If {f}(x)=3 x³+2 x² {f}^{}(1)+x {f}^{ }(2)+{f}^{ }(3) then {f}(x)= \_\_\_\_.

Correct Answer is : 17(21x3−90x2+72x+126)1/7≤ft(21 x³-90 x²+72 x+126)71(21x3−90x2+72x+126)

1/7≤ft(21 x³-90 x²+72 x+126)

Correct Answer is : 17(21x3−90x2+72x+126)1/7≤ft(21 x³-90 x²+72 x+126)71(21x3−90x2+72x+126)

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MHTCET 23 Apr 2025 Shift 2 Paper

Question : 143 of 150

Marks: +1, -0

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

\mathrm{f}(x)=

\_\_\_\_

Solution:

Concept:

The values

\mathrm{f}^{\prime}(1)

\mathrm{f}^{\prime \prime}(2)

, and

\mathrm{f}^{\prime \prime \prime}(3)

are constants, so differentiate the given expression and solve for these constants.

Explanation:

Let

a=\mathrm{f}^{\prime}(1)

b=\mathrm{f}^{\prime \prime}(2)

, and

c=\mathrm{f}^{\prime \prime \prime}(3)

Then

\mathrm{f}(x)=3x^3+2a x^2+b x+c

Differentiate step by step:

\mathrm{f}^{\prime}(x)=9x^2+4a x+b

\mathrm{f}^{\prime \prime}(x)=18x+4a

\mathrm{f}^{\prime \prime \prime}(x)=18

, hence

c=18

Put

x=1

\mathrm{f}^{\prime}(x)

a=9+4a+b \Rightarrow 3a+b=-9

Put

x=2

\mathrm{f}^{\prime \prime}(x)

b=36+4a

Solve the two equations:

3a+(36+4a)=-9 \Rightarrow 7a=-45 \Rightarrow a=-\frac{45}{7}

Then

b=36+4\left(-\frac{45}{7}\right)=\frac{72}{7}

Substitute

a

b

, and

c

into

\mathrm{f}(x)

\mathrm{f}(x)=3x^3+2\left(-\frac{45}{7}\right)x^2+\frac{72}{7}x+18

=\frac{1}{7}\left(21x^3-90x^2+72x+126\right)

Answer:

\mathrm{f}(x)=\frac{1}{7}\left(21x^3-90x^2+72x+126\right)

Hence, option B is correct.

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