Let f: R R be a function defined by f(x) = 4x4x + 2 and M = f(1-a)^f(a) x ^4(x(1 - x)) dx, N = f(1-a)^f(a) ^4(x(1 - x)) dx; a 12. If M = N,, N, then the least value of ^2 + ^2 is equal to:

Given: f(x) = 4x4x+2, M = f(1-a)^f(a) x ^4(x(1-x)) dx and N = f(1-a)^f(a) ^4(x(1-x)) dx. Find: The least value of ^2+ ^2 if M = N and, N. From the given function, the solution uses the identity f(a) + f(1-a) = 1. Hence the limits of integration are complementary about 12. Using the symmetry of ^4(x(1-x)) under the transformation x 1-x, the solution gives M = N - M. Therefore, 2M = N. So, 2M = N 2M = 1 N. Comparing with M = N, we get = 2, = 1. Now, ^2 + ^2 = 2^2 + 1^2 = 4 + 1 = 5. Therefore, the least value of ^2 + ^2 is 5, so the correct option is A.

Let f: R R be a function defined by f(x) = 4x4x + 2 and M = f(1-a… — Solution | JEE Mathematics

MCQMediumJEE 2024Properties of Definite Integrals

Let $f: \mathbb{R} \to \mathbb{R}$ be a function defined by $f(x) = \frac{4x}{4x + 2}$ and $M = \int_{f(1-a)}^{f(a)} x \sin^4(x(1 - x)) \, dx, \; N = \int_{f(1-a)}^{f(a)} \sin^4(x(1 - x)) \, dx; \; a \ne \frac{1}{2}.$ If $\alpha M = \beta N, \; \alpha, \beta \in \mathbb{N},$ then the least value of $\alpha^2 + \beta^2$ is equal to:

A
$5$
B
$10$
C
$8$
D
$7$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: $f(x) = \frac{4x}{4x+2}$ ,

M = \int_{f(1-a)}^{f(a)} x\sin^4(x(1-x)) \, dx

and

N = \int_{f(1-a)}^{f(a)} \sin^4(x(1-x)) \, dx.

Find: The least value of $\alpha^2+\beta^2$ if $\alpha M = \beta N$ and $\alpha,\beta \in \mathbb{N}$ .

From the given function, the solution uses the identity

f(a) + f(1-a) = 1.

Hence the limits of integration are complementary about $\frac{1}{2}$ .

Using the symmetry of $\sin^4(x(1-x))$ under the transformation $x \mapsto 1-x$ , the solution gives

M = N - M.

Therefore,

2M = N.

So,

2M = N \implies 2M = 1\cdot N.

Comparing with $\alpha M = \beta N$ , we get

\alpha = 2, \quad \beta = 1.

Now,

\alpha^2 + \beta^2 = 2^2 + 1^2 = 4 + 1 = 5.

Therefore, the least value of $\alpha^2 + \beta^2$ is $5$ , so the correct option is A.

Common mistakes

Assuming that $M$ and $N$ are equal because both integrals have the same limits. This is wrong because the integrand of $M$ contains the extra factor $x$ . Use the symmetry $x \mapsto 1-x$ carefully to relate $M$ to $N-M$ , not directly to $N$ .
Not using the identity $f(a)+f(1-a)=1$ . This is wrong because the symmetry argument depends on complementary limits. First establish this relation from the given function, then apply the substitution.
Taking $\alpha=1, \beta=2$ from $2M=N$ . This reverses the comparison. Match $2M=N$ with $\alpha M=\beta N$ correctly to get $\alpha=2$ and $\beta=1$ .

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