Skip to main contentSkip to solution

The value of 'λ\lambda' for which the function f(x)={1−cos⁡4x4x2,if x≠0λ,if x=0f(x) = \begin{cases}\frac{1-\cos 4x}{4x^2}, & \text{if } x \neq 0 \\ \lambda, & \text{if } x = 0\end{cases} is continuous at x=0x=0, is:

Solution

✅ Correct Option: 1

Using 1−cos⁡4x=2sin⁡22x1-\cos 4x = 2\sin^2 2x:

lim⁡x→02sin⁡22x4x2=lim⁡x→02(sin⁡2x2x)2=2\lim_{x\to 0}\frac{2\sin^2 2x}{4x^2} = \lim_{x\to 0} 2\left(\frac{\sin 2x}{2x}\right)^2 = 2.

For continuity at x=0x=0, λ\lambda must equal this limit, so λ=2\lambda = 2.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question