Q197 / 270 · NDA 2023 (Phase 2) · Mathematics

Given that $4 x^2+y^2=9$ What is the maximum value of y?

  • A
    $\frac{3}{2}$
  • Correct answer
    B
    3
  • C
    4
  • D
    6

Solution

Since, $4 x^2+y^2=9$ $\Rightarrow \mathrm{y}= \pm \sqrt{9-4 \mathrm{x}^2}$ So, max value of $y=3$

Hence, the correct answer is option: 'B'.

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