Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

A  (A -> B) -> B is a tautology

  ${\displaystyle \int \frac{\left(cosx-sinx\right)dx}{\sqrt[]{8-sin2x}}=asi{n}^{-1}\left(\frac{sinx+cosx}{b}\right)+c}$

${\displaystyle \int \frac{\left(cosx-sinx\right)dx}{\sqrt[]{9-{\left(sinx+cosx\right)}^2}}=}$           

Put (sin x + cos x) = t Þ (cos x – sin x) dx = dt

$={\displaystyle \int \frac{dt}{\sqrt[]{3^2-t^2}}=si{n}^{-1}\frac{t}{3}+c=si{n}^{-1}\left(\frac{sinx+cosx}{3}\right)+c}$        

= a = 1, b = 3

$\therefore \left(a,b\right)\equiv \left(1,3\right)$        

According to question,

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

$\frac{1}{a}+\frac{1}{b}=\frac{1}{2}........(i)$           

Equation of required line is $\frac{x}{a}+\frac{y}{b}=1$  

Obviously B (2, 2) satisfying condition (i)

  $\Delta -\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 2\hfill & \hfill -4\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 0\hfill & \hfill -8\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|,\left[R_2-R_1-2R_3\right]$

= -8 (-3 + k)

For inconsistent $\Delta =0\Rightarrow k=3$  

${\Delta }_x=\left|\begin{array}{ccc}\hfill 10\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 6\hfill & \hfill -4\hfill & \hfill -2\hfill \\ \hfill 5m\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=32-40m\ne 0\Rightarrow m\ne \frac{4}{5}$

y = x³

${\frac{dy}{dx}=3x^2\Rightarrow \frac{dy}{dx}|}_{\left(t,t^3\right)}=3t^2$           

Equation of tangent y – t³ = 3t² (x – t)  

Let again meet the curve at $Q\left(t_1,t_1^3\right)$  

$\Rightarrow t_1^3-t^3=3t^2\left(t_1-t\right)$           

$t_1^2+t{t}_1+t^2=3t^2\left[?t_1\ne t\right]$           

$t_1^2+t{t}_1-2t^2=0$            

->t₁ = -2t

Required ordinate = $\frac{2t^3+t_1^3}{3}=\frac{2t^3-8t^3}{3}=-2t^3$

$\frac{dp}{dt}=0.5p-450andP\left(0\right)=850$           

$\Rightarrow \frac{dp}{P-900}=0.5dt$         

${\displaystyle \underset{850}{\overset{0}{\int }}\frac{dp}{P-900}={\displaystyle \underset{0}{\overset{T}{\int }}0.5dt}}$           

$\Rightarrow ln\left(P-900\right){|}_{805}^0=0.5T$

$\frac{T}{2}=ln\left|\frac{900}{50}\right|=ln18$           

T = 2 ln 18

  $f\left(x\right)=\frac{4x^3-3x^2}{6}-2sinx+\left(2x-1\right)cosx$

$f\text{'}\left(x\right)=2x^2-x-2cosx+2cosx-\left(2x-1\right)sinx$

= $\left(2x-1\right)\left(x-sinx\right)\ge 0forx\ge 0$  

$f\text{'}\left(x\right)\ge 0\forall x\ge 1/2$           

$\therefore f\left(x\right)$ is increasing in $\left[\frac{1}{2},\infty \right)$  

$h=\frac{a\left(1+t^2\right)}{2}.......(i)$

k = at        ……… (ii)

From (i) & (ii) $\frac{2h}{a}=1+\frac{k^2}{a^2}$  

$\therefore$ required locus of Q is y² = 2a (x – a/2)

Equation of directrix x – a/2 = -a/2Þ x = 0

Let total number of throws = n

Probability of getting 2 times = Probability of getting an even number 3 times.

[as probability of getting odd number = probability of getting even number = ]

Probability of getting an odd number for odd number of times =

 

I (6)       F (8)

Case I       2            4

Case II      3            6

Case III     4            8

Total =