Assertion (A) & Reason (B) MCQ

Question 11 Mark

Assertion (A): Domain of $y=\cos ^{-1}(x)$ is $[-1,1]$.
Reason $(R)$ : The range of the principal value branch of $y=\cos ^{-1}(x)$ is $[0, \pi]-\left\{\frac{\pi}{2}\right\}$.

Answer

Domain of $\cos ^{-1}(x)$ is $[-1,1]$.
$\therefore \quad$ Assertion $( A )$ is true.
The range of principal value branch of $\cos ^{-1}(x)$ is $[0, \pi]$.
$\therefore \quad$ Reason $(R)$ is false.

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Question 21 Mark

Assertion (A): The principal value of $\cot ^{-1}(\sqrt{3})$ is $\frac{\pi}{6}$.
Reason $( R )$ : Domain of $\cot ^{-1} x$ is $R -\{-1,1\}$.

Answer

We know that, $\cot ^{-1}(x), x \in(0, \pi)$
$\cot ^{-1}(\sqrt{3})=\cot ^{-1}\left(\cot \frac{\pi}{6}\right)=\frac{\pi}{6}$
$\left[\because \cot ^{-1}(\cot \theta)=\theta\right]$
$\therefore \quad$ Assertion $( A )$ is true.
Domain of $\cot ^{-1} x$ is $R$.
So, reason (R) is false.

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Question 31 Mark

Assertion $(A)$ : The range of the function $f(x)=2 \sin ^{-1} x+\frac{3 \pi}{2}$, where $x \in[-1,1]$, is $\left[\frac{\pi}{2}, \frac{5 \pi}{2}\right]$.
Reason $(R)$ : The range of the principal value branch of $\sin ^{-1}(x)$ is $[0, \pi]$

Answer

Given, $f(x)=2 \sin ^{-1} x+\frac{3 \pi}{2}, x \in[-1,1]$
As, $-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2}$
$\Rightarrow-2 \times \frac{\pi}{2} \leq 2 \sin ^{-1} x \leq 2 \times \frac{\pi}{2} \Rightarrow-\pi \leq 2 \sin ^{-1} x \leq \pi$
$\Rightarrow \quad-\pi+\frac{3 \pi}{2} \leq 2 \sin ^{-1} x+\frac{3 \pi}{2} \leq \pi+\frac{3 \pi}{2}$
$\Rightarrow \frac{\pi}{2} \leq 2 \sin ^{-1} x+\frac{3 \pi}{2} \leq \frac{5 \pi}{2}$
$\therefore \quad$ The range of $f(x)$ is $\left[\frac{\pi}{2}, \frac{5 \pi}{2}\right]$
$\therefore \quad$ Assertion $(A)$ is true.
Now, the range of the principal branch of $\sin ^{-1}(x)$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
$\therefore \quad$ Reason (R) false.

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Question 41 Mark

Assertion (A) : All trigonometric functions have their inverses over their respective domains.
Reason $( R )$ : The inverse of $\tan ^{-1} x$ exists for some $x \in R$.

Answer

All trigonometric functions are periodic and hence not invertible over their respective domains but all trigonometric functions have inverse over their restricted domains.
Inverse of $\tan ^{-1} x$ is $\tan x$ which is defined for
$x \in R-(2 n+1) \frac{\pi}{2}, n \in Z$
$\therefore \quad$ Assertion is false and reason is true

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Question 51 Mark

Assertion (A): The domain of the function $\sec ^{-1} 2 x$ is $\left(-\infty,-\frac{1}{2}\right] \cup\left[\frac{1}{2}, \infty\right) .$ Reason $(R): \sec ^{-1}(-2)=-\frac{\pi}{4}$

Answer

$\sec ^{-1} x$ is defined if $x \leq-1$ or $x \geq 1$.
Hence, $\sec ^{-1} 2 x$ will be defined if $x \leq-\frac{1}{2}$ or $x \geq \frac{1}{2}$
The range of the function $\sec ^{-1} x$ is $[0, \pi]-\left\{\frac{\pi}{2}\right\}$
Hence, $A$ is true and $R$ is false.

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Question 61 Mark

Assertion (A) : Principal value of $\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$ is $\frac{\pi}{3}$.
Reason (R) : Principal value branch of $\sin ^{-1}$ function is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Answer

(a) : Let $y=\sin ^{-1}\left(\frac{2 \pi}{3}\right) \sin ^{-1}(\sin (\pi-\pi / 3))$
$=\sin ^{-1}\left(\sin \frac{\pi}{3}\right)=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
$=\frac{\pi}{3} \quad\left[\because \sin ^{-1}:[-1,1] \rightarrow\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\right]$

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Question 71 Mark

Assertion (A) : The domain of the function
$\sec ^{-1} 2 x$ is $\left(-\infty,-\frac{1}{2}\right] \cup\left[\frac{1}{2}, \infty\right)$.
Reason (R): $\sec ^{-1}(-2)=-\frac{\pi}{4}$.

Answer

(c) : $\sec ^{-1} x$ is defined if $x \leq-1$ or $x \geq 1$.
Hence, $\sec ^{-1} 2 x$ will be defined if $x \leq-\frac{1}{2}$ or $x \geq \frac{1}{2}$
The range of the function $\sec ^{-1} x$ is $[0, \pi]-\left\{\frac{\pi}{2}\right\}$.
Hence, Assertion is true and Reason is false.

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Question 81 Mark

Assertion (A) : Number of roots of the equation $\cot ^{-1} x+\cos ^{-1} 2 x+\pi=0$ is zero.
Reason (R) : Range of $\cot ^{-1} x$ and $\cos ^{-1} x$ is $(0, \pi)$ and $[0, \pi]$, respectively.

Answer

(a): Reason is correct, from which we can say $\cot ^{-1} x+\cos ^{-1} 2 x=-\pi$ is not possible. Hence, both the assertion, reason are correct, and reason is the correct explanation of assertion.

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Question 91 Mark

Assertion (A) : The domain for
$f(x)=\sin ^{-1}\left(\frac{1+x^2}{2 x}\right)$ is $\{0,1\}$.
Reason (R) : $\sin ^{-1} x$ is defined only if $x \in[-1,1]$.

Answer

(d) : $f(x)=\sin ^{-1}\left(\frac{1+x^2}{2 x}\right)$ is defined for
$
-1 \leq \frac{1+x^2}{2 x} \leq 1
$
or $\left|\frac{1+x^2}{2 x}\right| \leq 1$
$\Rightarrow\left|1+x^2\right| \leq|2 x|$, for all $x \in R$
$\Rightarrow 1+x^2 \leq|2 x|$, for all $x\left(\right.$ as $1+x^2>0$ )
$\Rightarrow x^2-2|x|+1 \leq 0$
$\Rightarrow|x|^2-2|x|+1 \leq 0\left(\right.$ as $\left.x^2=|x|^2\right)$
$\Rightarrow(|x|-1)^2 \leq 0$
But $(|x|-1)^2$ is always either positive or zero. Thus,
$
(|x|-1)^2=0
$
or $\quad|x|=1$ or $x= \pm 1$
Hence, domain for $f(x)$ is $\{-1,1\}$.

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Question 101 Mark

Assertion (A) : All trigonometric functions have their inverses over their respective domains.
Reason (R) : The inverse of $\tan ^{-1} x$ exists for some $x \in R$.

Answer

(d) : All trigonometric functions are periodic and hence not invertible over their respective domains but all trigonometric functions have inverse over their restricted domains.
Inverse of $\tan ^{-1} x$ is $\tan x$ which is defined for
$
x \in R-(2 n+1) \frac{\pi}{2}, n \in Z
$
$\therefore \quad$ Assertion is false and reason is true.

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Question 111 Mark

Assertion $(A) :$ Range of $f(x)=\cot ^{-1}\left(2 x-x^2\right)$ is $(0, \pi)$.
Reason $(R) : \cot ^{-1} x$ is defined for all $x \in R$.

Answer

$\text { (d) : Let } \theta=\cot ^{-1}\left(2 x-x^2\right) \text {, where } \theta \in(0, \pi)$
$\Rightarrow \cot \theta=2 x-x^2 \text {, where } \theta \in(0, \pi)$
$=1-\left(1-2 x+x^2\right) \text {, where } \theta \in(0, \pi)$
$=1-(1-x)^2 \text {, where } \theta \in(0, \pi)$
$\Rightarrow \cot \theta \leq 1$, where $\theta \in(0, \pi)$
$\Rightarrow \frac{\pi}{4} \leq \theta<\pi $
$\Rightarrow$ Range of $f(x)$ is $\left[\frac{\pi}{4}, \pi\right)$.

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Question 121 Mark

Assertion (A) : Range of $f(x)=\sin ^{-1} x$ $+\tan ^{-1} x+\sec ^{-1} x$ is $\left\{\frac{\pi}{4}, \frac{3 \pi}{4}\right\}$.
Reason (R) : $f(x)=\sin ^{-1} x+\tan ^{-1} x+\sec ^{-1} x$ is defined for all $x \in[-1,1]$.

Answer

(c) : $f(x)=\sin ^{-1} x+\tan ^{-1} x+\sec ^{-1} x$;
clearly, domain of $f(x)$ is $x= \pm 1$.
Thus, the range is $\{f(1), f(-1)\}$, i.e., $\left\{\frac{\pi}{4}, \frac{3 \pi}{4}\right\}$.

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MATHS Assertion (A) & Reason (B) MCQ

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