Exams › JEE Main › Maths

Let f be a positive function. If I1 = ∫[1−k to k] x f(x(1−x)) dx and I2 = ∫[1−k to k] f(x(1−x)) dx, where 2k − 1 > 0, then the ratio I1 : I2 is

1. 2
2. k
3. 1/2
4. 1

Correct answer: k

Solution

The ratio I1 : I2 equals k because I1 incorporates the additional factor of x in the integrand, which, when evaluated over the symmetric interval [1-k, k], results in a scaling of the integral by the factor of k, reflecting the average value of x in that range.

Related JEE Main Maths questions

• Suppose a function $f$ satisfies $f'(x)=-f(x)$. Define $g(x)=f^x$, and let \[ F(x)=\left(\int f\left(\frac{x}{2}\right)\right)^2+\left(\int g\left(\frac{x}{2}\right)\right)^2. \] If $F(5)=5$, what is the value of $F(10)$?
• Let $f(x)=\int_{2x}^{\sin x} \cos(t^3)\,dt$. Then the value of $f''(x)$ is
• If $\int_0^{a} f(2a-x)\,dx=m$ and $\int_0^{2a} f(x)\,dx=n$, then the value of $\int_0^{2a} f(x)\,dx$ is
• Evaluate the expression \[ \sum_{n=1}^{10}\int_{-2n}^{-n}\sin(27x)\,dx+\sum_{n=1}^{10}\int_{2n-1}^{2n}\sin(27x)\,dx. \]
• Evaluate the integral \[ \int \frac{\tan(\ln x)\,\tan\!\left(\ln\frac{x}{2}\right)\,\tan(\ln 2)}{x}\,dx. \]
• Evaluate the integral \[ \int \frac{\sqrt{5+x^2}}{x^4}\,dx. \]

⚔️ Practice JEE Main Maths free + battle 1v1 →