If the slope of tangent to the curve y = ax³ – bx² at x = 1 is equal to 3, where a $\in$  [1,2] then number of possible integral values of b is

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

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

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