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NDA 2 2025 Maths, question 72

Topic: Differential calculus

Passage · Q71 to Q72

Let f(x)={1−cos⁡2xx2,x<09,x=0x(16+x)−4,x>0f(x)=\left\{ \begin{matrix} \frac{1-\cos 2x}{x^{2}} & , x<0 \\ 9 & , x=0 \\ \frac{\sqrt{x}}{\sqrt{(16+\sqrt{x})}-4} & , x>0 \end{matrix} \right.

Q72

What is lim⁡x→0+f(x)\underset{x\to 0+}{\lim }f(x) equal to?

  1. (a) 66
  2. (b) 77
  3. (c) 88Correct answer
  4. (d) 99

Solution.

Answer: (c) 88

  1. Rationalize the denominator: √x/(√(16+√x)−4)=√(16+√x)+4 for x>0.
  2. As x→0⁺ this tends to 4+4=8.