Maths

f(x) =   $\left|2x^2+3x?2\right|+sinxcosx$

$=\left|\left(2x?1\right)\left(x+2\right)\right|+sinxcosx$

$f\left(x\right)=\{\begin{array}{cc}\hfill ?2x^2?3x+2+sinxcosx,\hfill & \hfill 0<x<\frac{1}{2}\hfill \\ \hfill 2x^2+3x?2+sinxcosx,\hfill & \hfill \frac{1}{2}?x<1\hfill \end{array}$

$f\text{'}\left(x\right)=\{\begin{array}{cc}\hfill ?4x?3+cos2x,\hfill & \hfill 0<x<\frac{1}{2}\hfill \\ \hfill 4x+3+cos2x,\hfill & \hfill \frac{1}{2}?x<1\hfill \end{array}$            

 f(1) = 3 + sin 1 cos 1 and   $f\left(\frac{1}{2}\right)=sin\frac{1}{2}cos\frac{1}{2}$

$?f\left(1\right)+f\left(\frac{1}{2}\right)=3+\frac{sin2}{2}+\frac{sin1}{2}=3+\frac{1}{2}\left(sin1+sin2\right)$

$=3+\frac{sin1}{2}\left(1+2cos1\right)$

f(x) =   $\left|2x^2+3x-2\right|+sinxcosx$

$=\left|\left(2x-1\right)\left(x+2\right)\right|+sinxcosx$

$f\left(x\right)=\{\begin{array}{cc}\hfill -2x^2-3x+2+sinxcosx,\hfill & \hfill 0<x<\frac{1}{2}\hfill \\ \hfill 2x^2+3x-2+sinxcosx,\hfill & \hfill \frac{1}{2}\le x<1\hfill \end{array}$

$f\text{'}\left(x\right)=\{\begin{array}{cc}\hfill -4x-3+cos2x,\hfill & \hfill 0<x<\frac{1}{2}\hfill \\ \hfill 4x+3+cos2x,\hfill & \hfill \frac{1}{2}\le x<1\hfill \end{array}$            

 f(1) = 3 + sin 1 cos 1 and   $f\left(\frac{1}{2}\right)=sin\frac{1}{2}cos\frac{1}{2}$

$\therefore f\left(1\right)+f\left(\frac{1}{2}\right)=3+\frac{sin2}{2}+\frac{sin1}{2}=3+\frac{1}{2}\left(sin1+sin2\right)$

$=3+\frac{sin1}{2}\left(1+2cos1\right)$

f (x) = 4log_e (x – 1) -2x² + 4x + 5, x > 1

    $(A)f\text{'}(x)=\frac{4}{x-1}-4x+4$

$\Rightarrow f\text{'}(x)>0\Rightarrow \frac{x\left(-x+2\right)}{x-1}>0$

option (A) is correct.

(B) f (x) = -1, has two solution

option (B) is correct

$\left(C\right)f"(x)=-\frac{4}{{\left(x-1\right)}^2}-4$                

$f\text{'}(e)-f"(2)=\frac{4e\left(2-e\right)}{e-1}+8>0$              

option (C) is not correct

(D) f (e) = 4log_e (e – 1) -2 (e² – 2e + 1) + 7 > 0

f (e + 1) = 4 – 2 (e + 1)² + 4 (e + 1) +5+ < 0

option (D) is correct   

S = $\left\{\sqrt[]{n}:1\le n\le 50\&n=odd\right\}$  

= $\left\{\sqrt[]{1},\sqrt[]{3},\sqrt[]{5},......,\sqrt[]{49}\left(25terms\right)\right\}$  

$\left|A\right|=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 4\hfill \\ \hfill -1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill -a\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right|$ = 1 + a²

^{$\therefore {\displaystyle \sum _{a\in S}^{}det\left(adjA\right)}=100\lambda \Rightarrow \sum {\left(1+a^2\right)}^2=100\lambda \Rightarrow \lambda =221$}

${\left(ta{n}^{-1}x\right)}^3+{\left(co{t}^{-1}x\right)}^3=k{\pi }^3,x\in R$

$\Rightarrow \left(ta{n}^{-1}x+co{t}^{-1}x\right)\left({\left(ta{n}^{-1}x+co{t}^{-1}x\right)}^2-3ta{n}^{-1}xco{t}^{-1}x\right)=k{\pi }^3$

$\Rightarrow \frac{1}{3}\le k<\frac{7}{8}$

$\alpha +\beta =-\frac{\lambda }{3},\alpha \beta =-\frac{1}{3}$

$\therefore \frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=15\Rightarrow \lambda =\pm 3$

$\therefore 6{\left({\alpha }^3+{\beta }^3\right)}^2=6{\left({\left(\alpha +\beta \right)}^3-3\alpha \beta \left(\alpha +\beta \right)\right)}^2=6*{\left(-\left(1+1\right)\right)}^2=24$

For inconsistent, $\Delta =0,\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill \alpha \hfill & \hfill 2a\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 3\alpha \hfill & \hfill 5\hfill \end{array}\right|=0$

a = 1

$\therefore {\Delta }_1=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 4\hfill & \hfill 3\hfill & \hfill 5\hfill \end{array}\right|=1\left(1\right)+1\left(12+5\right)+1\left(-3-8\right)=7\ne 0$              

Equation of tangent to the circle at (2, 4) is

(4 + A) x + (8 + B) y + 2A + 2B + 2C = 0……. (i)

Also equation of tangent to parabola y = x² at (2, 4) is

4x – y = 4 ………. (ii)

Comparing (i) and (ii) we get, A + C = 16

  $\frac{6}{11}$ = Required probability         

After solving, we get n = 4

Surface area, S = 4pr²

$?\frac{ds}{dt}=4?.2r\frac{dr}{dt}=8?\frac{dr}{dt}=constant=k\left(say\right)$                

$?\frac{ds}{dt}=k?s=kt+c$

$?4?r^2=kt+c$        

Initially t = 0, r = 3

c = 36 p

When t = 5, r = 7, k = 32p

When t = 9, r = r, r = 9