JEE MAIN - Mathematics (2022 - 24th June Morning Shift - No. 19)
Let $$f(\theta ) = \sin \theta + \int\limits_{ - \pi /2}^{\pi /2} {(\sin \theta + t\cos \theta )f(t)dt} $$. Then the value of $$\left| {\int_0^{\pi /2} {f(\theta )d\theta } } \right|$$ is _____________.
Answer
1
Explanation
$f(\theta)=\sin \theta\left(1+\int_{-\pi / 2}^{\pi / 2} f(t) d t\right)+\cos \theta\left(\int_{-\pi / 2}^{\pi / 2} t f(t) d t\right)$
Clearly $f(\theta)=a \sin \theta+b \cos \theta$
Where $a=1+\int_{-\pi / 2}^{\pi / 2}(a \sin t+b \cos t) d t \Rightarrow a=1+2 b\quad\quad...(i)$
and $b=\int_{-\pi / 2}^{\pi / 2}(a t \sin t+b t \cos t) d t \Rightarrow b=2 a\quad\quad...(ii)$
from (i) and (ii) we get
$$ a=-\frac{1}{3} \text { and } b=-\frac{2}{3} $$
So $f(\theta)=-\frac{1}{3}(\sin \theta+2 \cos \theta)$
$$ \Rightarrow\left|\int_{0}^{\pi / 2} f(\theta) d \theta\right|=\frac{1}{3}(1+2 \times 1)=1 $$
Clearly $f(\theta)=a \sin \theta+b \cos \theta$
Where $a=1+\int_{-\pi / 2}^{\pi / 2}(a \sin t+b \cos t) d t \Rightarrow a=1+2 b\quad\quad...(i)$
and $b=\int_{-\pi / 2}^{\pi / 2}(a t \sin t+b t \cos t) d t \Rightarrow b=2 a\quad\quad...(ii)$
from (i) and (ii) we get
$$ a=-\frac{1}{3} \text { and } b=-\frac{2}{3} $$
So $f(\theta)=-\frac{1}{3}(\sin \theta+2 \cos \theta)$
$$ \Rightarrow\left|\int_{0}^{\pi / 2} f(\theta) d \theta\right|=\frac{1}{3}(1+2 \times 1)=1 $$
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