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JEE Main 20206 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
6 September
Shift
Morning Shift
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Main 2020 · 6 September · Morning Shift · Mathematics · Continuity and Differentiability

Let f: R → R be defined as f(x) = \ cases x ^ 5 ((1)/(x)) + 5 x ^ 2...

Let f:R→Rf: R \to R be defined as f\left(x\right) = \left\{\begin{cases}{x}^{5}\sin \left(\dfrac{1}{x}\right) + 5{x}^{2} & , & x 0\end{cases}\right). The value of\lambdaforwhichfor which{f}^{"} ( 0 )$ exists, is___.

Correct answer

0

Step-by-step solution

f\left(x\right) = \left\{\begin{cases}5{x}^{4} \cdot \sin ( 1 / x ) - {x}^{3}\cos ( 1 / x ) + 10x, & x 0\end{cases}\right)<br />Differentiating two times.{f}^{"}\left(x\right) = \left\{{20x3⋅sin⁡(1/x)−5x2cos⁡(1/x)−3x2cos⁡(1/x)−xsin(1/x)+10,x0\begin{cases}20{x}^{3} \cdot \sin ( 1 / x ) - 5{x}^{2}\cos ( 1 / x ) - 3{x}^{2}\cos ( 1 / x ) - xsin ( 1 / x ) + 10, & x 0\end{cases}\right)
f"(0+)=f"(0−){f}^{"}\left({0}^{+}\right) = {f}^{"}\left({0}^{-}\right)
⇒2λ=10⇒λ=5⇒ 2\lambda = 10 ⇒ \lambda = 5
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