Limit Continuity and Differentiability 7 Question 20
20. Let $f: R \rightarrow R$ be a function defined by $f(x)=\max {x, x^{3} }$. The set of all points, where $f(x)$ is not differentiable, is
(a) ${-1,1}$
(b) ${-1,0}$
(c) ${0,1}$
(d) ${-1,0,1}$
(2001, 2M)
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Answer:
Correct Answer: 20. (a)
Solution:
- Since, $y^{2}=P(x)$
On differentiating both sides, we get
$$ 2 y y _1=P^{\prime}(x) $$
Again, differentiating, we get
$$ \begin{array}{rlrl} & & 2 y y _2+2 y _1^{2} & =P^{\prime \prime}(x) \\ \Rightarrow & 2 y^{3} y _2+2 y^{2} y _1^{2} & =y^{2} P^{\prime \prime}(x) \\ \Rightarrow & 2 y^{3} y _2 & =y^{2} P^{\prime \prime}(x)-2\left(y y _1\right)^{2} \\ \Rightarrow & & 2 y^{3} y _2 & =P(x) \cdot P^{\prime \prime}(x)-\frac{{P^{\prime}(x) }^{2}}{2} \end{array} $$
Again, differentiating, we get
$$ \begin{aligned} & \quad 2 \frac{d}{d x}\left(y^{3} y _2\right)=P^{\prime}(x) \cdot P^{\prime \prime}(x)+P(x) \cdot P^{\prime \prime \prime}(x) \\ &-\frac{2 P^{\prime}(x) \cdot P^{\prime \prime}(x)}{2} \\ & \Rightarrow \quad 2 \frac{d}{d x}\left(y^{3} y _2\right)=P(x) \cdot P^{\prime \prime \prime}(x) \\ & \Rightarrow \quad \frac{d^{2} y}{d x^{2}}=P(x) \cdot P^{\prime \prime \prime}(x) \end{aligned} $$
