Let a, b∈R and f : R→R be defined by fx=acos⁡x^3-x+bxsin⁡x^3+x . Then f is -

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

Q25 / 54 · JEE Advanced 2016 (Paper 2) · Mathematics

Let $a, b \in R$ and $f : R \to R$ be defined by $f (x) = a \cos (|x^{3} - x|) + b |x| \sin (|x^{3} + x|)$ . Then $f$ is -

Correct answer
A
Differentiable at $x = 0$ if $a = 0$ and $b = 1$
Correct answer
B
Differentiable at $x = 1$ if $a = 1$ and $b = 0$
C
NOT differentiable at $x = 0$ if $a = 1$ and $b = 0$
D
NOT differentiable at $x = 1$ if $a = 1$ and $b = 1$

Solution

x^{3} - x \geq 0 \Rightarrow \cos |x^{3} - x| = \cos (x^{3} - x)

x^{3} - x < 0 \Rightarrow \cos |x^{3} - x| = \cos (x^{3} - x)

Similarly

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

for all

x \in R

∴

f (x) = a \cos (x^{3} - x) + b x \sin (x^{3} + x)

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

Hence, the correct answer is options: 'A', 'B'.

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Let a, b∈R and f : R→R be defined by fx=acos⁡x3-x+bxsin⁡x3+x . Then f is -

Solution

Relevant Questions

Let $a, b \in R$ and $f : R \to R$ be defined by $f (x) = a \cos (|x^{3} - x|) + b |x| \sin (|x^{3} + x|)$ . Then $f$ is -