan^{-1}\Big(\tan\frac{5\pi}{4}\Big)$
$=\tan^{-1}\Big\{\tan\Big(\pi+\frac{\pi}{4}\Big)\Big\}$
$=\tan^{-1}\Big(\tan\frac{\pi}{4}\Big)$
$=\frac{\pi}{4}$
and
$\beta=\tan^{-1}\Big\{-\tan\Big(\frac{2\pi}{3}\Big)\Big\}$
$=\tan^{-1}\Big\{-\tan\Big(\pi-\frac{\pi}{3}\Big)\Big\}$
$=\tan^{-1}\Big\{\tan\Big(\frac{\pi}{3}\Big)\Big\}$
$=\frac{\pi}{3}$
$\therefore\ 4\alpha=\pi$
$3\beta=\pi$
$\therefore\ 4\alpha=3\beta$

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Question 301 Mark

The value of $\sin\Big(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\Big)$ is:

$\frac{1}{\sqrt2}$
$\frac{1}{\sqrt3}$
$\frac{1}{2\sqrt2}$
$\frac{1}{3\sqrt3}$

Answer

$\frac{1}{2\sqrt2}$

Solution:
$\sin\Big(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\Big)$
Let, $\sin^{-1}\frac{\sqrt{63}}{8}=\text{x}$
$\sin\text{x}=\frac{\sqrt{63}}{8}$
$\cos\text{x}\sqrt{1-\sin^2\text{x}}$
$\cos\text{x}=\sqrt{1-\frac{63}{64}}$
$\cos\text{x}=\frac{1}{8}$
Consider,
$\sin\Big(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\Big)$
$=\sin\Big(\frac{1}{4}\text{x}\Big)$
$=\sqrt{\frac{1-\cos\frac{\text{x}}{2}}{2}}$ $\Big(\because\ \sin\text{x}=\frac{1-\cos2\text{x}}{2}\Big)$
$=\sqrt{\frac{1-\sqrt{\frac{1+\cos\text{x}}{2}}}{2}}$ $\Big(\because\ \cos\text{x}=\frac{1+\cos2\text{x}}{2}\Big)$
$=\sqrt{\frac{1-\sqrt{1-\frac{1}{8}}}{2}}$
$=\sqrt{\frac{1-\frac{3}{4}}{2}}$
$=\sqrt{\frac{1}{8}}$
$=\frac{1}{2\sqrt2}$

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Question 311 Mark

If $\sin^{-1}\text{x}-\cos^{-1}\text{x}=\frac{\pi}{6},$ then x =

$\frac{1}{2}$
$\frac{\sqrt3}{2}$
$-\frac{1}{2}$
$\text{none of these}$

Answer

$\frac{\sqrt{3}}{2}$

Solution:
We know that $\sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}$
$\therefore\ \sin^{-1}\text{x}-\cos^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow\frac{\pi}{2}-\cos^{-1}\text{x}-\cos^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow-2\cos^{-1}\text{x}=\frac{\pi}{6}-\frac{\pi}{2}$
$\Rightarrow-2\cos^{-1}\text{x}=-\frac{\pi}{3}$
$\Rightarrow\cos^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow\text{x}=\cos\frac{\pi}{6}$
$\Rightarrow\text{x}=\frac{\sqrt3}{2}$

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Question 321 Mark

$\cot\Big(\frac{\pi}{4}-2\cot^{-1}3\Big)=$

7
6
5
none of these

Answer

7

Solution:
Let $2\cot^{-1}3=\text{y}$
Then, $\cot\frac{\text{y}}{2}=3$
$\cot\Big(\frac{\pi}{4}-2\cot^{-1}3\Big)=\cot\Big(\frac{\pi}{4}-\text{y}\Big)$
$=\frac{\cot\frac{\pi}{4}\cot\text{y}+1}{\cot\text{y}-\cot\frac{\pi}{4}}$
$=\frac{\cot\text{y}+1}{\cot\text{y}-1}$
$=\frac{\frac{\cot^2\frac{\text{y}}{2}-1}{2\cot\frac{\text{y}}{2}}+1}{\frac{\cot^2\frac{\text{y}}{2}-1}{2\cot\frac{\text{y}}{2}}-1}$
$=\frac{\cot^2\frac{\text{y}}{2}+2\cot\frac{\text{y}}{2}-1}{\cot^2\frac{\text{y}}{2}-2\cot\frac{\text{y}}{2}-1}$
$=\frac{9+6-1}{9-6-1}$
$=7$

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Question 331 Mark

The positive integral solution of the equation$\tan^{-1}\text{x}+\cos^{-1}\frac{\text{y}}{\sqrt{1+\text{y}^2}}=\sin^{-1}\frac{3}{\sqrt{10}}$ is:

x = 1, y = 2
x = 2, y = 1
x = 3, y = 2
x = -2, y = -1

Answer

x = 1, y = 2

Solution:
We have,
$\tan^{-1}\text{x}+\cos^{-1}\frac{\text{y}}{\sqrt{1+\text{y}^2}}=\sin^{-1}\frac{3}{\sqrt{10}}$
$\Rightarrow\tan^{-1}\text{x}+\tan^{-1}\begin{bmatrix}\frac{\sqrt{1-\bigg(\frac{\text{y}}{\sqrt{1+\text{y}^2}}\bigg)^2}}{\frac{\text{y}}{\sqrt{1+\text{y}^2}}}\end{bmatrix}=\tan^{-1}\begin{bmatrix}\frac{\frac{3}{\sqrt{10}}}{\sqrt{1-\Big(\frac{3}{\sqrt{10}}\Big)^2}}\end{bmatrix}$

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Question 341 Mark

If x > 1, then $2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ is equal to:

$4\tan^{-1}\text{x}$
$0$
$\frac{\pi}{2}$
$\pi$

Answer

$4\tan^{-1}\text{x}$

Solution:
$2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)=2\tan^{-1}\text{x}+2\tan^{-1}\text{x}$
$\Big[\because\ \sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)=2\tan^{-1}\text{x}\Big]$
$=4\tan^{-1}\text{x}$

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Question 351 Mark

If $\cos^{-1}\text{x}>\sin^{-1}\text{x},$ then:

$\frac{1}{\sqrt2}<\text{x}\leq1$
$0\leq\text{x}\leq\frac{1}{\sqrt2}$
$-1\leq\text{x}<\frac{1}{\sqrt2}$
$\text{x}>0$

Answer

$\frac{1}{\sqrt2}<\text{x}\leq1$

Solution:
$\cos^{-1}\text{x}>\sin^{-1}\text{x}$
$\Rightarrow\cos^{-1}\text{x}>\frac{\pi}{2}-\cos^{-1}\text{x}$
$\Rightarrow2\cos^{-1}\text{x}>\frac{\pi}{2}$
$\Rightarrow\cos^{-1}\text{x}>\frac{\pi}{4}$
$\Rightarrow\text{x}>\cos\frac{\pi}{4}$
$\Rightarrow\text{x}>\frac{1}{\sqrt2}$
We know that the maximum value of cosine function is 1.
$\therefore\ \frac{1}{\sqrt2}<\text{x}\leq1$
Hence, the correct ans is option (a).

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M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip