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Differentiation

Section: Mathematics

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Question : 2 of 79

Marks: +1, -0

Let

f(x)=x^3+x^2 f'(1)+2 x f''(2)+f'''(3), x\in\mathbb{R}

. Then the value of

f'(5)

[21 Jan 2026 Shift 2]

$\;\frac{657}{5}$
$\;\frac{117}{5}$
$\;\frac{2}{5}$
$\;\frac{62}{5}$

Solution:

Concept:

Treat

f'(1)

,

f''(2)

, and

f'''(3)

as constants. Differentiate the polynomial and substitute the given points to form equations.

Explanation:

Let

a = f'(1)

,

b = f''(2)

,

c = f'''(3)

.

Then

f(x) = x^3 + a x^2 + 2b x + c

.

Differentiate:

f'(x) = 3x^2 + 2a x + 2b

.

f''(x) = 6x + 2a

.

f'''(x) = 6

.

Apply conditions:

f'(1)=a \Rightarrow a = 3 + 2a + 2b \Rightarrow a + 2b = -3

…(1)

f''(2)=b \Rightarrow b = 12 + 2a \Rightarrow 2a - b = -12

…(2)

f'''(3)=c \Rightarrow c = 6

.

Solve (1) and (2):

From (2),

b = 12 + 2a

.

Substitute in (1):

a + 2(12+2a) = -3 \Rightarrow 5a = -27 \Rightarrow a = -\frac{27}{5}

.

Then

b = 12 + 2\left(-\frac{27}{5}\right) = \frac{60}{5} - \frac{54}{5} = \frac{6}{5}

.

Now

f'(5) = 3(25) + 2a(5) + 2b = 75 + 10a + 2b

.

Plug values:

75 + 10\left(-\frac{27}{5}\right) + 2\left(\frac{6}{5}\right) = 75 - 54 + \frac{12}{5} = 21 + \frac{12}{5} = \frac{105}{5} + \frac{12}{5} = \frac{117}{5}

.

Answer:

\frac{117}{5}

(Option B).

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