MATHS M.C.Q (1 Marks)

3. $\frac{\pi}{4}$

2. $2$

1. $\frac{\pi^2}{4}$

2. 0

2. $\frac{1}{2}$

3. $0$

2. $\frac{\pi}{4}$

3. 0

4. $\frac{\text{a}+\text{b}}{2}\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}$

2. $\text{e}-1$

2. $\frac{\pi}{60}$

3. $\pi$

1. 4

We have,
$\text{I}_{10}=\int\limits^\frac{\pi}{2}_0\text{x}^{10}\sin\text{x}\text{ dx}$
$=\big[\text{x}^{10}(-\cos\text{x})\big]^\frac{\pi}{2}_0-\int\limits^\frac{\pi}{2}_0\big[10\text{x}^9\int\sin\text{x}\text{ dx}\big]\text{dx}$
$=\big[-\text{x}^{10}\cos\text{x}\big]^\frac{\pi}{2}_0-10\int\limits^\frac{\pi}{2}_0\text{x}^9(-\cos\text{x})\text{dx}$
$=-\big[\text{x}^{10}\cos\text{x}\big]^\frac{\pi}{2}_0+10\int\limits^\frac{\pi}{2}_0\text{x}^9\cos\text{x}\text{ dx}$
$=-\big[\text{x}^{10}\cos\text{x}\big]^\frac{\pi}{2}_0+10\big[\text{x}^9\sin\text{x}\big]^\frac{\pi}{2}_0-10\int\limits^\frac{\pi}{2}_09\text{x}^8\sin\text{x}\text{ dx}$
$=-\Big[\Big(\frac{\pi}{2}\Big)^{10}\times0-0^{10}\cos0\Big]+10\Big[\Big(\frac{\pi}{2}\Big)^9\times1-0^9\times0\Big]\\-90\int\limits^\frac{\pi}{2}_0\text{x}^8\sin\text{x dx}$
$=10\Big[\Big(\frac{\pi}{2}\Big)^9\times1\Big]-90\text{I}_8$
$=10\Big(\frac{\pi}{2}\Big)^9-90\text{I}_8$
$\therefore\ \text{I}_{10}+90\text{I}_8$
$=10\Big(\frac{\pi}{2}\Big)^9$

3. $8$

1. $\frac{\pi}{4}$

3. $\big(\ln\text{x}\big)^{-1}\text{x}(\text{x}-1)$

3. $\frac{\pi}{2}$

1. $\frac{\pi}{12}$

4. $\frac{\pi}{8}$

1.  $1$

We have, 
$\text{I}=\int\limits^3_0\frac{3\text{x}+1}{\text{x}^2+9}\text{ dx}$
$\text{I}=\int\limits^3_0\frac{3\text{x}}{\text{x}^2+9}\text{ dx}+\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$
$\text{I}_1=\int\limits^3_0\frac{3\text{x}}{\text{x}^2+9}\text{ dx}$ and $\text{I}_2=\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$
Putting $\text{x}^2+9=\text{t}$ in $I_1$
$\Rightarrow 2\text{x}\text{ dx}=\text{dt}$
$\Rightarrow \text{x}\text{ dx}=\frac{\text{dt}}{2}$
when $\text{x}\rightarrow0;\text{t}\rightarrow9$
and $\text{x}\rightarrow3;\text{t}\rightarrow18$
$\therefore\text{I}=\int\limits^{18}_9\frac{3\text{ dt}}{2\text{ t}}+\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$
$=\frac{3}{2}\int\limits^{18}_9\frac{\text{dt}}{\text{t}}+\int\limits^3_0\frac{1}{\text{x}^2+3^2}\text{ dx}$
$=\frac{3}{2}\big[\log(\text{t})\big]^{18}_9+\frac{1}{3}\Big[\tan^{-1}\Big(\frac{\pi}{3}\Big)\Big]^3_0$
$=\frac{3}{2}\big[\log18-\log9\big]+\frac{1}{3}\Big(\frac{\pi}{4}-0\Big)$
$=\frac{3}{2}\Big[\log\frac{18}{9}\Big]+\frac{\pi}{12}$
$=\frac{3}{2}\big[\log2\big]+\frac{\pi}{12}$
$=\log(\sqrt{8})+\frac{\pi}{12}$
$=\log(2\sqrt{2})+\frac{\pi}{12}$
$=\frac{\pi}{12}+\log(2\sqrt{2})$

2.  $\log_\text{e}\sqrt{3}$

2. $\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$

4. 1

1. $\frac{\pi}{4}$

1.  $\frac{\pi}{\sqrt{\text{a}^2-\text{b}^2}}$

4. $\frac{\pi}{4}$

4. $\log\Big(\frac{4}{3}\Big)$

3. $\int\limits^{\text{a}}_0\text{f}(\text{x})\text{dx}+\int\limits^{\text{a}}_0\text{f}(2\text{a}-\text{x})\text{dx}$

4. $\frac{\pi}{4}$

3. $\frac{\pi}{12}$

3.  $\ln\Big(\frac{3}{2}\Big)$

3. 0

2. $2$

4. $-\frac{16}{3}$

$\int\limits^{\infty}_0\log\Big(\text{x}+\frac{1}{\text{x}}\Big)\frac{1}{1+\text{x}^2}\text{ dx}$
Substitute $\text{x}=\tan\theta$
$\text{dx}=\sec^2\theta\text{d}\theta$
When,
$\text{x}=0\Rightarrow\theta=0$
$\text{x}=\infty\Rightarrow\theta=\frac{\pi}{2}$
$\int\limits^{\frac{\pi}{2}}_0\Big(\tan\theta+\frac{1}{\tan\theta}\Big)\frac{1}{1+\tan^{2}}\times\sec^2\theta\text{d}\theta$
$\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{\tan^2\theta+1}{\tan\theta}\Big)\frac{1}{1+\tan^{2}\theta}\times\sec^2\theta\text{d}\theta$
$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{\sec^2\theta}{\tan\theta}\Big)\frac{1}{\sec^2\theta}\times\sec^2\theta\text{d}\theta$ $\big[\because1+\tan^2\theta=\sec^2\theta\big]$
$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{\sec^2\theta}{\tan\theta}\Big)\text{d}\theta$
$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{1}{\sin\theta\cdot\cos\theta}\Big)\text{d}\theta$
$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\big(\sin\theta\cos\theta\big)\text{d}\theta$
$\Rightarrow-\int\limits^{\frac{\pi}{2}}_0\big[\log\sin\theta+\log\cos\theta\big]\text{d}\theta$
$\Rightarrow-\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{d}\theta-\int\limits^{\frac{\pi}{2}}_0\log\cos\theta\text{d}\theta$
Let us conside,
$\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{d}\theta=\text{I}\ ....(\text{i})$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\Big(\sin\Big(\frac{\pi}{2}-\theta\Big)\Big)\text{d}\theta$
$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\cos\theta\text{d}\theta\ ...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{ d}\theta+\int\limits^{\frac{\pi}{2}}_0\log\cos\theta\text{ d}\theta$
$=\int\limits^{\frac{\pi}{2}}_0\log(\sin\theta\cdot\cos\theta)\text{d}\theta$
$=\int\limits^{\frac{\pi}{2}}_0\log\big(\sin2\theta\big)\text{d}\theta-\int\limits^{\frac{\pi}{2}}_0\log2\text{ d}\theta$
Let us consider $2\theta=\text{t}$
$2\text{d}\theta=\text{dt}$
$2\text{I}=\frac{1}{2}\int\limits^{\pi}_{0}\log(\sin\text{t})\text{dt}-\frac{\pi}{2}\log2$
$2\text{I}=\frac{2}{2}\int\limits^{\pi}_{0}\log(\sin\text{t})\text{dt}-\frac{\pi}{2}\log2$ $\big[\because\sin\theta$ is positive in both $1^{st}$ and $2^{nd}$ quadrants$\big]$
$2\text{I}=\text{I}-\frac{\pi}{2}\log2$
$2\text{I}-\text{I}=-\frac{\pi}{2}\log2$
$\text{I}=-\frac{\pi}{2}\log2,$ where $\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{ d}\theta$
Now,
$-\int\limits^{\frac{\pi}{2}}_0\log(\sin\theta)\text{ d}\theta-\int\limits^{\frac{\pi}{2}}_0\log\cos\theta\text{ d}\theta$
$-2\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{ d}\theta=-2\times\text{I}$
$=-2\times-\frac{\pi}{2}\log2$ $\Big[\text{Where I}=-\frac{\pi}{2}\log2\Big]$
$=\pi\log2$

1. $\sqrt{1-\pi^2}-1$

3. $2$

2. $\log2$

1. $2$