Q187 / 270 · NDA 2023 (Phase 1) · Mathematics

Paragraph: Let $x=\frac{\sin ^2 A+\sin A+1}{\sin A}$ where $0 \lt A \leq \frac{\pi}{2}$
Question: What is the minimum value of $x$ ?

  • A
    $1$
  • B
    $2$
  • Correct answer
    C
    $3$
  • D
    $4$

Solution

$x=\frac{\sin ^2 A+\sin A+1}{\sin A}=\sin A+\frac{1}{\sin A}+1$ Using $\mathrm{AM} \geq \mathrm{GM}$ $\frac{\sin A+\frac{1}{\sin A}}{2} \geq \sqrt{(\sin A) \cdot\left(\frac{1}{\sin A}\right)}$ $\sin A+\frac{1}{\sin A} \geq 2$ Minimum value of $\sin A+\frac{1}{\sin A}=2$ $x_{\operatorname{minimum}}=2+1=3$

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

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