JEE MAIN - Mathematics (2021 - 18th March Evening Shift - No. 8)
Let g(x) = $$\int_0^x {f(t)dt} $$, where f is continuous function in [ 0, 3 ] such that $${1 \over 3}$$ $$ \le $$ f(t) $$ \le $$ 1 for all t$$\in$$ [0, 1] and 0 $$ \le $$ f(t) $$ \le $$ $${1 \over 2}$$ for all t$$\in$$ (1, 3]. The largest possible interval in which g(3) lies is :
$$\left[ { - 1, - {1 \over 2}} \right]$$
$$\left[ { - {3 \over 2}, - 1} \right]$$
[1, 3]
$$\left[ {{1 \over 3},2} \right]$$
Explanation
Given, $g(x)=\int_0^x f(t) d t$
$\therefore g(3)=\int_0^3 f(t) d t=\int_0^1 f(t) d t+\int_1^3 f(t) d t$
$\Rightarrow \int_0^1 \frac{1}{3} d t+\int_1^3 0 \cdot d t \leq g(3) \leq \int_0^1 1 d t+\int_1^3 \frac{1}{2} d t$
$\Rightarrow \frac{1}{3} \leq g(3) \leq 1+1$
$\Rightarrow \frac{1}{3} \leq g(3) \leq 2$
$\therefore g(3)=\int_0^3 f(t) d t=\int_0^1 f(t) d t+\int_1^3 f(t) d t$
$\Rightarrow \int_0^1 \frac{1}{3} d t+\int_1^3 0 \cdot d t \leq g(3) \leq \int_0^1 1 d t+\int_1^3 \frac{1}{2} d t$
$\Rightarrow \frac{1}{3} \leq g(3) \leq 1+1$
$\Rightarrow \frac{1}{3} \leq g(3) \leq 2$
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