NDA 1 2024 Maths, question 79

Topic: Differential calculus

Q79

The non-negative values of $b$ for which the function $\frac{16x^{3}}{3}-4bx^{2}+x$ has neither maximum nor minimum in the range $x>0$ is

(a) $0<b<1$
(b) $1<b<2$
(c) $b>2$
(d) $0\le b<1$ Correct answer

Solution.

Answer: (d) $0\le b<1$

$f'(x) = 16x^{2} - 8bx + 1$ . Its discriminant $64b^{2} - 64$ is negative for $0 \leq b < 1$ , so $f' > 0$ and there is no maximum or minimum.
For $b > 1$ , $f'$ has two positive roots, giving a maximum and a minimum.
Of the options, only (d) covers the values with no extremum, starting from $b = 0$ .

Trap: At $b = 1$ , $f'(x) = (4x - 1)^{2} \geq 0$ , so there is still no extremum; the exact range is $0 \leq b \leq 1$ , and (d) is the closest option.