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For the next two(2) items that follow :

Consider

f'(x)=\frac{x^2}{2}-kx+1

such that

f(0)=0 \text{ and } f(3)=15

[NDA II 2015]

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Question : 21 of 24

Marks: +1, -0

f''\left(-\frac{2}{3}\right)

$-1$
$\frac{1}{3}$
$\frac{1}{2}$
1

Solution:

\because f'(x) = \;\;\frac{x^{2}}{2} + \;\;\frac{5}{3}x + 1

\Rightarrow f''(x) = x + \;\;\frac{5}{3}

f''\left(-\;\;\frac{2}{3}\right) = \;\;\frac{-2}{3} + \;\;\frac{5}{3}

=1

So option (d) is correct.

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