JEE Main 2026 — Mathematics Algebra

Given $f(x) + 3f\left(\dfrac{\pi}{2} - x\right) = \sin x$f(x)+3f(2π​−x)=sinx

Replacing $x$x with $\dfrac{\pi}{2} - x$2π​−x, we get:

$$f\left(\dfrac{\pi}{2} - x\right) + 3f(x) = \sin\left(\dfrac{\pi}{2} - x\right) = \cos x$$f(2π​−x)+3f(x)=sin(2π​−x)=cosx

Multiplying this equation by $3$3 and subtracting the first equation gives:

$$9f(x) - f(x) = 3\cos x - \sin x$$9f(x)−f(x)=3cosx−sinx

$$8f(x) = 3\cos x - \sin x$$8f(x)=3cosx−sinx

$$f(x) = \dfrac{3\cos x - \sin x}{8}$$f(x)=83cosx−sinx​

The maximum value of $f(x)$f(x) is $\alpha = \dfrac{\sqrt{3^2 + (-1)^2}}{8} = \dfrac{\sqrt{10}}{8}$α=832+(−1)2​​=810​​.

Thus, $\alpha^2 = \dfrac{10}{64} = \dfrac{5}{32}$α2=6410​=325​.

The points of intersection of the curves $g(x) = x^2$g(x)=x2 and $h(x) = \beta x^3$h(x)=βx3 are given by:

$$x^2 = \beta x^3 \Rightarrow x^2(1 - \beta x) = 0 \Rightarrow x = 0, x = \dfrac{1}{\beta}$$x2=βx3⇒x2(1−βx)=0⇒x=0,x=β1​

The area of the region bounded by the curves is:

$$\int_{0}^{1/\beta} (x^2 - \beta x^3) dx = \left[ \dfrac{x^3}{3} - \dfrac{\beta x^4}{4} \right]_{0}^{1/\beta}$$∫01/β​(x2−βx3)dx=[3x3​−4βx4​]01/β​

$$= \dfrac{1}{3\beta^3} - \dfrac{1}{4\beta^3} = \dfrac{1}{12\beta^3}$$=3β31​−4β31​=12β31​

Given that the area is $\alpha^2$α2, we have:

$$\dfrac{1}{12\beta^3} = \dfrac{5}{32}$$12β31​=325​

$$\beta^3 = \dfrac{32}{60} = \dfrac{8}{15}$$β3=6032​=158​

Therefore, $30\beta^3 = 30 \times \dfrac{8}{15} = 16$30β3=30×158​=16.

Answer: $16$16