JEE Main 2026 Jan 21 Shift 2, Mathematics Q3: Some Basic Applications Of Derivatives

Treat $f^{\prime }(1)$, $f^{\prime \prime }(2)$, $f^{\prime \prime \prime }(3)$ as constants. Differentiate:

$f^{\prime }(x)=3x^2+2x{f}^{\prime }(1)+2f^{\prime \prime }(2)$,   $f^{\prime \prime }(x)=6x+2f^{\prime }(1)$,   $f^{\prime \prime \prime }(x)=6$.

Pin down the constants:

From $f^{\prime \prime }(x)$ at $x=2$: $f^{\prime \prime }(2)=12+2f^{\prime }(1)$.

From $f^{\prime }(x)$ at $x=1$: $f^{\prime }(1)=3+2f^{\prime }(1)+2f^{\prime \prime }(2)=3+2f^{\prime }(1)+2(12+2f^{\prime }(1))=27+6f^{\prime }(1)$. So $-5f^{\prime }(1)=27\Rightarrow f^{\prime }(1)=-\frac{27}{5}$, hence $f^{\prime \prime }(2)=12-\frac{54}{5}=\frac{6}{5}$.

Thus $f^{\prime }(x)=3x^2-\frac{54}{5}x+\frac{12}{5}$.

$f^{\prime }(5)=75-54+\frac{12}{5}=21+\frac{12}{5}=\frac{117}{5}$.