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)} = a (cos α) (cos β), where a, α, β \in N, then the find value of (a + α + β) /2

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 $\sum_{r = 0}^{1} \frac{d^{2} g}{{dx}^{2}} |_{x = f' (r)}$ = a (cos α) (cos β), where a, α, β $\in$ N, then the find value of (a + α + β) /2

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f (x) = $\underset{ϕ (x)}{\underset{⏟}{\frac{x (π^{x} - 1) {(x - 1)}^{2}}{1 - x + x^{2}}}} + 1 + x + x^{2}$
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 $\equiv$ a cos α cos β
$∴$ a + α + β = 4 + 2 + 4 / 2 = 5