QJEEM20Ma5a51: JEE Main 2020 Mathematics - Class 12 JEEM20Ma5a51 Solution: Let f : R R be defined as f ( x ) = { x 5 sin ⁡ ( 1 x ) + 5 x 2 …

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''(0^–) = h''(0^–) + 2 $\lambda$ = f''(0⁺)

$\Rightarrow$ 2 $\lambda$ = 10

$\Rightarrow$ $\lambda$ = 5