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MHTCET 14 May 2023 Shift 1 Solved Paper

Section: Mathematics

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Question : 65 of 150

Marks: +1, -0

Let

f: R \longrightarrow R

be a function such that

f(x)=x^3+x^2 f'(1)+x f''(2)+6, x \in R

f(2)

equals

$30$
$-4$
$-2$
$8$

Solution:

\; f(x)=x^3+x^2 f'(1)+x f''(2)+6

\therefore \;\; f'(x)=3x^2+2x f'(1)+f''(2)

...(i)

\therefore f''(x)=6x+2 f'(1)

....(ii)

Substituting

x=1

in (i), we get

\; f'(1)=3(1)^2+2(1) f'(1)+f''(2)

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

...(iii)

Substituting

x=2

in (ii), we get

\; f''(2)=6(2)+2 f'(1)

\; \Rightarrow f''(2)=12+2 f'(1)

....(iv)

From (iii) and (iv), we get

\; f'(1)+12+2 f'(1)=-3

\; \Rightarrow 3 f'(1)=-15

\; \Rightarrow f'(1)=-5

\;\; \text{From (iii),}\; -5+f''(2)=-3

\; \Rightarrow f''(2)=2

\; \therefore \;\; f(2)=2^3+2^2(-5)+2(2)+6

\; =8-20+4+6

\; =-2

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