Let f(x) = \frac{\left(\right. x^{3} - 2 x^{2} + x \left.\right) \left(\right. \left(\pi\right)^{x} - 1 \left.\right) + \left(\right. 1 + x^{2} + x^{4} \left.\right)}{\left(\right. 1 - x + x^{2} \left.\right)} and g(x) = sin2 x. If \sum_{r = 0}^{1} \frac{d^{2} g}{\left(dx\right)^{2}} \left(\left|\right.\right)_{x = f \textrm{ } ' \textrm{ } \left(\right. r \left.\right)} = α (cos α) (cos β), where α, α, β\in N, then the value of (α + α + β) is equal to

Engineering

Mathematics

Successive Differentiation

Question

Let f(x) = $\frac{(x^{3} - 2 x^{2} + x) (π^{x} - 1) + (1 + x^{2} + x^{4})}{(1 - x + x^{2})}$ and g(x) = sin² x.
If $\sum_{r = 0}^{1} \frac{d^{2} g}{{dx}^{2}} |_{x = f' (r)}$ = α (cos α) (cos β), where α, α, β $\in$ N, then the value of (α + α + β) is equal to

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f (x) = $\underset{φ (x)}{\underset{⏟}{\frac{x (π^{x} - 1) {(x - 1)}^{2}}{1 - x + x^{2}}}} + 1 + x + x^{2}$             [TN, MOD, SB]
f ' (x) = $ϕ$ ' (x) + 1 + 2x       {φ ' (0) = φ ' (1) = 0}
f ' (0) = 1, f ' (1) = 3
g (x) = sin²x   ⇒   g ' (x) = sin 2x   ⇒   g '' (x) = 2 cos 2x
Now,   g '' (x) |_{x = f ' 0)}+ g '' (x) |_{x = f ' (1)}= g '' (1) + g '' (3)
= 2 cos 2 + 2 cos 6 = 2 (2 cos 4 · cos 2)
= 4 cos 2 cos 4 º α cos α cos β
$∴$    α + α + β = 4 + 2 + 4 = 10