Question

JEE Mains question is in the image.

Answer 1

question is -> $\int\limits^2_{-2}\frac{sin^2x}{\frac{1}{2}+\left[\frac{x}{\pi}\right]}\,dx$ is equal to ...

solution : we know,

$\int\limits^a_b{f(x)}\,dx=\int\limits^a_c{f(x)}\,dx+\int\limits^c_b{f(x)}\,dx$

let's use this concept here,

$\int\limits^2_{-2}\frac{sin^2x}{\frac{1}{2}+\left[\frac{x}{\pi}\right]}\,dx=\int\limits^2_{0}\frac{sin^2x}{\frac{1}{2}+\left[\frac{x}{\pi}\right]}\,dx+\int\limits^0_{-2}\frac{sin^2x}{\frac{1}{2}+\left[\frac{x}{\pi}\right]}\,dx$

here [x/π] , [.] denotes G.I.F

interval 2 < x < 0 , [x/π] = 0

and also interval -2 < x < 0 [x/π] = -1

now $\int\limits^2_{0}\frac{sin^2x}{\frac{1}{2}+\left[\frac{x}{\pi}\right]}\,dx=\int\limits^2_0{\frac{sin^2x}{\frac{1}{2}}}\,dx=2\int\limits^2_0sin^2x\,dx$

similarly, $\int\limits^0_{-2}\frac{sin^2x}{\frac{1}{2}+\left[\frac{x}{\pi}\right]}\,dx=\int\limits^0_{-2}\frac{sin^2x}{-\frac{1}{2}}\,dx=-2\int\limits^0_{-2}sin^2x\,dx$

$so,I=2\int\limits^2_0{sin^2x}\,dx-2\int\limits^0_{-2}{sin^2x}\,dx$

= 0

[ sin²x is an even function.

so, $\int\limits^2_{-2}\frac{sin^2x}{\frac{1}{2}+\left[\frac{x}{\pi}\right]}\,dx=2\int\limits^2_0{sin^2x}\,dx=2\int\limits^0_{-2}{sin^2x}\,dx$ ]

Therefore the value of $\int\limits^2_{-2}\frac{sin^2x}{\frac{1}{2}+\left[\frac{x}{\pi}\right]}\,dx$ is 0

Answer 2

Answer:

ztsdydckfzfacigzydfuxuxydzufchxufDaZhdExydzsyvidzytvuf