Ncert Solutions Maths class 12th

Let equation of normal PQ is

$y=-tx+3t+\frac{3}{2}{t}^3$ and it is passing through

$P\left(3,\frac{3}{2}\right)\therefore t=1$

$\therefore Q\left(\frac{3}{2},3\right)\Rightarrow a=\frac{3}{2}andb=3$     

2 (a + b) = 9

$\underset{x\to 0}{lim}\frac{\alpha x\left(1+x+\frac{x^2}{2}+....\right)-\beta \left(x-\frac{x^2}{2}+\frac{x^3}{3}+....\right)+\gamma x^2\left(1-x\right)}{x^3}=10$

For limit to exist a - b = 0 . (i), $\alpha +\frac{\beta }{2}+\gamma =0.........(ii)$

and $\frac{\alpha }{2}-\frac{\beta }{2}-\gamma =10$ . (iii)

Solving (i), (ii) and (iii) we get α = 6, β = 6 and $\lambda =-9\Rightarrow \alpha +\beta +\lambda =3$

  $P\left(\overline{A}\cap B\right)+P\left(A\cap \overline{B}\right)=1-k,$

$P\left(A\cap B\cap C\right)=k^2$             

$P\left(A\right)+P\left(B\right)-2P\left(A\cap B\right)=1-k$ .(i)

$P\left(B\right)+P\left(C\right)-2P\left(B\cap C\right)=1-k$ .(ii)

$P\left(A\right)+P\left(C\right)-2P\left(A\cap C\right)=1-2k$ .(iii)

Adding (i), (ii) and (iii) we get $P\left(A\cup B\cup C\right)=\frac{-4k+3}{2}+k^2$

$\Rightarrow P\left(A\cup B\cup C\right)=\frac{2k^2-4k+2+1}{2}=\frac{2{\left(k-1\right)}^2+1}{2}\Rightarrow P\left(A\cup B\cup C\right)>\frac{1}{2}$

 Let the smallest angle is C =?

Therefore angle A = 90° -?

And angle B = 90°

i.e. b > a > c

Here, a = 2 R cos? , b = 2R, c = 2R sin?

b² = a² + c²

according to question,

$\begin{array}{l}\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}\\ ?\frac{a^2{b}^2}{c^2}=a^2+b^2\end{array}$     

=> $sin?=?\frac{\sqrt[]{5}?1}{2}$

  $l={\displaystyle {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left(\left[x\right]+\left[-sinx\right]\right)dx}$ . (i)

$I={\displaystyle {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left(\left[-x\right]+\left[sinx\right]\right)dx}$ . (ii)

by using property $\left({\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx=}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx}\right)$

Adding (i) and (ii) we get 2l = ${\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left(-2\right)dx=-2\pi \Rightarrow l=-\pi }$

The projection of $\overrightarrow{BA}on\overrightarrow{BC}=cos\angle ABC=7\left|\frac{7^2+3^2-5^2}{2*7*3}\right|=\frac{11}{2}$

A = $-\frac{y}{x}+2sinx+2$  

$\therefore \frac{dy}{dx}+\frac{y}{x}=2sinx+2$ . (i)

$\Rightarrow l.F=e^{{\displaystyle \int !lnxdx}}$ = x from (i) d (xy) = 2 ${\displaystyle \int xsinxdx+2{\displaystyle \int xdx}}$

xy = 2 [-xcos x + sin x] + x² + c       . (ii)

according to question, we get c = 0 $\therefore y\left(\frac{\pi }{2}\right)=\frac{4}{\pi }+\frac{\pi }{2}$

$f\left(x\right)=x^3-3x^2-\frac{3f\text{'}\text{'}\left(2\right)}{2}x+f\text{'}\text{'}\left(1\right)$ . (i)

f' (x) = 3x² – 6x -   $\frac{3f\text{'}\text{'}\left(2\right)}{2}$ . (ii)

$f\text{'}\text{'}\left(x\right)=6x-6$ . (iii)

  $\therefore f\text{'}\text{'}\left(2\right)=12-6=6$           

 Hence f (x) is local maxima at x = -1

and f (x) is local minima at x = 3

$\therefore$ from (i) local minimum value at x = 3 is f (3) = -27

$\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill 4\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -3\hfill \\ \hfill 6\hfill & \hfill 5\hfill & \hfill k\hfill \end{array}\right|=0\Rightarrow k=-5$

Kindly consider the following figure