JEE MAIN - Mathematics (2020 - 6th September Morning Slot - No. 7)
Let f : R $$ \to $$ R be defined as
$$f\left( x \right) = \left\{ {\matrix{ {{x^5}\sin \left( {{1 \over x}} \right) + 5{x^2},} & {x < 0} \cr {0,} & {x = 0} \cr {{x^5}\cos \left( {{1 \over x}} \right) + \lambda {x^2},} & {x > 0} \cr } } \right.$$
The value of $$\lambda $$ for which f ''(0) exists, is _______.
$$f\left( x \right) = \left\{ {\matrix{ {{x^5}\sin \left( {{1 \over x}} \right) + 5{x^2},} & {x < 0} \cr {0,} & {x = 0} \cr {{x^5}\cos \left( {{1 \over x}} \right) + \lambda {x^2},} & {x > 0} \cr } } \right.$$
The value of $$\lambda $$ for which f ''(0) exists, is _______.
Answer
5
Explanation
If g(x) = x5sin$$\left( {{1 \over x}} \right)$$
and h(x) = x5cos$$\left( {{1 \over x}} \right)$$
then g''(0) = 0 and h''(0) = 0
So, f''(0+ ) = g''(0+ ) + 10 = 10
and f''(0ā) = h''(0ā) + 2$$\lambda $$ = f''(0+)
$$ \Rightarrow $$ 2$$\lambda $$ = 10
$$ \Rightarrow $$ $$\lambda $$ = 5
and h(x) = x5cos$$\left( {{1 \over x}} \right)$$
then g''(0) = 0 and h''(0) = 0
So, f''(0+ ) = g''(0+ ) + 10 = 10
and f''(0ā) = h''(0ā) + 2$$\lambda $$ = f''(0+)
$$ \Rightarrow $$ 2$$\lambda $$ = 10
$$ \Rightarrow $$ $$\lambda $$ = 5
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