Let f: \mathbbR arrow \mathbbR be a function defined by f(x)=\frac4^x4^x+2 and M=\int_f(a)^f(1-a) x \sin ^4(x(1-x)) d x, N=\int_f(a)^f(1-a) \sin ^4(x(1-x)) d x ; a eq \frac12 . If \alpha M=\beta N, \alpha, \beta \in \mathbbN , then the least value of \alpha^2+\beta^2 is equal to \underline\hspace2cm .

Definite Integration PYQ — JEE Main 2024 · 31 Jan · Shift 1 · Q60 with Solution

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Correct answer: 5

Given integrals

We have $f(x)=\frac{4^x}{4^x+2}$ and $M=\int_{f(a)}^{f(1-a)} x\sin^4(x(1-x))\,dx,$ $N=\int_{f(a)}^{f(1-a)} \sin^4(x(1-x))\,dx,$ with $a\ne \frac12$ .

We need natural numbers $\alpha,\beta$ such that $\alpha M=\beta N,$ and then find the least value of $\alpha^2+\beta^2.$

Find the limits and their relation

Let $u=f(a)=\frac{4^a}{4^a+2}.$ Now compute $f(1-a)$ : $f(1-a)=\frac{4^{1-a}}{4^{1-a}+2}.$ Let $t=4^a$ . Then $4^{1-a}=\frac{4}{t}$ . So $f(1-a)=\frac{4/t}{4/t+2}= rac{4}{4+2t}= rac{2}{t+2}.$ Also, $f(a)=\frac{t}{t+2}.$ Hence $f(a)+f(1-a)=\frac{t}{t+2}+\frac{2}{t+2}=1.$ So if the lower limit is $u=f(a)$ , then the upper limit is $f(1-a)=1-u.$

Thus, $M=\int_u^{1-u} x\sin^4(x(1-x))\,dx,$ $N=\int_u^{1-u} \sin^4(x(1-x))\,dx.$

Use symmetry of the integrand

Define $g(x)=\sin^4(x(1-x)).$ Then $g(1-x)=\sin^4((1-x)x)=\sin^4(x(1-x))=g(x).$ So $g(x)$ is symmetric about $x=\frac12$ .

Now consider $M=\int_u^{1-u} xg(x)\,dx.$ Using the substitution $x\mapsto 1-x$ , $M=\int_u^{1-u} (1-x)g(x)\,dx.$ Adding the two expressions,

=\int_u^{1-u} g(x)\,dx=N.$$ Therefore, $$M=\frac N2.$$ --- 4. **Relate $\alpha$ and $\beta$** Given $$\alpha M=\beta N,$$ and using $M=\frac N2$, $$\alpha\cdot \frac N2=\beta N.$$ Since $a\ne \frac12$, the limits are distinct, and $g(x)=\sin^4(x(1-x))\ge 0$, so the integral $N\ne 0$. Hence we can cancel $N$: $$\frac{\alpha}{2}=\beta$$ which gives $$\alpha=2\beta.$$ For natural numbers, the least choice is $$\beta=1,\quad \alpha=2.$$ Therefore, $$\alpha^2+\beta^2=2^2+1^2=4+1=5.$$ --- 5. **Final answer** $$\boxed{5}$$

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View written solutionFree

Correct answer: 5

Given integrals

We have $f(x)=\frac{4^x}{4^x+2}$ and $M=\int_{f(a)}^{f(1-a)} x\sin^4(x(1-x))\,dx,$ $N=\int_{f(a)}^{f(1-a)} \sin^4(x(1-x))\,dx,$ with $a\ne \frac12$ .

We need natural numbers $\alpha,\beta$ such that $\alpha M=\beta N,$ and then find the least value of $\alpha^2+\beta^2.$

Find the limits and their relation

Thus, $M=\int_u^{1-u} x\sin^4(x(1-x))\,dx,$ $N=\int_u^{1-u} \sin^4(x(1-x))\,dx.$

Use symmetry of the integrand

Define $g(x)=\sin^4(x(1-x)).$ Then $g(1-x)=\sin^4((1-x)x)=\sin^4(x(1-x))=g(x).$ So $g(x)$ is symmetric about $x=\frac12$ .

Now consider $M=\int_u^{1-u} xg(x)\,dx.$ Using the substitution $x\mapsto 1-x$ , $M=\int_u^{1-u} (1-x)g(x)\,dx.$ Adding the two expressions,

=\int_u^{1-u} g(x)\,dx=N.$$ Therefore, $$M=\frac N2.$$ --- 4. **Relate $\alpha$ and $\beta$** Given $$\alpha M=\beta N,$$ and using $M=\frac N2$, $$\alpha\cdot \frac N2=\beta N.$$ Since $a\ne \frac12$, the limits are distinct, and $g(x)=\sin^4(x(1-x))\ge 0$, so the integral $N\ne 0$. Hence we can cancel $N$: $$\frac{\alpha}{2}=\beta$$ which gives $$\alpha=2\beta.$$ For natural numbers, the least choice is $$\beta=1,\quad \alpha=2.$$ Therefore, $$\alpha^2+\beta^2=2^2+1^2=4+1=5.$$ --- 5. **Final answer** $$\boxed{5}$$

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