Question

Match $\text{List I}$ with the $\text{List II}$ and select the correct answer using the code given below the lists :

Let $[.]$ denote the greatest integer function and $\text{sgn}(.)$ denote the signum function.
$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{If}I=\underset{0}{\overset{\pi /4}{\int }}{\sin }^{3}x{\cos }^{3}xdx,\text{then}{I}^{-1}\text{is divisible by}& \text{(P)}& 1\\ \text{(B)}& \text{If}I=\underset{0}{\overset{2\pi }{\int }}\left[2\sin x\right]dx,\text{then}\left|\left[I\right]\right|\text{is divisible by}& \text{(Q)}& 2\\ \text{(C)}& \text{If}I=\underset{-2}{\overset{2}{\int }}\text{sgn}(x-1)dx,\text{then}|I|\text{is equal to}& \text{(R)}& 4\\ \text{(D)}& \text{If}I=\underset{0}{\overset{1}{\int }}\frac{x^3-1}{x+1}dx,\text{then}[3I+\ln 64]\text{is divisible by}& \text{(S)}& 8\\ & & \text{(T)}& 16\end{array}$

Which of the following is the only CORRECT combination ?

• A. $\text{(A)}\to \text{(P)},\text{(Q)},\text{(R)};\text{(B)}\to \text{(P)},\text{(Q)},\text{(R)}$
• B. $\text{(A)}\to \text{(P)},\text{(Q)},\text{(R)},\text{(S)};\text{(B)}\to \text{(Q)},\text{(R)},\text{(S)}$
• C. $\text{(A)}\to \text{(P)},\text{(Q)},\text{(R)},\text{(S)};\text{(B)}\to \text{(P)},\text{(Q)},\text{(R)}$
• D. $\text{(A)}\to \text{(R)},\text{(S)},\text{(T)};\text{(B)}\to \text{(P)},\text{(Q)},\text{(R)},\text{(T)}$

Answer 1

The correct option is C $\text{(A)}\to \text{(P)},\text{(Q)},\text{(R)},\text{(S)};\text{(B)}\to \text{(P)},\text{(Q)},\text{(R)}$

$\text{(A)}$
$I=\underset{0}{\overset{\pi /4}{\int }}{\sin }^{3}x{\cos }^{3}xdx$
$=\frac{1}{8}\underset{0}{\overset{\pi /4}{\int }}{\left(\sin 2x\right)}^{3}dx$
Put $2x=t\Rightarrow dx=\frac{dt}{2}$
$I=\frac{1}{16}\underset{0}{\overset{\pi /2}{\int }}{\left(\sin t\right)}^{3}dt$
$=\frac{1}{16}\times \frac{2}{3\times 1}=\frac{1}{24}$ [By Walli's theorem]

$\text{(A)}\to \text{(P)},\text{(Q)},\text{(R)},\text{(S)}$

$\text{(B)}$
We know, $\left[y\right]+[-y]=-1$ for all non-integral of $y.$
$I=\underset{0}{\overset{2\pi }{\int }}\left[2\sin x\right]dx$
$\Rightarrow I=\underset{0}{\overset{2\pi }{\int }}\left[2\sin \right(2\pi -x\left)\right]dx=\underset{0}{\overset{2\pi }{\int }}[-2\sin x]dx$
Adding both,
$2I=\underset{0}{\overset{2\pi }{\int }}(\left[2\sin x\right]+[-2\sin x])dx$
$\Rightarrow 2I=\underset{0}{\overset{2\pi }{\int }}-dx=-2\pi$
$\Rightarrow I=-\pi$
$\left|\right[I\left]\right|=|-4|=4$

$\text{(B)}\to \text{(P)},\text{(Q)},\text{(R)}$