Let f : R → R be defined as
f( x ) = { x⁵sin ( 1/x ) + 5x²,, x < 0; 0,, x = 0; x⁵cos ( 1/x ) + λ x²,, x > 0
The value of λ for which f ''(0) exists, is _______.

Solution

Maths sir · 4 steps

The x⁵ terms vanish twice over at 0, so only the x² terms decide whether f''(0) exists.

1. First derivative. For x < 0, f'(x) = 5x⁴sin 1/x - x³cos 1/x + 10x. For x > 0, f'(x) = 5x⁴cos 1/x + x³sin 1/x + 2λ x.
2. Check f'(0). From the left, (f(x) - 0)/x = x⁴sin 1/x + 5x → 0. From the right, x⁴cos 1/x + λ x → 0. Sine and cosine stay between -1 and 1, so these powers of x force the limits to 0. So f'(0) = 0 for every λ.
3. Second derivative. Now find f''(0) as the limit of (f'(x) - f'(0))/x = f'(x)/x from each side. Every term carrying sin 1/x or cos 1/x still has a power of x in front, so it goes to 0.

   lim(x → 0⁻)(5x³sin 1/x - x²cos 1/x + 10) = 10, lim(x → 0⁺)(5x³cos 1/x + x²sin 1/x + 2λ) = 2λ

4. So. f''(0) exists only when the two sides agree: 2λ = 10, so λ = 5.

The common slip: 2.5. Setting 2λ equal to 5, the coefficient of x², gives 2.5. The left side's second derivative is 10, which comes from differentiating 5x² twice.

The idea to keep. A term like xⁿ sin 1/x dies out at 0 as long as a power of x is left in front after differentiating. Find the derivative at 0 from the definition, one order at a time.

Answer from the NTA final answer key. Solution by PrepWiser.