Maths NCERT Exemplar Solutions Class 12th Chapter Seven

Three consecutive integers belong to 98 sets and four consecutive integers belongs to 97 sets.

Þ Number of permutations of b₁ b₂ b₃ b₄ = number of permutations when b₁ b₂ b₃ are consecutive + number of permutations when b₂, b₃, b₄ are consecutive – b₁ b₂ b₃ b₄ are consecutive = 98 * 97 * 98 * 97 – 97 = 18915

 $co{s}^{-1}\frac{1}{\sqrt[]{5}}=co{t}^{-1}\frac{1}{2}$

$tan\left(2\left(ta{n}^{-1}\frac{1}{2}+co{t}^{-1}\frac{1}{2}\right)\right)+ta{n}^{-1}\left(\frac{1}{2}\right)=tan\left(\pi +ta{n}^{-1}\frac{1}{2}\right)=\frac{1}{2}$

$tan2\alpha =2\sqrt[]{2}\Rightarrow \frac{2tan\alpha }{1-ta{n}^2\alpha }=2\sqrt[]{2}$

$\sqrt[]{2}ta{n}^2\alpha +tan\alpha -\sqrt[]{2}=0$

$tan\alpha =\frac{1}{\sqrt[]{2}},\left(-\sqrt[]{2}\right)\to rejected$

 $I={\displaystyle \underset{0}{\overset{5}{\int }}cos}\left(\pi x-\pi \left[\frac{x}{2}\right]\right)dx$

$I={\displaystyle \underset{0}{\overset{2}{\int }}cos\left(\pi x\right)}dx+{\displaystyle \underset{2}{\overset{4}{\int }}cos}\left(\pi x-\pi \right)dx+{\displaystyle \underset{4}{\overset{5}{\int }}cos\left(\pi x-2\pi \right)dx}$

$I={\frac{sin\pi x}{\pi }|}_0^2+{\frac{sin\left(\pi x-\pi \right)}{\pi }|}_2^4+{\frac{sin\left(\pi x-2\pi \right)}{\pi }|}_4^5=0$

Last two digit must be in form

$\begin{array}{l}23,4,516\\ 324\\ 32\\ 52\end{array}\}3*4=12$

Total number of required number = 12 + 18 = 30

  ${\displaystyle \int \frac{\left(1-\frac{1}{\sqrt[]{3}}\right)\left(cosx-sinx\right)}{\left(1+\frac{2}{\sqrt[]{3}}sin2x\right)}}dx={\displaystyle \int \frac{\left(\frac{\sqrt[]{3}-1}{\sqrt[]{3}}\right)\sqrt[]{2}sin\left(\frac{\pi }{4}-x\right)}{\left(\frac{2}{\sqrt[]{3}}\right)\left(sin\frac{\pi }{3}+sin2x\right)}dx}$

$={\displaystyle \int \frac{\left(\frac{\sqrt[]{3}-1}{\sqrt[]{2}}\right)sin\left(\frac{\pi }{4}-x\right)}{\left(sin\frac{\pi }{3}+sin2x\right)}}dx={\displaystyle \int \frac{\left(\frac{\sqrt[]{3}-1}{2\sqrt[]{2}}\right)sin\left(\frac{\pi }{4}-x\right)}{sin\left(\frac{\pi }{6}+x\right)cos\left(\frac{\pi }{6}-x\right)}dx}$

$=\frac{1}{2}\left[lo{g}_e\left|tan\left(\frac{x}{2}+\frac{\pi }{12}\right)\right|-lo{g}_e\left|tan\left(\frac{x}{2}+\frac{\pi }{6}\right)\right|\right]+C=\frac{1}{2}lo{g}_e\left|\frac{tan\left(\frac{x}{2}+\frac{\pi }{12}\right)}{tan\left(\frac{x}{2}+\frac{\pi }{6}\right)}\right|+C$

 $l={\displaystyle \underset{0}{\overset{20\pi }{\int }}{\left(\left|sinx\right|+\left|cosx\right|\right)}^{2}dx=20{\displaystyle \underset{0}{\overset{\pi }{\int }}\left(1+\left|sin2x\right|\right)dx=40{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1+\left|sin2x\right|\right)dx=40{\left(x-\frac{cos2x}{2}\right)}_0^{\frac{\pi }{2}}}}}$

$=40\left(\frac{\pi }{2}+\frac{1}{2}+\frac{1}{2}\right)$ = 20 (p + 2)

 $Letf\left(x\right)=lo{g}_{cosx}cosecx=\frac{logcosecx}{logcosx}$

$f\text{'}\left(x\right)=\frac{logcosx.sinx\left(-cosecxcotx-\left(logcosecx\right)\frac{1}{cosx}.\left(sinx\right)\right)}{{\left(logcosx\right)}^2}$

$Atx=\frac{\pi }{4}f\text{'}\left(\frac{\pi }{4}\right)=\frac{-log\left(\frac{1}{\sqrt[]{2}}\right)+log\sqrt[]{2}}{{\left(log\frac{1}{\sqrt[]{2}}\right)}^2}=\frac{2}{log\sqrt[]{2}}\therefore atx=\frac{\pi }{4},lo{g}_e\left(2f\text{'}\left(x\right)\right)=4$

 $l=60{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(\frac{sin6x-sin4x}{sinx}+\frac{sin4x-sin2x}{sinx}+\frac{sin2x}{sinx}\right)dx}$

$l=60{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(2cos5x+2cos3x+2cosx\right)dx}$

${l=60\left(\frac{2}{5}sin5x+\frac{2}{3}sin3x+2sinx\right)|}_0^{\pi /2}$ = 104

Kindly consider the following figure

 $l_n\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}\frac{dt}{{\left(t^2+s\right)}^n}}$

Applying integral by parts

$l_n\left(x\right)={\left[\frac{t}{{\left(t^2+5\right)}^n}\right]}_0^x-{\displaystyle \underset{0}{\overset{x}{\int }}n{\left(t^2+5\right)}^{-n-1}.2t^2}$

$10n{l}_{n+1}\left(x\right)+\left(1-2n\right)l_n\left(x\right)=\frac{x}{{\left(x^2+5\right)}^n}$

Put n = 5