Ncert Solutions Maths class 12th

For $x^2+\alpha x=\beta >0\forall x\in R$  to hold, we should have   ${\alpha }^2-4\beta <0$

If $\alpha =1,\beta$  can be 1, 2, 3, 4, 5, 6 i.e., 6 choices

If a  = 4, b can be 5 or 6 i.e., 2 choices

hence total favourable outcomes

= 6 + 5 + 4 + 2 + 0 + 0 = 17

Required probability =  $\frac{17}{36}$

If n is number of trials, p is probability of success and q is probability of unsuccess then,

Mean = np and variance = npq.

Here np + npq = 24         …. (i)

np. npq = 128                   …. (ii)

and q = 1 – p                    …… (iii)

From eq. (i), (ii) and (iii) :

$=\left(32+\frac{32*31}{2}\right).\frac{1}{2^{32}}$      

$=\frac{33}{2^{28}}$

  $?\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$ …….(i)

then

$\overrightarrow{a}+\overrightarrow{c}=-\overrightarrow{b}$

then

$\left(\overrightarrow{a}+\overrightarrow{c}\right)*\overrightarrow{b}=-\overrightarrow{b}*\overline{b}$

$\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}$  $\therefore \overrightarrow{a}*\overrightarrow{b}+\overrightarrow{c}*\overrightarrow{b}=\overrightarrow{0}$ …….(ii)

Now (S1) : $\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\text{exception 25:}$

$\left|\overrightarrow{a}*\overrightarrow{b}+\overrightarrow{c}*\overrightarrow{b}\right|-\left|\overrightarrow{c}\right|=6\left(2\sqrt[]{2}-1\right)$

$\therefore cos\left(\angle ACB\right)=\sqrt[]{\frac{2}{3}}$

$\therefore \angle ACB=co{s}^{-1}\sqrt[]{\frac{2}{3}}$

$$ $\therefore S\left(2\right)$  is correct

Let a, b, c be direction ratios of plane containing lines

$\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$

and

$\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$

Equation of plane P is : 1 (x – 3) 1 (y + 4) + 2 (z – 7) = 0

$\Rightarrow x-y+2z-21=0$

Distance from point (2, 5, 11) is

$d=\frac{\left|2+5+22-2\right|}{\sqrt[]{6}}$

$\therefore d^2=\frac{32}{3}$

Radius of circle S touching x-axis and centre  $\left(\alpha ,\beta \right)$ $$ is |. According to given conditions

${\alpha }^2+{\left(\beta -1\right)}^2={\left(\left|\beta \right|+1\right)}^2$

${\alpha }^2=4\beta as\beta >0$

$$ $\therefore$  Required locus is L : x2 = 4y

The area of shaded region = $2{\displaystyle {\int }_0^{4}2\sqrt[]{y}dy}$ ${\displaystyle {}_{}^{}\text{exception 25:}}$

= $4.{\left[\frac{\frac{y^{\frac{3}{2}}}{3}}{2}\right]}_0^4$ ${\frac{\frac{{}^{\frac{}{}}}{}}{}}_{}^{}$

= $\frac{64}{3}$ $\frac{}{}$ square units.

$f\left(x\right)=\{\begin{array}{l}x-\left[x\right]if\left[x\right]isodd\\ 1+\left[x\right]-x,if\left[x\right]iseven\end{array}$

Graph of f (x)

So

$=2{\pi }^2{\displaystyle \underset{0}{\overset{1}{\int }}\left(1-x\right)cos\pi xdx}$

=4

$A=\left\{\left(x,y\right):x^2\le y\le min\{x+2,4-3x\right\}$

So, area of the required region

$A={\displaystyle \underset{-1}{\overset{\frac{1}{2}1}{\int }}\left(x+2-x^2\right)dx+{\displaystyle \underset{\frac{1}{2}}{\overset{1}{\int }}\left(4-3x-x^2\right)}dx}$

$={\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]}^{\frac{1}{2}}+{\left[4x-\frac{3x^2}{2}-\frac{x^3}{3}\right]}^1$

= $\frac{17}{6}$

$f\left(x\right)=y=a{x}^3+b{x}^2+cx+5$ ……. (i)

$\frac{dy}{dx}=3a{x}^2+2bx+c$ ……. (ii)

Touches x-axis at P (-2, 0)

$\Rightarrow y{|}_{x=-2}=0\Rightarrow -8a+4b-2c+5=0$ ……… (iii)

Touches x –axis at P (-2, 0) also implies

$\frac{dy}{dx}{|}_{x=-2}=0\Rightarrow 12a-4b+c=0$ ……… (iv)

y = f (x) cuts y-axis at (0, 5)

Given,

$\frac{dy}{dx}{|}_{x=0}=c=3$ ……. (v)

From (iii), (iv) and (v)

f (x) = 0 at x = -2 and x = 1

Local maximum value of f (x) is at x = 1

i.e., $\frac{27}{4}$

Given,

$f\left(x\right)=\underset{f_1\left(x\right)}{\underset{?}{\left(x^2-2x+7\right)}}\underset{f_2\left(x\right)}{\underset{?}{\left(e^{\left(4x^3-12x^2-180x+31\right)}\right)}}$        

f₁ (x) = x² – 2x + 7

So f (x) is decreasing in [-3, 0]

and positive also

$\therefore$ absolute maximum value of f (x) occurs at x = -3

$\therefore \alpha =-3$      

$S_n=\left\{z\in C:\left|z-3+2i\right|=\frac{n}{4}\right\}$

represents a circle with centre C1 (3, 2) and radius

$r_1=\frac{n}{4}$

Similarly Tn represents circle with centre C2 (2, 3) and radius

$r_1=\frac{1}{n}$

$\sqrt[]{2}<\left|\frac{n}{4}-\frac{1}{n}\right|$

n take infinite values.