Ncert Solutions Maths class 12th

S = $\underset{n\to \infty }{Lt}{\displaystyle \sum _{r=1}^n\frac{n^2}{\left(n^2+r^2\right)\left(n+r\right)}}$

$\frac{\pi }{8}+\frac{1}{4}\mathcal{l}n2$

$l={\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{dx}{\left(1+e^x\right)\left(si{n}^{6}x+co{s}^{6}x\right)}}$ ……. (i)

$=2{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{dt}{4+t^2}=2{\left(ta{n}^{-1}\left(\frac{t}{2}\right)\right)}_0^{\infty }}=\pi$

$f\left(x\right)=\{\begin{array}{l}\frac{sin\left(x+2\right)}{x+2},x\in \left(-2,-1\right)\\ 0,x\in \left(-1,0\right]\\ 2x,x\in \left(0,1\right)\\ 1,otherwise\end{array}$

$LHD=\underset{h\to 0}{Lt}\frac{f\left(0-h\right)-f\left(0\right)}{-h}=0$

$RHD=\underset{h\to 0}{Lt}\frac{f\left(0+h\right)-f\left(0\right)}{h}=2$

Hence f (x) is not differentiable at x = 1, 0, 1

$\therefore m=2,n=3$

$e^{2x}-4=0or6e^{2x}-5e^x+1=0$

$\Rightarrow x=\mathcal{l}n2x=\mathcal{l}n\left(\frac{1}{3}\right),\mathcal{l}n\left(\frac{1}{2}\right)$

= $-\mathcal{l}n3,-\mathcal{l}n2$

  $\left(1,1\right)\left(1,4\right)\left(4,1\right)\left(2,4\right)\left(4,2\right)\left(3,4\right)\left(4,3\right)\left(4,4\right)$  all have only one image.

(2, 1) (1, 2), (2, 2) each element has 3 choice.

(3, 2) (2, 3) (3, 1) (1, 3) (3, 3) each element has two choices.

$\therefore$ total function = 3 * 3 * 2 * 2 * 2 = 72

Case I

None of the pre image have 3 as image, total functions = 2 * 2 * 1 * 1 * 1 = 4

Case II

None of the pre images have 2 as image then number of function = 2⁵ = 32

Case III

None of the pre image have either 3 or 2 as image

Total function = 1⁵ = 1

$\therefore$ Total number of onto function

= 72 – 4 – 32 + 1 = 37

cota = 1        & secβ =   $\frac{-5}{3}$

< < $\frac{}{}$       $\frac{3\pi }{2}$  secβ =   $\frac{-5}{3}$

$\alpha =\left(\pi +\frac{\pi }{4}\right)$  cosβ = $\frac{-3}{5}=cos\left(180-53\right)$  

tanb =  $\frac{-4}{3}$  

tan(α + β) =   $\frac{tan\alpha +tan\beta }{1-tan\alpha tan\beta }$

$\therefore A-\frac{1}{4}\&4thquadrant.$

$=\frac{1-\frac{4}{3}}{1+\frac{4}{3}}=\frac{-1}{7}$

$\therefore -\frac{1}{4}\&4thquadrant.$

ax + by + cz = d

2a + 3b – 5c = d               2a + 3b – 5c = d …………….(v)

Putting (i) & (iv) in (v) we get a = -9d.

In conditions

$\left(2\widehat{i}+\widehat{j}-5\widehat{k}\right),\left(a\widehat{i}+b\widehat{j}+c\widehat{k}\right)=0$

2b = d ………….(i) d > 0

2a + b – 5c = 0                  2a + 3b – 5c = d   $\left|a\right|,\left|b\right|,\left|c\right|,d\to g.c.d$           

=   $\frac{\alpha }{2}$

$\therefore \frac{\alpha }{2}=1\Rightarrow \alpha =2$

$\left(3\widehat{i}+5\widehat{j}-7\widehat{k}\right).\left(a\widehat{i}+b\widehat{j}+c\widehat{k}\right)=0$

3a + 5b – 7c = 0) *   $\frac{4}{7}...................\left(iii\right)$

$\therefore a=-18,b=1$

c = -7, d = 2

(iii)…(ii) ® 2a + 3b   $\frac{-33c}{7}=0$

2a + 3b =  $\frac{33}{7}c$ $c=\frac{-7}{2}d.................\left(iv\right)$

Putting values

a + 7b + c + 20d = 22