⇒ Definite Integration

Correct Option is (A)

$\begin{array}{rl}& \text{ Let }\mathrm{I}=\underset{1}{\overset{2}{\int }}\frac{\mathrm{d}x}{{(x^2-2x+4)}^{\frac{3}{2}}}\\ & =\underset{1}{\overset{2}{\int }}\frac{\mathrm{d}x}{{[(x-1{)}^2+3]}^{\frac{3}{2}}}\\ & \text{ Put }x-1=\sqrt{3}\tan \theta \\ & \mathrm{d}x=\sqrt{3}{\sec }^2\theta \mathrm{d}\theta \\ & \text{ When }x=1,\theta =0\\ & \text{ When }x=2,\theta =\frac{\pi }{6}\\ & \end{array}$

$\begin{array}{rl}\therefore \mathrm{I}& =\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{\sqrt{3}{\sec }^2\theta }{{[3{\tan }^2\theta +3]}^{\frac{3}{2}}}\mathrm{d}\theta \\ & =\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{\sqrt{3}{\sec }^2\theta }{{\left[3(1+{\tan }^2\theta )\right]}^{\frac{3}{2}}}\\ & =\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{\sqrt{3}{\sec }^2\theta }{3\cdot \sqrt{3}{({\sec }^2\theta )}^{\frac{3}{2}}}\\ & =\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{1}{3}\cdot \frac{\sec {}^2\theta }{{\sec }^3\theta }\\ & =\frac{1}{3}\underset{0}{\overset{\frac{\pi }{6}}{\int }}\cos \theta \\ & =\frac{1}{3}[\sin \theta {]}_0^{\frac{\pi }{6}}\\ \mathrm{I}& =\frac{1}{3}[\sin \frac{\pi }{6}-\sin 0]\\ \mathrm{I}& =\frac{1}{6}\\ \therefore & \frac{k}{k+5}=\frac{1}{6}\\ & 6k=k+5\\ \therefore & k=1\end{array}$

Correct Option is (C)

As ${\mathrm{f}}^{\prime }(x)=\mathrm{f}(x)$

$\frac{{\mathrm{f}}^{\prime }(x)}{\mathrm{f}(x)}=1$

Integrating on both sides, we get

$\log \mathrm{f}(x)=x+\mathrm{c}$ ..... (i)

As $\mathrm{f}(0)=1$

$\begin{array}{ll}\therefore & \text{ (i) }\Rightarrow \mathrm{c}=0\\ \therefore & \log \mathrm{f}(x)=x\\ \therefore & \mathrm{f}(x)={\mathrm{e}}^x\end{array}$

As $\mathrm{f}(x)+\mathrm{g}(x)=x^2$

$\mathrm{g}(x)=x^2-{\mathrm{e}}^x$

$\begin{array}{rl}\therefore \mathrm{f}(x)\mathrm{g}(x)& ={\mathrm{e}}^x(x^2-{\mathrm{e}}^x)\\ & =\underset{0}{\overset{1}{\int }}({\mathrm{e}}^x{x}^2-{\mathrm{e}}^{2x})\mathrm{d}x\\ & ={\left[(x^2-2x+2){\mathrm{e}}^x\right]}_0^1-\frac{1}{2}{\mathrm{e}}^2+\frac{1}{2}\\ & =\mathrm{e}-\frac{1}{2}{\mathrm{e}}^2-\frac{3}{2}\end{array}$

Correct Option is (D)

Let

$\begin{array}{rl}I& =\underset{5}{\overset{10}{\int }}\frac{dx}{(x-1)(x-2)}\\ \\ & =\underset{5}{\overset{10}{\int }}[\frac{1}{x-1}-\frac{1}{x-2}](-1)dx=-\underset{5}{\overset{10}{\int }}[\frac{1}{x-1}-\frac{1}{x-2}]dx\\ \\ & =-[\log |x-1|{]}_5^{10}+[\log |x-2|{]}_5^{10}=-[\log |9|-\log |4|]+[\log |8|-\log 31]\\ \\ & =[\log \left|\frac{8}{3}\right|]-[\log \left|\frac{9}{4}\right|]=\log |\frac{8}{3}\times \frac{4}{9}|=\log \left|\frac{32}{27}\right|\end{array}$

Correct Option is (A)

$2\mathrm{f}(\mathrm{x})-3\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)$ ..... (1) and replacing x by $\frac{1}{x}$, we get

$2\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)-3\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}}$ ..... (2)

$[2\times$ equation (1) $]+[3\times$ equation (2)] gives,

$\begin{array}{rl}& 4f(x)-9f(x)=2x+\frac{3}{x}\Rightarrow -5f(x)=2x+\frac{3}{x}\\ & \begin{array}{rl}\therefore f(x)=\frac{-2}{5}x& -\frac{3}{5x}\\ \therefore {\int }_1^{e}f(x)dx& ={\int }_1^e(\frac{-2}{5}x-\frac{3}{5x})dx=\frac{-2}{5}{\int }_1^{e}xdx-\frac{3}{5}{\int }_1^e\frac{1}{x}dx\\ & =\frac{-2}{5}{\left[\frac{x^2}{2}\right]}_1^e-\frac{3}{5}[\log x{]}_1^e=\frac{-1}{5}(e^2-1)-\frac{3}{5}(\log e-\log 1)=\frac{-1}{5}{e}^2+\frac{1}{5}-\frac{3}{5}\\ & =-\left(\frac{2+e^2}{5}\right)\end{array}\end{array}$

Correct Option is (B)

$a+\frac{b}{\log 2}=\underset{2}{\overset{e}{\int }}[\frac{1}{\log x}-\frac{1}{(\log x{)}^2}]dx$

Put $\log x=t\Rightarrow \frac{1}{x}dx=dt\Rightarrow dx=e^{t}dt$

When $x=e,t=1$ and when $x=2,t=\log 2$

$\begin{array}{rl}& \therefore a+\frac{b}{\log 2}=\underset{\log 2}{\overset{1}{\int }}(\frac{1}{t}-\frac{1}{t^2}){\mathrm{e}}^{\mathrm{t}}\cdot \mathrm{dt}={[\mathrm{e}t\cdot \frac{1}{\mathrm{t}}]}_{\log 2}^1=e-\frac{{\mathrm{e}}^{\log 2}}{\log 2}=\mathrm{e}-\frac{2}{\log 2}\\ & \therefore a=e,b=-2\end{array}$