QJEEAdv16Mad527: JEE Advanced 2016 Mathematics - Class 12 JEEAdv16Mad527 Solution: Let a, b R and f : R R be defined by . Then f is

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

If ${x^3} - x \ge 0$

$\Rightarrow \cos \left| {{x^3} - x} \right| = \cos ({x^3} - x)$

and if ${x^3} - x \le 0$

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

$\therefore$ $\cos (|{x^3} - x|) = \cos ({x^3} - x),\,\forall x \in R$ ..... (i)

Again, if ${x^3} + x \ge 0$

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

and if ${x^3} + x \le 0$

$\Rightarrow |x|\sin (|{x^3} + x|) = - x\sin \{ - ({x^3} + x)\}$

$\therefore$ $|x|\sin (|{x^3} + x|) = x\sin ({x^3} + x),\,\forall x \in R$ ...... (ii)

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

$\therefore$ $f(x) = a\cos ({x^3} - x) + bx\sin ({x^3} + x)$ ..... (iii)

For a function to be differentiable at x = 0, the function must be continuous.

$f(0) = a\cos (0 - 0) + b(0)\sin (0) = a$

Therefore,

$f({0^ + }) = \mathop {\lim }\limits_{h \to \infty } [a\cos ({h^3} - h) + bh\sin ({h^3} + h)] = 0$

$f({0^ - }) = \mathop {\lim }\limits_{h \to \infty } [a(\cos ( - {h^3} + h) + b( - h)\sin ( - {h^3} - h)]$

$= \mathop {\lim }\limits_{h \to \infty } [a\cos ({h^3} - h) + bh\sin ({h^3} + h)] = 0$

which is continuous at x = 0; hence, f(x) is differentiable for all values of a and b.

Therefore,

$f(1) = a\cos (1 - 1) + 1\sin (1 + 1) = a + b\sin 2$

$f({1^ + }) = \mathop {\lim }\limits_{h \to 0} a\cos {(1 + h)^3} - (1 + h) + b(1 + h)\sin {(1 + h)^3} + (1 + h)$

$= f(1)$

$f({1^ - }) = \mathop {\lim }\limits_{h \to 0} a\cos {(1 - h)^3} - (1 - h) + b(1 - h)\sin {(1 - h)^3} + (1 - h)$

$= f(1)$

Thus, f(x) is continuous and we can also see that f is differentiable at x = 0 and x = 1.