Assertion (A) & Reason (B) MCQ

Question 11 Mark

Assertion $(A): \int_2^8 \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x=3$
Reason $(R): \int_a^b f(x) d x=\int_a^b f(a+b-x) d x$

Answer

Let $I=\int_2^8 \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x$
$=\int_2^8 \frac{\sqrt{10-(10-x)}}{\sqrt{10-x}+\sqrt{10-(10-x)}} d x\left(\because \int_a^b f(x) d x=\frac{b}{a} f(a+b-x) d x\right)$
$=\int_2^8 \frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}} d x$
Adding $(i)$ and $(ii)$, we get
$2 I=\int_2^8 \frac{\sqrt{10-x}+\sqrt{x}}{\sqrt{x}+\sqrt{10-x}} d x=\int_2^8 1 d x=[x]_2^8$
$\Rightarrow I =\frac{1}{2}(8-2)=\frac{6}{2}=3
$Hence, both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ is the correct explanation of assertion $(A).$

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Question 21 Mark

Assertion (A) : $\int_0^{2 \pi} \sin ^3 x d x=0$
Reason (R) : $\sin ^3 x$ is an odd function.

Answer

(b) : Let $I=\int_0^{2 \pi} \sin ^3 x d x=\int_0^{2 \pi}\left(1-\cos ^2 x\right) \sin x d x$
Putting $\cos x=t \Rightarrow \sin x d x=-d t$
When $x=0, t=1$ and $x=2 \pi, t=1$
$\therefore \quad I=\int_1^1\left(1-t^2\right)(-d t)=0$

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Question 31 Mark

Assertion (A) : The value of$\int_{-3}^3\left(a x^5+b x^3+c x+k\right) d x,$ where $a, b, c, k$ are constants, depends on only $k$.
Reason (R) : $\int_{-a}^a f(x) d x=0$, if $f(-x)=-f(x)$ i.e., $f$ is an odd function.

Answer

(a) : Clearly, Reason is true.
Let $I=\int_{-3}^3\left(a x^5+b x^3+c x+k\right) d x$
$
=a \int_{-3}^3 x^5 d x+b \int_{-3}^3 x^3 d x+c \int_{-3}^3 x d x+k \int_{-3}^3 1 d x
$
Since, $x^5, x^3, x$ are odd function
$
\therefore \quad I=0+0+0+k[x]_{-3}^3=6 k \text {, }
$
which is dependent only on $k$.

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Question 41 Mark

Assertion $(A) : I=\int_0^1 \frac{d x}{\sqrt[3]{1+x^3}}=\int_0^{2^{-1 / 3}} \frac{d t}{1-t^3}$
Reason $(R) :$ The integrand of the integral $I$ becomes rational by the substitution $t=\frac{x}{\sqrt[3]{1+x^3}}$.

Answer

$\text { (a) : Let } t=\frac{x}{\sqrt[3]{1+x^3}} \Rightarrow d t=\frac{d x}{\left(1+x^3\right)^{\frac{4}{3}}}$
$\therefore \quad\left(1+x^3\right) t^3=x^3 \Rightarrow t^3+x^3 t^3=x^3$
$\Rightarrow t^3=x^3\left(1-t^3\right) \Rightarrow x^3=\frac{t^3}{1-t^3} \Rightarrow 1+x^3=\frac{1}{1-t^3}$When $x=0, t=0$ and $x=1, t=2^{-1 / 3}$
$\Rightarrow I=\int_0^{2^{-1 / 3}} \frac{d t}{1-t^3}$

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Question 51 Mark

Let $F(x)$ be an indefinite integral of $\sin ^2 x$.
Assertion (A) : The function $F(x)$ satisfies $F(x+\pi)=F(x)$ for all real $x$.
Reason (R) : $\sin ^2(x+\pi)=\sin ^2 x$ for all real $x$.

Answer

$
\begin{aligned}
(d): F(x) & =\int \sin ^2 x d x=\int \frac{1}{2}(1-\cos 2 x) d x \\
& =\frac{x}{2}-\frac{\sin 2 x}{4}+C
\end{aligned}
$
$
\because \quad F(x+\pi)-F(x)=\frac{\pi}{2} \neq 0
$
$\therefore \quad$ Assertion is false.
$
\sin ^2(x+\pi)=(-\sin x)^2=\sin ^2 x
$
$\therefore \quad$ Reason is true.

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Question 61 Mark

Assertion $(A) : \int \sin 3 x \cos 5 x d x=\frac{-\cos 8 x}{16}+\frac{\cos 2 x}{4}+C$
Reason $(R) : 2 \cos A \sin B=\sin (A+B)-\sin (A-B)$

Answer

$(a)$ : We have, $\int \sin 3 x \cos 5 x d x$
$=\frac{1}{2} \int 2 \cos 5 x \sin 3 x d x$
$=\frac{1}{2} \int(\sin 8 x-\sin 2 x) d x$
$=\frac{1}{2}\left[\int \sin 8 x d x-\int \sin 2 x d x\right]$
$=\frac{1}{2}\left[\frac{-\cos 8 x}{8}\right]-\frac{1}{2}\left[\frac{-\cos 2 x}{2}\right]+C$
$=\frac{-\cos 8 x}{16}+\frac{\cos 2 x}{4}+C$
$\therefore$ Both assertion and reason are true and reason is the correct explanation of assertion. $\text{ZXZ}$

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MATHS Assertion (A) & Reason (B) MCQ

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