Question

Match the List I with List II.
$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{P.}& \text{The value of the integral}& 1.& 0\\ & {\int }_{-\pi }^{\pi }\frac{x\sin x}{e^x+1}dx\text{is}& & \\ \text{Q.}& \text{The value of}& 2.& xy=e^x(x-1)+c\\ & {\int }_{-\frac{2}{3}}^{\frac{2}{3}}\frac{x^2.\log (x+\sqrt{1+x^2})}{\cos x}dx& & \\ & \text{is}& & \\ \text{R.}& \text{The solution of}& 3.& y{x}^2+e^x=cy\\ & x\frac{dy}{dx}+y=x{e}^x\text{is}& & \\ \text{S.}& \text{The solution of}& 4.& \pi \\ & y(2xy+e^x)dx-e^{x}dy=0& & \\ & \text{is}& & \end{array}$

$\begin{array}{ccccc}& \text{P}& \text{Q}& \text{R}& \text{S}\\ \text{(A)}& 4& 1& 2& 3\\ \text{(B)}& 1& 4& 2& 3\\ \text{(C)}& 1& 4& 3& 2\\ \text{(D)}& 4& 1& 3& 2\end{array}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

(P) Let $l={\int }_{-\pi }^{\pi }\frac{x\sin x}{e^x+1}dx$
Put $x=-t\Rightarrow dx=-dt$
$l={\int }_{-\pi }^{\pi }\frac{x\sin x.e^x}{1+e^x}dx$
adding $2l={\int }_{-\pi }^{\pi }x\sin xdx\Rightarrow l=\pi$
(Q) $l={\int }_{-\frac{2}{3}}^{\frac{2}{3}}\frac{x^2\log (x+\sqrt{1+x^2})}{\cos x}dx=0$
As given integr and i.e., $f\left(x\right)=\frac{x^2\log (x+\sqrt{1+x^2})}{\cos x}$ is an odd function.
(R) $\frac{dy}{dx}+\frac{y}{x}=e^x$
Its integrating factor $=e^{\int \frac{1}{x}dx}=x$
So, its solution is $y.x=\int x{e}^{x}dx$
$\Rightarrow xy=e^x(x-1)+c$
(S) $\frac{1}{y^2}.\frac{dy}{dx}+\left(\frac{-1}{y}\right)=2x{e}^{-x}$
Since, the given equation in the form of $\frac{dy}{dx}+Py=Q{y}^n$. Therefore solution becomes $e^x+x^{2}y=cy$, where $c$ is the arbitrary parameter.