Ncert Solutions Maths class 11th

After giving 2 apples to each child 15 apples left now 15 apples can be distributed in
^15+3–1C₂ = ¹⁷C₂ ways

  $=\frac{17*16}{2}=136$          

${\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{x^{2}sinx\cdot cosx}{si{n}^{4}x+co{s}^{4}x}dx}$

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

$={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinx\cdot cosx\left(2\pi x-{\pi }^2\right)}{si{n}^{4}x+co{s}^{4}x}x}$

$=2\pi \cdot \frac{\pi }{4}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinxcosx}{si{n}^{4}x+co{s}^{4}x}dx}-{\pi }^2{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinxcosx}{si{n}^{4}x+co{s}^{4}x}dx}$

$=-\frac{{\pi }^2}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinxcosx}{si{n}^{4}x+co{s}^{4}x}dx}$

$=-\frac{{\pi }^2}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\frac{1}{2}sin2x}{1-\frac{1}{2}si{n}^{2}2x}dx}$

$=-\frac{{\pi }^2}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sin2x}{2-si{n}^{2}2x}dx}$

$=-\frac{{\pi }^2}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sin2x}{1+co{s}^{2}2x}dx}$

Let cot2x = t

$=-\frac{{\pi }^2}{2}{\displaystyle \underset{1}{\overset{-1}{\int }}\frac{-\frac{1}{2}dt}{1+t^2}}$

$=-\frac{{\pi }^2}{4}{\displaystyle \underset{-1}{\overset{1}{\int }}\frac{dt}{1+t^2}}$

$=-\frac{{\pi }^2}{4}\cdot \frac{\pi }{2}=-\frac{{\pi }^2}{8}$

$\frac{120}{{\pi }^3}\left|-\frac{{\pi }^3}{8}\right|=15$

$=8*7*6*5*4+\frac{9*8}{2}$

= 6756

(y – 2) = m (x – 8)

⇒   x-intercept

⇒     $\left(\frac{-2}{m}+8\right)$

⇒   y-intercept

⇒   (–8m + 2)

⇒   OA + OB = $\frac{-2}{m^2}$  + 8 – 8m + 2

$f\text{'}\left(m\right)=\frac{2}{m^2}-8=0$  

-> $m^2=\frac{1}{4}$

-> $m=\frac{-1}{2}$

-> $f\left(\frac{-1}{2}\right)=18$

->Minimum = 18

z₁ + z₂ = 5

$z_1^3+z_2^3=20+15i$            

  $z_1^3+z_2^3={\left(z_1+z_2\right)}^3-3z_1{z}_2\left(z_1+z_2\right)$           

$z_1^3+z_2^3=125-3z_1\cdot z_2\left(5\right)$            

 ⇒ 20 + 15i = 125 – 15z₁z₂

⇒ 3z₁z₂ = 25 – 4 – 3i

3z₁z₂ = 21– 3i

z₁⋅z₂ = 7 – i

(z₁ + z₂)² = 25

$z_1^2+z_2^2=25-27\left(7-i\right)$     

= 11 + 2i

  ${\left(z_{\text{?}1}^2+z_2^2\right)}^2$         = 121 − 4 + 44i

⇒   $z_1^4+z_2^4+2{\left(7-i\right)}^2=117+44i$

⇒   $z_1^4+z_2^4$ = 117 + 44i − 2(49 −1−14i )

= 21 + 72i

⇒   $\left|Z_1^4+Z_2^4\right|=75$

a + d, a + 7d and a + 43d are 1^st, 2^nd, 3^rd term of G.P.

$\frac{a+7d}{a+d}=\frac{a+43d}{a+7d}$              

⇒ (a + 7d)² = (a + d) (a + 43d)

⇒ a² + 49d² + 14d = a² + 44ad + 43d³

⇒ 6d² = 30ad

⇒ d² = 5d

⇒ d = 0, 5

a = 1, d = 5

  $S_{20}=\frac{20}{2}\left[2+\left(19\right)5\right]$          

= 10 [95 + 2]

= 970

  $\frac{a+b+68+44+40+60}{6}=55$

212 + a + b = 330

⇒ a + b = 118

$\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{n}-{\left(\overline{x}\right)}^2=194$          

$\frac{a^2+b^2+{\left(68\right)}^2+{\left(44\right)}^2+{\left(40\right)}^2+{\left(60\right)}^2}{6}={\left(55\right)}^2=194$

= 3219

11760 + a² + b² = 19314

⇒ a² + b² = 19314 – 11760

= 7554

(a + b)² –2ab = 7554

From here b = 41.795

a + b = 118

⇒ a + b + 2b = 118 + 83.59

= 201.59

a = sin⁻¹ (sin5) = 5 − 2π

and b = cos⁻¹ (cos5) = 2π − 5

∴    a² + b² = (5 − 2π)² + (2π − 5)²

= 8π² − 40π + 50