Let the functions f: R → R and g: R → R be defined as :
f(x) = {x+2, x<0; x², x≥0} and g(x) = {x³, x<1; 3x-2, x≥1}
Then, the number of points in R where (fog)(x) is NOT differentiable is equal to :

$?$ gof is differentiable at x = 0

 So R.H.D = L.H.D.

$\frac{d}{dx}\left(4e^x+k_2\right)=\frac{d}{dx}\left({\left(-\left|x+3\right|\right)}^2-k_1\left|x+3\right|\right)$  

⇒ 4 = 6 – k₁ Þ k₁ = 2

Now g (f (-4) + g (f (4)

=2 (2e⁴ – 1)

$f\left(x\right)=[\begin{array}{l}-\frac{1}{x}x\in \left(-\infty ,-1\right)\\ a{x}^2+bx\in \left(-1,1\right)\\ \frac{1}{x}x\in \left[1,\infty \right)\end{array}$

cont at x = 1, a + b = 1

diff at x = 1, 2a = 1 $\Rightarrow a=-\frac{1}{2}\Rightarrow b=\frac{3}{2}$

Differentiable ⇒ continuous
Continuous at x = 2 ⇒ 2a + b = 0
Continuous at x = 3
0 = 9p + 3q + 1
Differentiable at x=2
a = 2×2 - 5 ⇒ a = -1
Differentiable at x=3
2×3 - 5 = 2p×3 + q ⇒ 6p + q = 1
a = -1, b = 2, p = 4/9, q = -5/3

g (x) = (f (nf (x) – n)?
g' (x) = n (f (nf (x) – n)? ¹ . f' (nf (x) – n) . n . f' (x)
∴ g' (0) = 0
⇒ 4 = n (f (nf (0) – n)? ¹ . f' (nf (0) – n) . nf' (0)
⇒ 4 = n (f (0)? ¹ . f' (0) . nf' (0)
⇒ 4 = n . 1 . (-1) . n (-1)
n² = 4
⇒ n = 2

$f\left(x\right)=\frac{\left(2^x+2^{-x}\right)tanx\sqrt[]{ta{n}^{-1}\left(2x^2-3x+1\right)}}{{\left(7x^2-3x+1\right)}^3}$

$f\left(x\right)=\left(2^x+2^{-x}\right).tanx.\sqrt[]{ta{n}^{-1}\left(2x^2-3x+1\right)}.{\left(7x^2-3x+1\right)}^{-3}$

$f\text{'}\left(x\right)=\left(2^x+2^{-x}\right).se{c}^{2}x.\sqrt[]{ta{n}^{-1}\left(2x^2-3x+1\right)}.{\left(7x^2-3x+1\right)}^{-3}+tanx.\left(Q\left(x\right)\right)$

$f\text{'}\left(0\right)=2.1\sqrt[]{\frac{\pi }{4}}.1$

$=\sqrt[]{\pi }$