Ncert Solutions Maths class 11th

Total 4 digit number

4 digit divisible by 7

1001, 1008, -9996

9996 = 1001 + (n₁ – 1) 7

n₁ = 1286

4 digit no divisible by 3

1002, 1005, -9999

9999 = 1002 + (n₂ – 1)3

n₂ = 3000

4 digit number visible by 21

1008, 1031, -9996

n₃ = 429

4 digit number divisible by 7 or 3

= 9000 – 1286 – 3000 + 429

= 5143

$AP=\sqrt[]{9+1-6+4+1}$

AP = 3 = AQ

$r=\sqrt[]{1+4-1}=2$        

$tan\theta =\frac{3}{2}$  

$\frac{Areaof\Delta APQ}{Areaof\Delta BPQ}=\frac{AR}{RB}=\frac{3sin\theta }{2cos\theta }=\frac{9}{4}$

$f\left(2\right)=f\left(4\right)=0andf\left(x\right)=x^3-6x^2+ax+b$  

$\Rightarrow f\left(x\right)=\left(x-2\right)\left(x-4\right)x=x^3-6x^2+8x$               

->a = 8, b = 0

$f\text{'}\left(x\right)=0\Rightarrow x=2\pm \frac{2}{\sqrt[]{3}},x_4=2+\frac{2}{\sqrt[]{3}},f\left(x_4\right)=\frac{-16}{3\sqrt[]{3}}$           

$f\text{'}\left(x_3\right)=\frac{\sqrt[]{3}}{2}f\left(x_4\right)=\frac{-8}{3}$               

The tangent to the parabola

$y^2=8xat\left(2,-4\right)is-4y=4\left(x+2\right)$  

x + y + 2 = 0

$OA=\sqrt[]{a}$         

$\Rightarrow \left|\frac{0+0+2}{\sqrt[]{2}}\right|=\sqrt[]{a}$  

$\sqrt[]{2}=\sqrt[]{a}$      

a = 2

Required number =   $\frac{4*2+24*3+36*4}{2}=112$

Let point on ellipse (2sinθ, 3cosθ) and the mid point of line segment joining (-3, -5) and

(2 sin θ, 3cosθ) will be (h, k)

$\frac{2sin\theta -3}{2}=h\frac{3cos\theta -5}{2}=k$

sin² θ + cos² θ = 1

${\left(\frac{2h+3}{2}\right)}^2+{\left(\frac{2k+5}{3}\right)}^2$ = 1

$36x^2+16y^2+108x+80y+145=0$

$\frac{dy}{dx}=\frac{2^{x}y+2^y.2^x}{2^x+2^{x+y}lo{g}_{e}2}$

${\displaystyle \int \frac{1+2^{y}lo{g}_{e}2}{y+2^y}dy={\displaystyle \int dx}}$        

=> ln|y + 2^y| = x + c

y (0) = 0

ln |y + 2^y| = x

y = 1

x = ln 3

$x\in \left(1,2\right)$

Angle bisectors are

$\frac{x-2y-2z+1}{3}=\pm \frac{2x-3y-6z+1}{7}$              

->x – 5y + 4z + 4 = 0, (i)

3x – 23y – 32z + 10 = 0 . (ii)

As distance of a point (-1, 0,0) on x – 2y – 2z + 1 = 0

from (i) is greater than that form (ii)

(ii) is the acute angle bisector.

$l={\pi }^2{\displaystyle \underset{0}{\overset{2}{\int }}\left(sin\left(\frac{\pi x}{2}\right){\left(x-\left[x\right]\right)}^{\left[x\right]}\right)dx}$

$l={\pi }^2{\displaystyle \underset{0}{\overset{1}{\int }}sin\left(\frac{\pi x}{2}\right).x^{0}dx+{\pi }^2}{\displaystyle \underset{1}{\overset{2}{\int }}sin\left(\frac{\pi x}{2}\right){\left(x-1\right)}^{1}dx}$

$l={\pi }^2\left(\frac{2}{\pi }\right)+\frac{2{\pi }^2}{\pi }\left[1-0\right]+2\pi *\frac{2}{\pi }{\left[sin\frac{\pi x}{2}\right]}_1^2$

$\begin{array}{l}l=2\pi +2\pi +4\left(0-1\right)\\ l=4\pi -4\end{array}$

Required number =  - number of solution of x₁ + x₂ + x₃ = 15, where x₁ < x₂ < x₃

For x₂ = x₁ + a, a   $\ge 1$

->3x₁ + 2a + b = 15

Coefficient of x¹⁵ in

$\left(x^3+x^6+x^9+x^{12}+x^{15}\right)$

$\left(x^1+x^2+x^3+.....+x^{10}\right)=12$

Required number = 455 – 12 = 443