Can I prepare for JEE using Shiksha's NCERT Applications of Derivatives Solutions?

${\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left(\frac{x^{2}cosx}{1+{\pi }^2}+\frac{1+si{n}^{2}x}{1+e^{{\left(sinx\right)}^{2023}}}\right)dx}=\frac{\pi }{4}\left(\pi +\alpha \right)-2$

${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left\{\left(\frac{x^{2}cosx}{1+{\pi }^x}+\frac{1+si{n}^{2}x}{1+e^{{\left(sinx\right)}^{2023}}}\right)+\left(\frac{x^{2}cosx}{1+{\pi }^{-x}}+\frac{1+si{n}^{2}x}{1+e^{-{\left(sinx\right)}^{2023}}}\right)\right\}dx}$

$=\frac{\pi }{4}\left(\pi +\alpha \right)-2$

${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(x^{2}cosx+1+si{n}^{2}x\right)dx}=\frac{\pi }{4}\left(\pi +\alpha \right)-2$

${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}{x}^{2}cosxdx}+{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1+si{n}^{2}x\right)dx}=\frac{\pi }{4}\left(\pi +\alpha \right)-2$ ....(1)

Let $I_1={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1+si{n}^{2}x\right)dx}$

$I_1={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}1\cdot dx}+{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(\frac{1-cos2x}{2}\right)dx}$

$I_1=\frac{\pi }{2}+\frac{1}{2}\left[\frac{\pi }{2}+0\right]$

$I_1=\frac{3\pi }{4}$

Let $I_2={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}{x}^{2}cosxdx}$

$I_2={\left[x^2\left(sinx\right)-{\displaystyle \int 2x}{\displaystyle \int cosxdx}\right]}_0^{\frac{\pi }{2}}$

$I_2={\left[x^2\left(sinx\right)-2{\displaystyle \int xsinx}\right]}_0^{\frac{\pi }{2}}$

$I_2={\left[x^{2}sinx-2\left(x\left(-cosx\right)+{\displaystyle \int cosx}\right)\right]}_0^{\frac{\pi }{2}}$

$I_2={\left[x^{2}sinx-2\left(-xcosx+sinx\right)\right]}_0^{\frac{\pi }{2}}$

$I_2=\left(\frac{{\pi }^2}{4}-2\right)$

Put l₁ and l₂ in (1)

$\frac{{\pi }^2}{4}-2+\frac{3\pi }{4}$

$\frac{{\pi }^2}{4}+\frac{3\pi }{4}-2$

$\frac{\pi }{4}\left(\pi +3\right)-2$

α = 3

P (2 obtained on even numbered toss) = k (let)

P (2) = $\frac{1}{6}$  

P (  $\overline{2}$ )= $\frac{5}{6}$  

$k=\frac{5}{6}\times \frac{1}{6}+{\left(\frac{5}{6}\right)}^3\times \frac{1}{6}+{\left(\frac{5}{6}\right)}^5\times \frac{1}{6}+...$

$=\frac{\frac{5}{6}\times \frac{1}{6}}{1-{\left(\frac{5}{6}\right)}^2}$

$=\frac{5}{11}$

$I=9{\displaystyle \underset{0}{\overset{9}{\int }}\left[\sqrt[]{\frac{10x}{x+1}}\right]dx}$

$=9\left[{\displaystyle \underset{0}{\overset{1/9}{\int }}0dx}+{\displaystyle \underset{1/9}{\overset{2/3}{\int }}dx}+{\displaystyle \underset{2/3}{\overset{9}{\int }}2dx}\right]$

$=9\left[\frac{2}{3}-\frac{1}{9}+2\left[9-\frac{2}{3}\right]\right]$

$=9\left[\frac{5}{9}+2\times \frac{25}{3}\right]$

= 5 + 6 × 25

= 5 + 150

= 155

If x = 0, y = 6, 7, 8, 9, 10

If x = 1, y = 7, 8, 9, 10

If x = 2, y = 8, 9, 10

If x = 3, y = 9, 10

If x = 4, y = 10

If x = 5, y = no possible value

Total possible ways = (5 + 4 + 3 + 2 + 1) * 2

= 30

Required probability  $=\frac{30}{11*11}=\frac{30}{121}$

Given $\left|\overrightarrow{a}\right|=1$ , $\left|\overrightarrow{b}\right|=4$ , $\overrightarrow{a}\cdot \overrightarrow{b}=2$

$\overrightarrow{c}=2\left(\overrightarrow{a}\times \overrightarrow{b}\right)-3\overrightarrow{b}$  

Dot product with  $\overrightarrow{a}$ on both sides

$\overrightarrow{c}\cdot \overrightarrow{a}=-6$ ... (1)

Dot product with  $\overrightarrow{b}$  on both sides

$\overrightarrow{b}\cdot \overrightarrow{c}=-48$ ... (2)

$\overrightarrow{c}\cdot \overrightarrow{c}=4{\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2+9{\left|\overrightarrow{b}\right|}^2$

${\left|\overrightarrow{c}\right|}^2=4\left[{\left|\overrightarrow{a}\right|}^2\cdot {\left|\overrightarrow{b}\right|}^2-{\left(\overrightarrow{a}\cdot \overrightarrow{b}\right)}^2\right]+9{\left|\overrightarrow{b}\right|}^2$

${\left|\overrightarrow{c}\right|}^2=4\left[\left(1\right){\left(4\right)}^2-\left(4\right)\right]+9\left(16\right)$

${\left|\overrightarrow{c}\right|}^2=4\left[12\right]+144$

${\left|\overrightarrow{c}\right|}^2=48+144$

${\left|\overrightarrow{c}\right|}^2=192$

$cos\theta =\frac{\overrightarrow{b}\cdot \overrightarrow{c}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|}$

$cos\theta =\frac{-48}{\sqrt[]{192}\cdot 4}$

$cos\theta =\frac{-48}{8\sqrt[]{3}\cdot 4}$

$cos\theta =\frac{-3}{2\sqrt[]{3}}$

$cos\theta =\frac{-\sqrt[]{3}}{2}$ $\theta =co{s}^{-1}\left(\frac{-\sqrt[]{3}}{2}\right)$