Let a, b ∈ ℝ and f: ℝ → ℝ be defined by f (x) = acos (| x ^ 3 - x|) + b|x| (| x ^ 3 + x|) . Then f is -

Correct answer: A. Differentiable at x = 0 if a = 0 and b = 1 If x ^ 3 - x ≥ 0 ⇒ | x ^ 3 - x| = ( x ^ 3 - x) x ^ 3 - x Similarly b|x| | x ^ 3 + x| = bxsin ( x ^ 3 + x) for all x ∈ R ∴ f(x) = acos ( x ^ 3 - x) + bxsin ( x ^ 3 + x) Which is composition and sum of differentiable functions therefore always continuous and differentiable.

JEE Advanced 2016Paper 2

Exam: JEE Advanced
Year: 2016
Paper: Paper 2
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Advanced 2016 · Paper 2 · Mathematics · Continuity and Differentiability

Let a, b ∈ ℝ and f: ℝ → ℝ be defined by f (x) = acos (| x ^ 3 - x|)...

Let

a, b \in \mathbb{R}

and

f: \mathbb{R} \to \mathbb{R}

be defined by

f

\left(x\right) = acos

\left(\left|{x}^{3} - x\right|\right) + b\left|x\right|\sin \left(\left|{x}^{3} + x\right|\right)

. Then

f

is -

Options

ADifferentiable at $x = 0$ if $a = 0$ and $b = 1$
BDifferentiable at $x = 1$ if $a = 1$ and $b = 0$
CNOT differentiable at $x = 0$ if $a = 1$ and $b = 0$
DNOT differentiable at $x = 1$ if $a = 1$ and $b = 1$

Correct answer

A. Differentiable at $x = 0$ if $a = 0$ and $b = 1$

Step-by-step solution

{x}^{3} - x \ge 0 ⇒ \cos \left|{x}^{3} - x\right| = \cos \left({x}^{3} - x\right)

{x}^{3} - x < 0 ⇒ \cos \left|{x}^{3} - x\right| = \cos \left({x}^{3} - x\right)

Similarly

b\left|x\right|\sin \left|{x}^{3} + x\right| = bxsin \left({x}^{3} + x\right)

for all

x \in R

∴

f\left(x\right) = acos \left({x}^{3} - x\right) + bxsin \left({x}^{3} + x\right)

Which is composition and sum of differentiable functions therefore always continuous and differentiable.

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