JEE MAIN - Mathematics (2021 - 17th March Morning Shift - No. 18)
If [ . ] represents the greatest integer function, then the value of
$$\left| {\int\limits_0^{\sqrt {{\pi \over 2}} } {\left[ {[{x^2}] - \cos x} \right]dx} } \right|$$ is ____________.
$$\left| {\int\limits_0^{\sqrt {{\pi \over 2}} } {\left[ {[{x^2}] - \cos x} \right]dx} } \right|$$ is ____________.
Answer
1
Explanation
$$\int\limits_0^{\sqrt {{\pi \over 2}} } {\left[ {[{x^2}] - \cos x} \right]} dx$$
$$ = \int\limits_0^1 {[ - \cos x]dx} + \int\limits_1^{\sqrt {{\pi \over 2}} } {[1 - \cos x]dx} $$
= $$\int\limits_0^1 {\left[ { - \cos x} \right]} dx + \int\limits_1^{\sqrt {{\pi \over 2}} } {1dx} + \int\limits_1^{\sqrt {{\pi \over 2}} } {\left[ { - \cos x} \right]} dx$$
When 0 $$ \le $$ x $$ \le $$ 1 then -1 $$ \le $$ -cosx $$ \le $$ 0
$$ \therefore $$ $${\left[ { - \cos x} \right]}$$ = -1
When 1 $$ \le $$ x $$ \le $$ $${\sqrt {{\pi \over 2}} }$$ = 1.24 then -0.33 $$ \le $$ -cosx $$ \le $$ 0
$$ \therefore $$ $${\left[ { - \cos x} \right]}$$ = -1 (Integer value present in the left side of -0.33)
$$ = - \int\limits_0^1 {dx + \int\limits_1^{\sqrt {{\pi \over 2}} } {dx - \int\limits_1^{\sqrt {{\pi \over 2}} } {dx} } } $$
$$ = - (x)_0^1 = - 1$$
$$ \therefore $$ $$\left| {\int\limits_0^{\sqrt {{\pi \over 2}} } {\left[ {[{x^2}] - \cos x} \right]dx} } \right|$$ = 1
$$ = \int\limits_0^1 {[ - \cos x]dx} + \int\limits_1^{\sqrt {{\pi \over 2}} } {[1 - \cos x]dx} $$
= $$\int\limits_0^1 {\left[ { - \cos x} \right]} dx + \int\limits_1^{\sqrt {{\pi \over 2}} } {1dx} + \int\limits_1^{\sqrt {{\pi \over 2}} } {\left[ { - \cos x} \right]} dx$$
When 0 $$ \le $$ x $$ \le $$ 1 then -1 $$ \le $$ -cosx $$ \le $$ 0
$$ \therefore $$ $${\left[ { - \cos x} \right]}$$ = -1
When 1 $$ \le $$ x $$ \le $$ $${\sqrt {{\pi \over 2}} }$$ = 1.24 then -0.33 $$ \le $$ -cosx $$ \le $$ 0
$$ \therefore $$ $${\left[ { - \cos x} \right]}$$ = -1 (Integer value present in the left side of -0.33)
$$ = - \int\limits_0^1 {dx + \int\limits_1^{\sqrt {{\pi \over 2}} } {dx - \int\limits_1^{\sqrt {{\pi \over 2}} } {dx} } } $$
$$ = - (x)_0^1 = - 1$$
$$ \therefore $$ $$\left| {\int\limits_0^{\sqrt {{\pi \over 2}} } {\left[ {[{x^2}] - \cos x} \right]dx} } \right|$$ = 1
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