For x > 0, if f(x) = ${\displaystyle \underset{1}{\overset{x}{\int }}\frac{lo{g}_{e}t}{\left(1+t\right)}}dt,$  then f(e) + f $\left(\frac{1}{e}\right)$  is equal to:

y (x) = 2^x – x²

y? (x) = 2x log 2 – 2x

M = 3

N = 2

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

$\frac{dy}{dx}=3a{x}^2-2bx$

${\frac{dy}{dx}|}_{x=1}=3a-2b=3$

$a=\frac{2b+3}{3}\in \left[1,2\right]$

$2b+3\in \left[3,6\right]$

$2b\in \left[0,3\right]$

$b\in \left[0,\frac{3}{2}\right]$

f' (x) = cosx + sinx − k ≤ 0∀x ∈ R
k ≥ √2

f (x) = x? /20 - x? /12 + 5
f' (x) = x? /4 - x³/3 = x³ (x/4 - 1/3)
Local maxima at 0, Local minima at 4/3
f' (x) = x³ - x² = x² (x-1)
x = 1 point of inflection