If {f}(x)=3 x²+2 x {f}^{}(1)+{f}^{ }(2), then {f}(x)= .

Correct Answer is : 3x2−12x+63 x²-12 x+63x2−12x+6

Correct Answer is : 3x2−12x+63 x²-12 x+63x2−12x+6

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MHTCET 26 Apr 2025 Shift 2 Paper

Question : 149 of 150

Marks: +1, -0

\mathrm{f}(x)=3 x^{2}+2 x \mathrm{f}^{\prime}(1)+\mathrm{f}^{\prime \prime}(2)

\mathrm{f}(x)=\ldots \ldots

Solution:

Concept:

Differentiate the given function and substitute

x=1

to find the constant coefficients.

Explanation:

Given:

\mathrm{f}(x)=3x^{2}+2x\,\mathrm{f}'(1)+\mathrm{f}''(2)

Here,

\mathrm{f}'(1)

and

\mathrm{f}''(2)

are constants.

Differentiate both sides with respect to

x

\mathrm{f}'(x)=6x+2\mathrm{f}'(1)

Differentiate again:

\mathrm{f}''(x)=6

Now put

x=1

in the first derivative equation:

\mathrm{f}'(1)=6(1)+2\mathrm{f}'(1)

\Rightarrow \mathrm{f}'(1)=6+2\mathrm{f}'(1)

\Rightarrow \mathrm{f}'(1)=-6

Also,

\mathrm{f}''(2)=6

Substitute these values into the original equation:

\mathrm{f}(x)=3x^{2}+2x(-6)+6

\mathrm{f}(x)=3x^{2}-12x+6

Answer:

\mathrm{f}(x)=3x^{2}-12x+6

Hence, the correct option is C.

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