Question

$\begin{array}{cc}ColumnI& ColumnII\\ \text{a. If}\int \frac{2^x}{\sqrt{1-4^x}}dx=k{\sin }^{-1}\left(f\right(x\left)\right)+c,& \text{p.}0\\ \text{then}k\text{is greater than}& \\ \text{b. If}\int \frac{(\sqrt{x}{)}^5}{(\sqrt{x}{)}^7+x^6}dx=a\ln \frac{x^k}{x^k+1}+c,& \text{q.}1\\ \text{then}ak\text{is less than}& \\ \text{c. If}\int \frac{x^4+1}{x(x^2+1{)}^2}dx=k\ln |x|+\frac{m}{1+x^2}+n,& \text{r.}3\\ \text{where}n\text{is the constant of integration, then}& \\ mk\text{is greater than}& \\ \text{d. If}\int \frac{dx}{5+4\cos x}dx=k{\tan }^{-1}\left(m\tan \frac{x}{2}\right)+c,& \text{s.}4\\ \text{then}k/m\text{is greater than}& \end{array}$
Which of the following is the correct combination?

• A. $\left(a\right)\to \left(q\right);\left(b\right)\to \left(r\right),\left(s\right);\left(c\right)\to \left(p\right);\left(d\right)\to \left(r\right),\left(q\right)$
• B. $\left(a\right)\to \left(p\right);\left(b\right)\to \left(p\right),\left(q\right);\left(c\right)\to \left(q\right);\left(d\right)\to \left(r\right),\left(s\right)$
• C. $\left(a\right)\to \left(p\right),\left(q\right);\left(b\right)\to \left(p\right),\left(q\right);\left(c\right)\to \left(p\right);\left(d\right)\to \left(r\right),\left(s\right)$
• D. $\left(a\right)\to \left(p\right),\left(q\right);\left(b\right)\to \left(r\right),\left(s\right);\left(c\right)\to \left(p\right);\left(d\right)\to \left(p\right),\left(q\right)$

Answer 1

The correct option is D $\left(a\right)\to \left(p\right),\left(q\right);\left(b\right)\to \left(r\right),\left(s\right);\left(c\right)\to \left(p\right);\left(d\right)\to \left(p\right),\left(q\right)$

$\left(a\right).$
Let $I=\int \frac{2^x}{\sqrt{1-4^x}}dx$
Putting $2^x=t\Rightarrow 2^x\log 2dx=dt,$
$\Rightarrow I=\frac{1}{\log 2}\int \frac{1}{\sqrt{1-t^2}}dt$
$\Rightarrow I=\frac{1}{\log 2}{\sin }^{-1}\left(\frac{t}{1}\right)+c=\frac{1}{\log 2}{\sin }^{-1}\left(2^x\right)+c$
$\therefore k=\frac{1}{\log 2}$
$\left(a\right)\to \left(p\right),\left(q\right)$

$\left(b\right).\int \frac{(\sqrt{x}{)}^5}{(\sqrt{x}{)}^7+x^6}dx=\int \frac{dx}{(\sqrt{x}{)}^2+(\sqrt{x}{)}^7}=\int \frac{dx}{(\sqrt{x}{)}^7(1+\frac{1}{(\sqrt{x}{)}^5})}$
Putting $\frac{1}{(\sqrt{x}{)}^5}=y,\frac{dy}{dx}=-\frac{5}{2(\sqrt{x}{)}^7}$
$\therefore I=\int \frac{-2dy}{5(1+y)}=-\frac{2}{5}\ln |1+y|+c=\frac{2}{5}\ln \left(\frac{1}{1+\frac{1}{(\sqrt{x}{)}^5}}\right)=\frac{2}{5}\ln \left(\frac{(\sqrt{x}{)}^5}{1+(\sqrt{x}{)}^5}\right)$
$\Rightarrow a=\frac{2}{5},k=\frac{5}{2}\Rightarrow ak=\frac{2}{5}\times \frac{5}{2}=1$
$\left(b\right)\to \left(r\right),\left(s\right)$

$\left(c\right).$
$I=\int \frac{x^4+1}{x(x^2+1{)}^2}dx\Rightarrow I=\int \frac{x^4+2x+1-2x^2}{x(x^2+1{)}^2}dx\Rightarrow I=\int \frac{(x^2+1{)}^2-2x^2}{x(x^2+1{)}^2}dx\Rightarrow I=\int \frac{1}{x}-\frac{2x}{(x^2+1{)}^2}dx\Rightarrow I=\ln |x|+\frac{1}{1+x^2}+c$
$k=1,m=1\Rightarrow km=1$
$\left(c\right)\to \left(p\right)$

$\left(d\right).I=\int \frac{dx}{5+4\cos x}$
$=\int \frac{dx}{5({\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2})+4({\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2})}$
$=\int \frac{dx}{9{\cos }^2\frac{x}{2}+{\sin }^2\frac{x}{2}}=\int \frac{{\sec }^2\frac{x}{2}}{9+{\tan }^2\frac{x}{2}}dx$
Let $t=\tan \frac{x}{2}$
$\Rightarrow dt=\frac{1}{2}{\sec }^2\frac{x}{2}dx$
$\therefore I=\int \frac{2dt}{9+t^2}=\frac{2}{3}{\tan }^{-1}\left(\frac{t}{3}\right)+c$
$=\frac{2}{3}{\tan }^{-1}\left(\frac{\tan \left(\frac{x}{2}\right)}{3}\right)+c$
$\therefore k=\frac{2}{3},m=\frac{1}{3}\Rightarrow k/m=2$
$\left(d\right)\to \left(p\right),\left(q\right)$