Let f: [- 2 , 2 ] R be a continuous function such that f(0)=1 and… - Vidyadip

MCQ

Let $f:\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow R$ be a continuous function such that

$f(0)=1 \text { and } \int_0^{\frac{\pi}{3}} f( t ) dt =0$

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

$(A)$ The equation $f( x )-3 \cos 3 x =0$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$

$(B)$ The equation $f( x )-3 \sin 3 x =-\frac{6}{\pi}$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$

$(C)$ $\lim _{x \rightarrow 0} \frac{x \int_0^x f(t) d t}{1- e ^{x^2}}=-1$

$(D)$ $\lim _{ x \rightarrow 0} \frac{\sin x \int_0^{ x } f( t ) dt }{ x ^2}=-1$

✓
$A,B,C$
B
$A,B,D$
C
$A,B$
D
$A,C$

✓

Answer

Correct option: A.

$A,B,C$

a
$(A)$ Let $g(x)=f(x)-3 \cos 3 x$

Now $\int_0^{\pi / 3} g(x) d x=\int_0^{\pi / 3} f(x) d x-3 \int_0^{\pi / 3} \cos 3 x d x=0$

Hence $g ( x )=0$ has a root in $\left(0, \frac{\pi}{3}\right)$

$(B)$ Let $h ( x )=f( x )-3 \sin 3 x +\frac{6}{\pi}$

Now $\int_0^{\pi / 3} h(x) d x=\int_0^{\pi / 3} f(x) d x-3 \int_0^{\pi / 3} \sin 3 x d x+\int_0^{\pi / 3} \frac{6}{\pi} d x$ $=0-2+2=0$

Hence $h(x)=0$ has a root in $\left(0, \frac{\pi}{3}\right)$

$(C)$ $\lim _{x \rightarrow 0} \frac{x \int_0^x f(t) d t}{1- e ^{x^2}}=\lim _{x \rightarrow 0} \underbrace{\left.\frac{x^2}{1-e^{x^2}}\right)}_{-1} \underbrace{\frac{\int_0^x f(t) d t}{x}}_{\text {Apply }}$

$=-1 \lim _{x \rightarrow 0} \frac{f( x )}{1}=-1$

$(D)$ $\lim _{ x \rightarrow 0} \frac{(\sin x ) \int_0 f( t ) dt }{ x ^2}$

$=\lim _{x \rightarrow 0} \underbrace{\left(\frac{\sin x}{x}\right)}_1\underbrace{\frac{\int_0^x f(t) d t}{x}}_{\text {Apply }}$

$=1 \lim _{x \rightarrow 0} \frac{f(x)}{1}=1$

Ans. $A,B,C$

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