Let f(x) be a positive function and I_1 = \int\limits_-\frac12^1 2x \, f(2x(1-2x)) \, dx and I_2 = \int\limits_-1^2 f(x(1-x)) \, dx . Then the value of \fracI_2I_1 is equal to \underline\hspace2cm

Definite Integration PYQ — JEE Main 2025 · 8 Apr · Shift 2 · Q40 with Solution

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Correct answer: D

Given integrals

We have $I_1=\int_{-\frac12}^{1} 2x\, f\big(2x(1-2x)\big)\,dx$ and $I_2=\int_{-1}^{2} f\big(x(1-x)\big)\,dx.$

We need to find $\frac{I_2}{I_1}.$

Evaluate $I_1$ by substitution

Let $u=2x(1-2x)=2x-4x^2.$ Then $\frac{du}{dx}=2-8x=2(1-4x).$ This does not directly match the factor $2x\,dx$ , so instead we rewrite using a more suitable substitution.

Let $t=2x \Rightarrow dx=\frac{dt}{2}.$ Then as $x$ goes from $-\frac12$ to $1$ , $t$ goes from $-1$ to $2$ .

So,

=\frac12 \int_{-1}^{2} t\, f\big(t(1-t)\big)\,dt.$$ Hence, $$I_1=\frac12\int_{-1}^{2} x\,f\big(x(1-x)\big)\,dx$$ (rename dummy variable back to $x$). --- 3. **Use symmetry in $I_2$** Now consider $$I_2=\int_{-1}^{2} f\big(x(1-x)\big)\,dx.$$ Use the substitution $$x=1-t \Rightarrow dx=-dt.$$ When $x=-1$, $t=2$; when $x=2$, $t=-1$. Thus, $$I_2=\int_{2}^{-1} f\big((1-t)t\big)(-dt)=\int_{-1}^{2} f\big(t(1-t)\big)\,dt,$$ which is the same integral, as expected. More importantly, define $$J=\int_{-1}^{2} x\, f\big(x(1-x)\big)\,dx.$$ Apply the substitution $x=1-t$: $$J=\int_{-1}^{2} (1-t)f\big(t(1-t)\big)\,dt.$$ So, $$J=\int_{-1}^{2} f\big(t(1-t)\big)\,dt - \int_{-1}^{2} t f\big(t(1-t)\big)\,dt.$$ That is, $$J=I_2-J.$$ Therefore, $$2J=I_2 \Rightarrow J=\frac{I_2}{2}.$$ --- 4. **Relate $I_1$ and $I_2$** From step 2, $$I_1=\frac12 J.$$ Using $J=\frac{I_2}{2}$, $$I_1=\frac12\cdot \frac{I_2}{2}=\frac{I_2}{4}.$$ Hence, $$\frac{I_2}{I_1}=4.$$ --- 5. **Check options** The correct option is $$\boxed{4}.$$ So **Option D** is correct.

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View written solutionFree

Correct answer: D

Given integrals

We have $I_1=\int_{-\frac12}^{1} 2x\, f\big(2x(1-2x)\big)\,dx$ and $I_2=\int_{-1}^{2} f\big(x(1-x)\big)\,dx.$

We need to find $\frac{I_2}{I_1}.$

Evaluate $I_1$ by substitution

Let $u=2x(1-2x)=2x-4x^2.$ Then $\frac{du}{dx}=2-8x=2(1-4x).$ This does not directly match the factor $2x\,dx$ , so instead we rewrite using a more suitable substitution.

Let $t=2x \Rightarrow dx=\frac{dt}{2}.$ Then as $x$ goes from $-\frac12$ to $1$ , $t$ goes from $-1$ to $2$ .

So,

=\frac12 \int_{-1}^{2} t\, f\big(t(1-t)\big)\,dt.$$ Hence, $$I_1=\frac12\int_{-1}^{2} x\,f\big(x(1-x)\big)\,dx$$ (rename dummy variable back to $x$). --- 3. **Use symmetry in $I_2$** Now consider $$I_2=\int_{-1}^{2} f\big(x(1-x)\big)\,dx.$$ Use the substitution $$x=1-t \Rightarrow dx=-dt.$$ When $x=-1$, $t=2$; when $x=2$, $t=-1$. Thus, $$I_2=\int_{2}^{-1} f\big((1-t)t\big)(-dt)=\int_{-1}^{2} f\big(t(1-t)\big)\,dt,$$ which is the same integral, as expected. More importantly, define $$J=\int_{-1}^{2} x\, f\big(x(1-x)\big)\,dx.$$ Apply the substitution $x=1-t$: $$J=\int_{-1}^{2} (1-t)f\big(t(1-t)\big)\,dt.$$ So, $$J=\int_{-1}^{2} f\big(t(1-t)\big)\,dt - \int_{-1}^{2} t f\big(t(1-t)\big)\,dt.$$ That is, $$J=I_2-J.$$ Therefore, $$2J=I_2 \Rightarrow J=\frac{I_2}{2}.$$ --- 4. **Relate $I_1$ and $I_2$** From step 2, $$I_1=\frac12 J.$$ Using $J=\frac{I_2}{2}$, $$I_1=\frac12\cdot \frac{I_2}{2}=\frac{I_2}{4}.$$ Hence, $$\frac{I_2}{I_1}=4.$$ --- 5. **Check options** The correct option is $$\boxed{4}.$$ So **Option D** is correct.

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