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Differentiation question

2019 · 10 Jan · Shift 1 · Q43

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JEE Main
Mathematics
Differentiation
2019 · 10 Jan · Shift 1 · Q43

Differentiation question

2019 · 10 Jan · Shift 1 · Q43

JEE MainMathematicsDifferentiationMCQ+4 / −1

Let f : R

\to

R be a function such that f(x) = x3 + x2f'(1) + xf''(2) + f'''(3), x

\in

R. Then f(2) equals -

A
30
B
$-$ 2
C
$-$ 4
D
8

View written solutionFree

Correct answer: B

We are given $f(x)=x^3+x^2f'(1)+x f''(2)+f'''(3), \,\forall x\in \mathbb R.$

We must find $f(2)$ .

Since the right-hand side is a polynomial in $x$ , let us differentiate it.

Given $f(x)=x^3+x^2f'(1)+x f''(2)+f'''(3),$ where $f'(1), f''(2), f'''(3)$ are constants.

So, $f'(x)=3x^2+2x f'(1)+f''(2).$

Differentiating again, $f''(x)=6x+2f'(1).$

Differentiating third time, $f'''(x)=6.$

Now use the self-referential conditions.

From $f'''(x)=6,$ we get $f'''(3)=6.$

From $f''(x)=6x+2f'(1),$ put $x=2$ : $f''(2)=12+2f'(1).$

From $f'(x)=3x^2+2x f'(1)+f''(2),$ put $x=1$ : $f'(1)=3+2f'(1)+f''(2).$

Substitute $f''(2)=12+2f'(1)$ : $f'(1)=3+2f'(1)+12+2f'(1)$ $f'(1)=15+4f'(1)$ $-3f'(1)=15$ $f'(1)=-5.$

Then $f''(2)=12+2(-5)=2.$

Also, $f'''(3)=6.$

Now compute $f(2)$ using the original equation: $f(2)=2^3+2^2f'(1)+2f''(2)+f'''(3).$

Substitute the values: $f(2)=8+4(-5)+2(2)+6$ $f(2)=8-20+4+6$ $f(2)=-2.$

Therefore, the correct option is $\boxed{-2}$ which is option $\boxed{\text{B}}$ .

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