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Limits Continuity and Differentiability question

2020 · 6 Sep · Shift 1 · Q26

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JEE Main
Mathematics
Limits Continuity and Differentiability
2020 · 6 Sep · Shift 1 · Q26

Limits Continuity and Differentiability question

2020 · 6 Sep · Shift 1 · Q26

JEE MainMathematicsLimits Continuity and DifferentiabilityNumerical+4 / −1

Let f : R

\to

R be defined as

f\left( x \right) = \left\{ {\begin{matrix} {{x^5}\sin \left( {{1 \over x}} \right) + 5{x^2},} & {x \lt 0} \\ {0,} & {x = 0} \\ {{x^5}\cos \left( {{1 \over x}} \right) + \lambda {x^2},} & {x \gt 0} \\ \end{matrix} } \right.

The value of

\lambda

for which f ''(0) exists, is

\underline{\hspace{2cm}}

.

Numerical answer

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