JEE 2025 — Mathematics PYQ

$$f(x)=7{\tan }^{4}x+7{\tan }^{2}x-3{\tan }^{4}x-3{\tan }^{2}x$$
$$=7{\tan }^{6}x({\tan }^{2}x+1)-3{\tan }^{2}x({\tan }^{2}x+1)$$
$$={\sec }^{2}x(7{\tan }^{6}x-3{\tan }^{2}x)$$
$$I_1={\int }_0^{\frac{\pi }{4}}f(x)dx$$
$$={\int }_0^{\frac{\pi }{4}}{\sec }^{2}x(7{\tan }^{6}x-3{\tan }^{2}x)dx$$

Let, tanx = t ⇒ sec²x dx = dt

$$\begin{array}{rl}I& amp;={\int }_0^1(7t^4-3t^2)dt\\ & amp;={\left[\frac{7t^5}{5}-\frac{3t^3}{3}\right]}_0^1\\ & amp;=0\\ I_1& amp;={\int }_0^{\frac{\pi }{2}}f(x)-\frac{2}{\pi }{\int }_0^{\frac{\pi }{2}}f(x)dxdx\\ & amp;=0-\frac{2}{\pi }{\int }_0^{\frac{\pi }{2}}{\tan }^{4}x({\tan }^{4}x-1)dx\\ & amp;=-\frac{2}{\pi }{\int }_0^{\frac{\pi }{2}}{\tan }^{4}x({\tan }^{4}x-1){\sec }^{2}xdx\\ & amp;=-\frac{2}{\pi }{\int }_0^{\frac{\pi }{2}}{\tan }^{4}x({\tan }^{4}x-1){\sec }^{2}xdx\\ & amp;=\pi \\ I_2& amp;={\int }_0^1{\tan }^{4}xdxdt={\int }_0^1(t^4-t^2)dt\\ & amp;={\left[\frac{t^5}{5}-\frac{t^3}{3}\right]}_0^1\\ & amp;=\frac{1}{5}-\frac{1}{6}\\ & amp;=\frac{1}{30}\\ & amp;\therefore 7_1+12I_2=1\end{array}$$