Limit Continuity and Differentiability 7 Question 28
28. Let $f: R \rightarrow R, g: R \rightarrow R$ and $h: R \rightarrow R$ be differentiable functions such that $f(x)=x^{3}+3 x+2, g(f(x))=x$ and $h(g(g(x)))=x$ for all $x \in R$. Then,
(2016 Adv.)
(a) $g^{\prime}(2)=\frac{1}{15}$
(b) $h^{\prime}(1)=666$
(c) $h(0)=16$
(d) $h(g(3))=36$
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Answer:
Correct Answer: 28. (b)
Solution:
- $y=\frac{a x^{2}}{(x-a)(x-b)(x-c)}+\frac{b x}{(x-b)(x-c)}+\frac{c}{(x-c)}+1$
$$ =\frac{a x^{2}}{(x-a)(x-b)(x-c)}+\frac{b x}{(x-b)(x-c)}+\frac{x}{(x-c)} $$
$$ \begin{aligned} & =\frac{a x^{2}}{(x-a)(x-b)(x-c)}+\frac{x}{(x-c)} \frac{b}{x-b}+1 \\ & =\frac{a x^{2}}{(x-a)(x-b)(x-c)}+\frac{x}{(x-c)} \cdot \frac{x}{(x-b)} \\ & =\frac{x^{2}}{(x-c)(x-b)} \frac{a}{x-1}+1 \Rightarrow y=\frac{x^{3}}{(x-a)(x-b)(x-c)} \end{aligned} $$
$\Rightarrow \quad \log y=\log x^{3}-\log (x-a)(x-b)(x-c)$
$\Rightarrow \log y=3 \log x-\log (x-a)-\log (x-b)-\log (x-c)$
On differentiating, we get
$$ \begin{array}{rlrl} \frac{y^{\prime}}{y} & =\frac{3}{x}-\frac{1}{x-a}-\frac{1}{x-b}-\frac{1}{x-c} \\ \Rightarrow & \frac{y^{\prime}}{y} & =\frac{1}{x}-\frac{1}{x-a}+\frac{1}{x}-\frac{1}{x-b}+\frac{1}{x}-\frac{1}{x-c} \\ \Rightarrow & \frac{y^{\prime}}{y} & =\frac{-a}{x(x-a)}-\frac{b}{x(x-b)}-\frac{c}{x(x-c)} \\ \Rightarrow & \frac{y^{\prime}}{y} & =\frac{a}{x(a-x)}+\frac{b}{x(b-x)}+\frac{c}{x(c-x)} \\ \Rightarrow & \frac{y^{\prime}}{y} & =\frac{1}{x} \frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x} \end{array} $$
