QJEEAdv21Ma7536: JEE Advanced 2021 Mathematics - Class 12 JEEAdv21Ma7536 Solution: Let be a continuous function such that and . Then which of the…

Question

Verified Solution • 2025–26 Session moderate

Let

f:\left[ { - {\pi \over 2},{\pi \over 2}} \right] \to R

be a continuous function such that

f(0) = 1

and

\int_0^{{\pi \over 3}} {f(t)dt = 0}

. Then which of the following statements is(are) TRUE?

A The equation $f(x) - 3\cos 3x = 0$ has at least one solution in $\left( {0,{\pi \over 3}} \right)$
B The equation $f(x) - 3\sin 3x = - {6 \over \pi }$ has at least one solution in $\left( {0,{\pi \over 3}} \right)$
C $\mathop {\lim }\limits_{x \to 0} {{x\int_0^x {f(t)dt} } \over {1 - {e^{{x^2}}}}} = - 1$
D $\mathop {\lim }\limits_{x \to 0} {{\sin x\int_0^x {f(t)dt} } \over {{x^2}}} = - 1$

Accepted Answer

Given, f(0) = 1 and

\int_0^{\pi /3} {f(t)\,dt = 0}

For option (a)

Consider a function

g(x) = \int_0^\pi {f(t)\,dt - \sin 3x}

g(x) is continuous and differentiable function and g(0) = g(

\pi

/3) = 0

\therefore

By Rolle's theorem, g'(x) = 0 has at least one solution in (0,

\pi

/3).

i.e. g'(x) = f(x)

\times

1

-

3cos 3x = 0 for some

x \in \left( {0,{\pi \over 3}} \right)

For option (b)

Consider the function

\phi (x) = \int_0^x {f(t)dt + \cos 3x + {6 \over \pi }x}

\phi

(x) is continuous and differentiable function as well as

\phi

(0) =

\phi

(

\pi

/3) = 1

Hence, by Rolle's theorem,

\phi

'(x) = 0 has at least one solution in (0,

\pi

/ 3).

i.e.

\phi

'(x) = f(x)

\times

1

-

3sin 3x +

{6 \over \pi } = 0

for some

x \in \left( {0,{\pi \over 3}} \right)

.

For option (c)

Let

L = \mathop {\lim }\limits_{x \to 0} {{x\int_0^x {f(t)dt} } \over {1 - {e^{{x^2}}}}}

(form

{0 \over 0}

)

\Rightarrow L = \mathop {\lim }\limits_{x \to 0} {{xf(x) + \int_0^x {f(t)dt} } \over { - 2x{e^{{x^2}}}}}

(Using L-Hospital Rule)

Again, using L'-Hospital Rule (

\therefore

form

{0 \over 0}

)

L = \mathop {\lim }\limits_{x \to 0} {{xf'(x) + f(x) + f(x)} \over { - 4{x^2}{e^{{x^2}}} - 2{e^{{x^2}}}}} = {{0 + 2f(0)} \over { - 0 - 2}} = - 1

(

\because

f(0) = 1)

For option (d)

Let

P = \mathop {\lim }\limits_{x \to 0} {{\sin x.\int_0^x {f(t)dt} } \over {{x^2}}}

(form

{0 \over 0}

)

Applying L-Hospital Rule,

P = \mathop {\lim }\limits_{x \to 0} {{\sin x.f(x) + \cos x.\int_0^x {f(t)dt} } \over {2x}}

(form

{0 \over 0}

)

Again using L-Hospital Rule,

P = \mathop {\lim }\limits_{x \to 0} {{[\cos x.f(x) + \sin x.f'(x) + \cos x.f(x) - \sin x\int_0^x {f(t)dt]} } \over 2}

\Rightarrow P = {{1 \times f(0) + 0 \times f'(0) + 1 \times f(0) - 0 \times 0} \over 2}

\Rightarrow P = {{1 + 0 + 1 - 0} \over 2} = 1

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