Let f: R → R and g: R → R be respectively given by f(x) = |x| + 1 and g(x) = x ^ 2 + 1 . Define h: R → R by h(x) = \ cases max \ f(x), g(x)\ & if & x ≤ 0 min \ f(x), g(x)\ & if & x > 0 cases ) Then number of points at

Correct answer: 0 h(x) = \ array ll x ^ 2 + 1, x ∈ (- ∞, - 1] - x + 1, & x ∈ [- 1, 0] x ^ 2 + 1, & x ∈ [0, 1] array cases x + 1, & x ∈ [1, ∞) cases ) Hence, not differentiable at x = - 1, 0, 1

JEE Advanced 2014Paper 1

Exam: JEE Advanced
Year: 2014
Paper: Paper 1
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Advanced 2014 · Paper 1 · Mathematics · Continuity and Differentiability

Let f: R → R and g: R → R be respectively given by f(x) = |x| + 1...

Let

f: R \to R and

g: R \to R

be respectively given by

f\left(x\right) = \left|x\right| + 1

and

g\left(x\right) = {x}^{2} + 1 .

Define

h: R \to R

by

h\left(x\right) = \left\{

\begin{cases}max \left\{f\left(x\right), g\left(x\right)\right\} & if & x \le 0 \\ min \left\{f\left(x\right), g\left(x\right)\right\} & if & x > 0\end{cases}

\right)
Then number of points at which

h\left(x\right)

is not differentiable is_________

Correct answer

Step-by-step solution

h\left(x\right) = \left\{

\begin{array}{ll}{x}^{2} + 1, x \in \left(- \infty, - 1\right] \\ - x + 1, & x \in \left[- 1, 0\right] \\ {x}^{2} + 1, & x \in \left[0, 1\right]\end{array}

\begin{cases}x + 1, & x \in \left[1, \infty\right)\end{cases}

\right) <br >Hence, not differentiable at

x = - 1, 0, 1

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