Visual SolutionPYQ 2024 · Apr Shift 2Standard

Question

If the value of the integral

\int_{-1}^{1} \frac{\cos \alpha x}{1+3^x} \, dx

\frac{2}{\pi}

, then a value of

\alpha

is:

(A)

\frac{\pi}{3}

(B)

\frac{\pi}{6}

(C)

\frac{\pi}{4}

(D)

\frac{\pi}{2}

Solution Path

King's property removes

3^x

. Integral becomes

\sin(\alpha)/\alpha = 2/\pi

. Solution:

\alpha = \pi/2

01Question Setup

1/4

Find

\alpha

such that

\int_{-1}^{1} \frac{\cos \alpha x}{1+3^x}\,dx = \frac{2}{\pi}

\int_{-1}^{1} \frac{\cos \alpha x}{1+3^x}\,dx = \frac{2}{\pi}

02King's Property

2/4

Replace

x

with

-x

: the

3^x

in the denominator flips to

3^{-x}

. Adding the original and transformed integrals cancels the

3^x

term.

I + I' = \int_{-1}^{1} \cos(\alpha x)\,dx

03Sinc FunctionKEY INSIGHT

3/4

Since

2I = \int_{-1}^{1} \cos(\alpha x)\,dx = \frac{2\sin\alpha}{\alpha}

, we get

I = \frac{\sin\alpha}{\alpha}

. Set equal to

\frac{2}{\pi}

I = \frac{\sin \alpha}{\alpha} = \frac{2}{\pi}

04Final Answer

4/4

Verify:

\sin(\pi/2)/(\pi/2) = 1/(\pi/2) = 2/\pi

. Confirmed.

\boxed{\alpha = \frac{\pi}{2}}

Concepts from this question1 concepts unlocked

★

EASY

For definite integrals, replace x with (a+b-x) to exploit symmetry

\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx

If f(a+b-x) = -f(x), the integral is zero. If f(a+b-x) = f(x), the integral doubles

Symmetry integralsTrigonometric integralsPolynomial integrals

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