Let $f: R \rightarrow R$ be defined as $f(x)=\left\{\begin{array}{cc}x^{5} \sin \left(\frac{1}{x}\right)+5 x^{2}, & x 0\end{array}\right.$ The value of $\lambda$ for which $f^{\prime \prime}(0)$ exists, is ______.

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Continuity and Differentiability

Let

f : R \to R

be defined as

f (x) = ⎩ ⎨ ⎧ x^{5} sin (\frac{1}{x}) + 5 x^{2}, 0, x^{5} cos (\frac{1}{x}) + λ x^{2}, x < 0 x = 0 x > 0

The value of

λ

for which

f^{''} (0)

exists, is ______.

Answer: 5

Near

x = 0

the terms

x^{5} sin \frac{1}{x}

and

x^{5} cos \frac{1}{x}

together with their first two derivatives all vanish, so the second derivative at

0

is governed entirely by the quadratic parts. From the left branch

5 x^{2}

the second derivative at

0

10

, and from the right branch

λ x^{2}

it is

2 λ

. For

f^{''} (0)

to exist these must agree,

10 = 2 λ

, so

λ = 5

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Integer answer