Let f(x)=x 3 +x 2 f prime (1)+xf "(2) +f primeprimeprime (3), xin R Then f prime (10) is equal to

Mathematics · Methods of Differentiation

JEE Main 2024 — 27 January, Shift 1 — Question 26

Let $f(x)=x^{3}+x^{2}f^{\prime}(1)+xf$ "(2) $+f^{\prime\prime\prime}(3),$ $x\in R$ Then $\mathrm{f}^{\prime}(10)$ is equal to

Numerical answer — enter this value.

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

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

$f^{\prime \prime}(x)=6 x+2 f^{\prime}(1)$

$f^{\prime \prime \prime}(x)=6$

$\mathrm{f}^{\prime}(1)=-5, \mathrm{f}^{\prime \prime}(2)=2, \mathrm{f}^{\prime \prime \prime}(3)=6$

$f(x)=x^{3}+x^{2} \cdot(-5)+x \cdot(2)+6$

$\mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}-10 \mathrm{x}+2$

$\mathrm{f}^{\prime}(10)=300-100+2=202$

Answer key and solution verified before publishing.

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