JEE Main 2022 · 28 Jul · Shift 1 · Mathematics Q17

MathematicsSingle correct+4 / -1Question 17 of 90

If the minimum value of $f(x)=\dfrac{5x^2}2+\dfrac\alpha{x^5}$ , $x>0$ , is 14, then the value of $\alpha$ is equal to :

128

256

Answer and worked solution

Official answer: C

Stationary point. $5x=\frac{5\alpha}{x^6}$ , so $x^7=\alpha$ and $f=\frac52x^2+x^2=\frac72x^2$ .

Minimum 14 gives $x=2$ , $\alpha=128$ . (C)

JEE Main 2022 · 28 Jul · Shift 1, Mathematics question 17