Let f:≤ft[-π/2,π/2] arrow {R} be a continuous function such that f(0)=1 \; {and}\; ₀^{π/3} f(t) \, dt = 0 Then which of the following statements is (are) TRUE? [JEE Adv 2021 P2]

The equation f(x)-3 3x=0 has at least one solution in ≤ft(0,π/3)

The equation f(x)-3 3x = -6/π has at least one solution in ≤ft(0,π/3)

ₓ → ∞} {x ₀^{x} f(t) \, dt}{1-e^{x²} = -1 `

ₓ → ∞} { x ₀^{x} f(t) \, dt}{x² = -1

Test Index

Definite Integration

Section: Mathematics

Question : 11 of 46

Marks: +1, -0

Let

f:\left[-\frac{\pi}{2},\frac{\pi}{2}\right] \rightarrow \mathbb{R}

f(0)=1 \; \text{and}\; \int\limits_{0}^{\frac{\pi}{3}} f(t) \, dt = 0

The equation $f(x)-3\cos 3x=0$ has at least one solution in $\left(0,\frac{\pi}{3}\right)$
The equation $f(x)-3\sin 3x = -\frac{6}{\pi}$ has at least one solution in $\left(0,\frac{\pi}{3}\right)$
$\lim\limits_{x \to \infty} \frac{x \int\limits_{0}^{x} f(t) \, dt}{1-e^{x^2}} = -1$ `
$\lim\limits_{x \to \infty} \frac{\sin x \int\limits_{0}^{x} f(t) \, dt}{x^2} = -1$

Solution:

(a) Let

g(x)=\int\limits_{0}^{\frac{\pi}{3}} \int (x) \, dx - 3 \int\limits_{0}^{\frac{\pi}{3}} \cos 3x \, dx = 0

g(x)

is continuous and differentiable function and

g(0)=0

g\left(\frac{\pi}{3}\right)=0

By Rolle's theorem

g'(x)=f(x)-3\cos x=0

has least one solution in

\left(0,\frac{\pi}{3}\right)

g'(x)=f(x)-3\cos x=0

has least one solution in

\left(0,\frac{\pi}{3}\right)

(b) Let

h(x)=f(x)-3\sin 3x + \frac{6}{x}

Let

h(x)=\int\limits_{0}^{\frac{\pi}{3}} f(x) \, dx - 3 \int\limits_{0}^{\frac{\pi}{3}} \sin 3x \, dx + \int\limits_{0}^{\frac{\pi}{3}} \frac{6}{\pi} \, dx

=0-2+2=0

h(x)

is continuous and differentiable function and

h(0)=0

and

h\left(\frac{\pi}{3}\right)=0

ByRolle's theorem

h'(x)=f(x)-3\sin 3x + \frac{6}{n} = 0

has least one solution in

\left(0,\frac{\pi}{3}\right)

(c)

\lim\limits_{x \to 0} \frac{x \int\limits_{0}^{x} f(t) \, dt}{1-e^{x^2}} = \lim\limits_{x \to 0} \left( \frac{x^2}{1-e^{x^2}} \right) \frac{\int\limits_{0}^{x} f(t) \, dt}{x}

By L'Hospital's Rule

=-1 \lim\limits_{x \to 0} \frac{f(x)}{1} = -1 \quad [\because f(0)=1]

(d)

\lim\limits_{x \to 0} \frac{(\sin x) \int\limits_{0}^{x} f(t) \, dt}{x^2} = \lim\limits_{x \to 0} \left( \frac{\sin x}{x} \right) \frac{\int\limits_{0}^{x} f(t) \, dt}{x}

By L'Hospital's Rule

=1 \lim\limits_{x \to 0} \frac{f(x)}{1} = 1 \quad [\because f(0)=0]

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