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MHTCET 13 May 2023 Shift 2 Solved Paper

Section: Mathematics

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Question : 81 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)+f'''(3), x \in R

f(2)

equals

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

Solution:

f(x)=x^3+x^2 f'(1)+x f''(2)+f'''(3)

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

...(i)

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

..(ii)

\Rightarrow \;\; f'''(x)=6

...(iii)

Put

x=1

, in Eq. (i)

f'(1)=3+2 f'(1)+f''(2)

...(iv)

Put

x=2

in Eq. (ii)

f''(2)=12+2 f'(1)

....(v)

From Eqs. (iv) and (v),

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

\Rightarrow 3 f'(1) = -15

\Rightarrow f'(1) = -5

\Rightarrow f''(2)=2 \;\; [\; \text{ from Eq. (v)] } \;

Put

x=3

in Eq. (iii), we get

\; f'''(3)=6

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

\; f(2)=8-20+4+6=-2

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