If the surface area of a cube is increasing at a rate of 4.8cm²/sec. retaining its shape, then the rate of change of its volume (in cm³/sec.) when the length of a side of the cube is 15 cm is:

y (x) = 2^x – x²

y? (x) = 2x log 2 – 2x

M = 3

N = 2

M + N = 5

Area of ?

$=\frac{1}{2}\left|\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill x\hfill & \hfill y\hfill & \hfill 1\hfill \\ \hfill -x\hfill & \hfill y\hfill & \hfill 1\hfill \end{array}\right|$

-> $\left|\frac{1}{2}\left(xy+xy\right)\right|=\left|xy\right|$

->Area (D) = |xy| = |x (– 2x² + 54x)|

$\frac{d\left(\Delta \right)}{dx}=\left|\left(-6x^2+108x\right)\right|\Rightarrow \frac{d\Delta }{dx}=0$  at x = 0 and 18

->at x = 0, minima

and at x = 18 maxima

Area (D) = |18 (– 2 (18)² + 54 × 18)| = 5832

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

Given curves $\frac{x^2}{9}+\frac{y^4}{4}=1.........\left(i\right)$  

&       $x^2+y^2=\frac{31}{4}...........\left(ii\right)$   

Equation of any tangent to (i) be y = mx +    $\sqrt[]{9m^2+4}............\left(iii\right)$

For common tangent (iii) also should be tangent to (ii) so by condition of common tangency

$9m^2+4=\frac{31}{4}\left(1+m^2\right)$          

OR 36m² + 16 = 31 + 31m²

->m² = 3

Given f(X) = ${\displaystyle {\int }_1^x\frac{lo{g}_{e}t}{\left(1+t\right)}dt............\left(i\right)}$  

So   $f\left(\frac{1}{x}\right)={\displaystyle {\int }_1^{1/x}\frac{\lambda lo{g}_{e}t}{1+t}dt}............\left(ii\right)$

put $t=\frac{1}{z}thendt=-\frac{1}{z^2}dz$  

$f\left(\frac{1}{x}\right)={\displaystyle {\int }_1^x\frac{lo{g}_{e}t}{t\left(1+t\right)}}dt............\left(iii\right)$

(i) + (iii), f(x) + $f\left(\frac{1}{x}\right)={\displaystyle {\int }_1^x\left(\frac{lo{g}_{e}t}{1+t}+\frac{lo{g}_{e}t}{t\left(1+t\right)}\right)dt}$

$={\left[\frac{{\left(lo{g}_{e}t\right)}^2}{2}\right]}_1^x=\frac{{\left(lo{g}_{x}x\right)}^2}{2}$

Hence f(e) + $f\left(\frac{1}{e}\right)=\frac{1}{2}$