Calculus – Derivatives – JEE Main 21 Jan 2026 Shift 2

Solution:

Differentiate $f(x)$ with respect to $x$:
$$f^{\prime}(x)=3x^{2}+2xf^{\prime}(1)+2f^{\prime\prime}(2)$$

Differentiate again with respect to $x$:
$$f^{\prime\prime}(x)=6x+2f^{\prime}(1)$$

Put $x=2$ in $f^{\prime\prime}(x)$:
$$f^{\prime\prime}(2)=12+2f^{\prime}(1)$$

Substitute the value of $f^{\prime\prime}(2)$ into the equation for $f^{\prime}(x)$:
$$f^{\prime}(x)=3x^{2}+2xf^{\prime}(1)+2(12+2f^{\prime}(1))$$

$$f^{\prime}(x)=3x^{2}+2xf^{\prime}(1)+24+4f^{\prime}(1)$$

Put $x=1$ in the above equation to find $f^{\prime}(1)$:
$$f^{\prime}(1)=3(1)^{2}+2(1)f^{\prime}(1)+24+4f^{\prime}(1)$$

$$f^{\prime}(1)=3+6f^{\prime}(1)+24$$

$$f^{\prime}(1)-6f^{\prime}(1)=27$$

$$-5f^{\prime}(1)=27 \Rightarrow f^{\prime}(1)=-\frac{27}{5}$$

Now substitute $f^{\prime}(1)$ back into the simplified expression for $f^{\prime}(x)$:
$$f^{\prime}(x)=3x^{2}+2x\left(-\frac{27}{5}\right)+24+4\left(-\frac{27}{5}\right)$$

$$f^{\prime}(x)=3x^{2}-\frac{54}{5}x+24-\frac{108}{5}$$

$$f^{\prime}(x)=3x^{2}-\frac{54}{5}x+\frac{120-108}{5}$$

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

Calculate $f^{\prime}(5)$:
$$f^{\prime}(5)=3(5)^{2}-\frac{54}{5}(5)+\frac{12}{5}$$

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

$$f^{\prime}(5)=21+\frac{12}{5}=\frac{105+12}{5}=\frac{117}{5}$$

Ans. (4)