Assertion (A) & Reason (B) MCQ

MCQ 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 $[0, \pi]-\left\{\frac{\pi}{2}\right\}$.

A
Both Assertion (A) and Reason (R) are true and Reason $(R)$ is the correct explanation of the Assertion (A).
B
Both Assertion (A) and Reason (R) are true, but Reason $(R)$ is not the correct explanation of the Assertion (A).
C
Assertion (A) is true, but Reason (R) is false.
D
Assertion (A) is false, but Reason (R) is true.

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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MCQ 21 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 $[0, \pi]$

A
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
B
Both Assertion (A) and Reason (R) are true, but Reason $(R)$ is not the correct explanation of the Assertion (A).
C
Assertion (A) is true and Reason (R) is false.
D
Assertion (A) is false and Reason (R) is true.

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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MCQ 31 Mark

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

A
Both Assertion (A) and Reason (R) are true and Reason $(R)$ is the correct explanation of the Assertion (A).
B
Both Assertion (A) and Reason (R) are true, but Reason $(R)$ is not the correct explanation of the Assertion (A).
C
Assertion (A) is true and Reason (R) is false.
D
Assertion (A) is false and Reason (R) is true.

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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MCQ 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 $x \in R$.

A
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
B
Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
C
Assertion (A) is true but Reason (R) is false.
D
Assertion (A) is false but Reason (R) is true.

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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MCQ 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 $(R): \sec ^{-1}(-2)=-\frac{\pi}{4}$

A
Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.
B
Both $A$ and $R$ are true but $R$ is not the correct explanation of $A$.
C
$A$ is true but $R$ is false.
D
$A$ is false but $R$ is true.

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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MCQ 61 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): $\sec ^{-1}(-2)=-\frac{\pi}{4}$.

A
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
✓
(A) is true but (R) is false.
D
(A) is false but (R) is true.

Answer

Correct option: C.

(A) is true but (R) is false.

(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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MCQ 71 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.

✓
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
C
(A) is true but (R) is false.
D
(A) is false but (R) is true.

Answer

Correct option: A.

Both (A) and (R) are true and (R) is the correct explanation of (A).

(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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MCQ 81 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 $x \in[-1,1]$.

A
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
C
(A) is true but (R) is false.
✓
(A) is false but (R) is true.

Answer

Correct option: D.

(A) is false but (R) is true.

(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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MCQ 91 Mark

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

A
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
C
(A) is true but (R) is false.
✓
(A) is false but (R) is true.

Answer

Correct option: D.

(A) is false but (R) is true.

(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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MCQ 101 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 $x \in[-1,1]$.

A
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
✓
(A) is true but (R) is false.
D
(A) is false but (R) is true.

Answer

Correct option: C.

(A) is true but (R) is false.

(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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MCQ 111 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 $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

✓
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
C
(A) is true but (R) is false.
D
(A) is false but (R) is true.

Answer

Correct option: A.

Both (A) and (R) are true and (R) is the correct explanation of (A).

(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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MCQ 121 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 $x \in R$.

A
Both (A) and (R) are true and (R) is the correct explanation of (A).
B
Both (A) and (R) are true but (R) is not the correct explanation of (A).
C
(A) is true but (R) is false.
D
(A) is false but (R) is true.

Answer

$
\begin{array}{l}
\text { (d) : Let } \theta=\cot ^{-1}\left(2 x-x^2\right) \text {, where } \theta \in(0, \pi) \\
\Rightarrow \quad \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)
\end{array}
$
$\Rightarrow \cot \theta \leq 1$, 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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