Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
The function $f:[0,1] \to [0,1]$ is continuous and has the property $f(f(x)) = 1-x$ for all $x \in [0,1]$ and $\alpha = \int_0^1 f(x)dx$, then:
f\left(\frac{1}{4}\right) + f\left(\frac{3}{4}\right) = 1
the value of $\alpha$ equals to $\frac{1}{2}$
f\left(\frac{1}{3}\right)f\left(\frac{2}{3}\right) = 1
$\int_0^{\pi/2} \frac{\sin x dx}{(\sin x + \cos x)^2}$ has the same value as $\alpha$
Step-by-Step Solution
Key Concept: The product of exponential functions with exponents containing $x$ and $x^2$ is solved via substitution of $5^{x^2}$.
The determinant of matrix $A$ is $\det(A) = 5^x \cdot 5^{5x} \cdot 5^{5x^2}$. We compute the integral $I = \int 5^x \cdot 5^{5x} \cdot 5^{5x^2} dx$. Substituting $5^{x^2} = t$ gives $5^{x^2} \cdot 2x\ln(5) dx = dt$. The integral becomes $I = \int \frac{dt}{(\ln 5)^3} = \frac{5^{x^2}}{(\ln 5)^3} + c$.
Correct Answer: 1,2,4