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QJEEAdv12Ma772f: JEE Advanced 2012 Mathematics - Class 12 JEEAdv12Ma772f Solution: Let f ( x ) = { x 2

Let f(x)={x2|cos⁡πx|,x≠00,x=0

x ∈\in R, then f is

Multiple Choice Options

  • A differentiable both at x = 0 and at x = 2.
  • B differentiable at x = 0 but not differentiable at x = 2.
  • C not differentiable at x = 0 but differentiable at x = 2.
  • D differentiable neither at x = 0 nor at x = 2.

Step-by-Step Verified Solution

To check differentiable at x = 0

R{f′(0)}=lim⁡h→0f(0+h)−f(0)hR\{ f'(0)\} = \mathop {\lim }\limits_{h \to 0} {{f(0 + h) - f(0)} \over h}

=lim⁡h→0h2∣cos⁡πh∣−0h= \mathop {\lim }\limits_{h \to 0} {{{h^2}\left| {\cos {\pi \over h}} \right| - 0} \over h}

=lim⁡h→0h. ∣cos⁡πh∣=0= \mathop {\lim }\limits_{h \to 0} h.\,\left| {\cos {\pi \over h}} \right| = 0

L{f′(0)}=lim⁡h→0f(0−h)−f(0)−hL\{ f'(0)\} = \mathop {\lim }\limits_{h \to 0} {{f(0 - h) - f(0)} \over { - h}}

=lim⁡h→0h2∣cos⁡(−πh)∣−0−h= \mathop {\lim }\limits_{h \to 0} {{{h^2}\left| {\cos \left( { - {\pi \over h}} \right)} \right| - 0} \over { - h}}

=0= 0

So, f(x) is differentiable at x = 0

To check differentiability at x = 2

R{f′(2)}=lim⁡h→0f(2+h)−f(2)hR\{ f'(2)\} = \mathop {\lim }\limits_{h \to 0} {{f(2 + h) - f(2)} \over h}

=lim⁡h→0(2+h)2∣cos⁡(π2+h)∣−0h= \mathop {\lim }\limits_{h \to 0} {{{{(2 + h)}^2}\left| {\cos \left( {{\pi \over {2 + h}}} \right)} \right| - 0} \over h}

=lim⁡h→0(2+h)2.cos⁡(π2+h)h= \mathop {\lim }\limits_{h \to 0} {{{{(2 + h)}^2}.\cos \left( {{\pi \over {2 + h}}} \right)} \over h}

=lim⁡h→0(2+h)2.sin⁡(π2−π2+h)h= \mathop {\lim }\limits_{h \to 0} {{{{(2 + h)}^2}.\sin \left( {{\pi \over 2} - {\pi \over {2 + h}}} \right)} \over h}

=lim⁡h→0(2+h)2.sin⁡(πh2(2+h))h.π2(2+h).π2(2+h)= \mathop {\lim }\limits_{h \to 0} {{{{(2 + h)}^2}.\sin \left( {{{\pi h} \over {2(2 + h)}}} \right)} \over {h.{\pi \over {2(2 + h)}}.{\pi \over {2(2 + h)}}}}

⇒R(f′(2))=π\Rightarrow R\left( {f'(2)} \right) = \pi

L(f′(2))=lim⁡h→0f(2−h)−f(2)−hL\left( {f'(2)} \right) = \mathop {\lim }\limits_{h \to 0} {{f(2 - h) - f(2)} \over { - h}}

=lim⁡h→0(2−h)2.∣cos⁡π2−h∣−22.∣cos⁡π2∣−h= \mathop {\lim }\limits_{h \to 0} {{{{(2 - h)}^2}.\left| {\cos {\pi \over {2 - h}}} \right| - {2^2}.\left| {\cos {\pi \over 2}} \right|} \over { - h}}

=lim⁡h→0(2−h)2−(−cos⁡π2−h)−0−h= \mathop {\lim }\limits_{h \to 0} {{{{(2 - h)}^2} - \left( { - \cos {\pi \over {2 - h}}} \right) - 0} \over { - h}}

=lim⁡h→0−(2−h)2.sin⁡(π2−π2−h)h= \mathop {\lim }\limits_{h \to 0} {{ - {{(2 - h)}^2}.\sin \left( {{\pi \over 2} - {\pi \over {2 - h}}} \right)} \over h}

=lim⁡h→0(2−h)2.sin⁡(−πh2(2−h))h×−π2(2−h)×−π2(2−h)= \mathop {\lim }\limits_{h \to 0} {{{{(2 - h)}^2}.\sin \left( { - {{\pi h} \over {2(2 - h)}}} \right)} \over {h \times {{ - \pi } \over {2(2 - h)}} \times {{ - \pi } \over {2(2 - h)}}}}

⇒L(f′(2))=−π\Rightarrow L(f'(2)) = - \pi

Thus, f(x) is differentiable at x = 0 but not at x = 2.

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