Ncert Solutions Maths class 12th

 $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{2}=\frac{-2\left(1+4+2-16\right)}{1+2^2+2^2}$

(x, y, z) = (3, 6, 5)

now point Q and line both lies in the plane.

So, equation of plane is

$\left|\begin{array}{ccc}\hfill x\hfill & \hfill y\hfill & \hfill z+1\hfill \\ \hfill 3\hfill & \hfill 6\hfill & \hfill 6\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right|=0$

2x – z = 1

option (B) satisfies.

Let AB $\equiv x-2y+1=0$

AC $\equiv 2x-y+1=0$

So vertex A = (1, 1)

altitude from B is perpendicular to AC and passing through orthocentre.

So, BH = x + 2y – 7 = 0

CH = 2x + y – 7 = 0

now solve AB & BH to get B (3, 2) similarly CH and AC to get C (2, 3) so centroid is at (2, 2)

Line $\perp$ to the normal

⇒ 3p + 2q – 1 = 0

$\left(2,-1,-3\right)$ lies in the plane 2p + q = 8

From here p = 15, q = -22

Equation of plane 15x – 22y + z – 5 = 0

Distance from origin = $\left|\frac{5}{\sqrt[]{15^2+{\left(-22\right)}^2+1^2}}\right|$$=$√5/142

$\Rightarrow {\displaystyle \underset{}{\overset{}{\int }}\frac{dy}{1+y^2}=-2{\displaystyle \underset{}{\overset{}{\int }}\frac{e^{x}dx}{1+{\left(e^x\right)}^2}+C}}$

$\Rightarrow ta{n}^{-1}y=-2\cdot ta{n}^{-1}{e}^x+C$

x = 0, y = 0

$\Rightarrow 0=2ta{n}^{-1}+C$

$C=+\frac{\pi }{2}$

now at x = $\mathcal{l}n\sqrt[]{3}$

$ta{n}^{-1}y=-2ta{n}^{-1}\left(e^{\mathcal{l}n\sqrt[]{3}}\right)+\frac{\pi }{2}$

$6\left(y\text{'}\left(0\right)+{\left(y\left(\mathcal{l}n\sqrt[]{3}\right)\right)}^2\right)=6\left(-1+\frac{1}{3}\right)=-4$

 ${\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2xdx}}-{\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2x-x^{2}dx}}={\displaystyle \underset{0}{\overset{1}{\int }}dy}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-y^2}dy-{\displaystyle \underset{0}{\overset{2}{\int }}\frac{y^2}{2}dy}+{\displaystyle \underset{1}{\overset{2}{\int }}2dy+I}}$

$\Rightarrow \frac{8}{3}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}=1-\frac{8}{6}+2+I$

$I=1-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}$

$f\left(x\right)=x+x{\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt}-{\displaystyle \underset{0}{\overset{1}{\int }}{t}_{0}f\left(t\right)dt}$

Let $1+{\displaystyle \underset{0}{\overset{1}{\int }}f}\left(t\right)dt=\alpha$

${\displaystyle \underset{0}{\overset{1}{\int }}tf\left(t\right)dt=\beta }$

So, f (x) = αx – β

Now, $\alpha ={\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt+1}$

$\alpha ={\displaystyle \underset{0}{\overset{1}{\int }}\left(at-\beta \right)dt+1}$

$\beta ={\displaystyle \underset{0}{\overset{1}{\int }}t\cdot f\left(t\right)dt}$

$\beta =\frac{4}{13},\alpha =\frac{18}{13}$

f (x) = αx – β

$=\frac{18x-4}{13}$

option (D) satisfies

$f\text{'}\left(x\right)=\frac{n_1\cdot f\left(x\right)}{x-3}+n_2\cdot \frac{f\left(x\right)}{x-5}$

$=\frac{f\left(x\right)\cdot \left(n_1+n_2\right)}{\left(x-3\right)\left(x-5\right)}\left(x-\frac{\left(5n_1+3n_2\right)}{n_1+n_2}\right)$

$f\text{'}\left(x\right)={\left(x-3\right)}^{n_1-1}\cdot {\left(x-5\right)}^{n_2-1}\cdot \left(n_1+n_2\right)\left(x-\frac{\left(5n_1+3n_2\right)}{n_1+n_{2l}}\right)$

option (C) is incorrect, there will be minima.

Case – I $\Delta \equiv \nabla \equiv \wedge$

$\left(p\wedge q\right)\to \left(\left(p\wedge q\right)\wedge r\right)$

it can be false if r is false,

so not a tautology

Case – II If $\Delta \equiv \nabla \equiv \vee$

$\left(p\vee q\right)\text{?}\to \left(\left(p\vee q\right)\vee r\right)\equiv$ tautology

then $\left(p\vee q\right)\vee r\equiv \left(p\Delta r\right)\vee q$

Case – III If $\Delta \equiv v,\nabla \equiv \wedge$

then $\left(p\wedge q\right)\to \left\{\left(p\vee q\right)\wedge r\right\}$

Not a tautology

Case – IV If $\Delta \equiv \wedge ,\nabla \equiv \vee$

$\left(p\wedge q\right)\to \left\{\left(p\wedge q\right)\vee r\right\}$

Not a tautology

P (H) = x . P (T) = 1 – x

P (4H. 1T) = P (5H)

               

6x = 5 = 0     $x=\frac{5}{6}$        

P (atmost 2H)

$P\left(OH,5T\right)+P\left(1H,4T\right)+P\left(2H,3T\right)$

$=\frac{1}{6^5}\left(1+25+250\right)=\frac{276}{6^5}=\frac{4^6}{6^4}$                   

Consider the equation of plane,

$P:\left(2x+3y+z+20\right)+\lambda \left(x-3y+5z-8\right)=0$

$?$ Plane P is perpendicular to 2x + 3y + z + 20 = 0

So,  $4+2\lambda +9-9\lambda +1+5\lambda =0$

0 $\lambda =7$

P : 9x – 18y + 36z – 36 = 0

Or P : x – 2y + 4z = 4

If image of

$\left(2,-\frac{1}{2},2\right)$

In plane P is (a, b, c) then

$\frac{a-2}{1}=\frac{b+\frac{1}{2}}{-2}=\frac{c-2}{4}$

and $\left(\frac{a+2}{2}\right)-2\left(\frac{b-\frac{1}{2}}{2}\right)+4\left(\frac{c+2}{2}\right)=4$

clearly

$a=\frac{4}{3},b=\frac{5}{6}andc=-\frac{2}{3}$

So, a : b : c = 8 : 5 : 4