`("lim")_(xvecpi/2)(sin(xcosx)/("cos"(xsinx)i se q u a lto` 0 (b) `p/2` (c) `p` (d) `2p`

Options

A. `a=3" and "b=9//2`
B. `a=3" and "b=9//2`
C. `a=-3" and "b=-9//2`
D. `a=3" and "b=-9//2`

Answer

B

Explanation

`L=underset(xto pi//2)lim(sin(xcosx))/(sin((pi)/(2)-xsinx))`
`L=underset(xto (pi)/(2))lim(sin(xcosx))/((xcosx))(((pi)/(2)-xsinx))/(sin((pi)/(2)-xsinx))(xcosx)/(((pi)/(2)-xsinx))`
`=1xx1underset(xto pi//2)lim(xcosx)/(((pi)/(2)-xsinx))`
Put `x=pi//2+h." Then "`
`L=underset(hto0)lim(((pi)/(2)+h)cos((pi)/(2)+h))/((pi)/(2)-((pi)/(2)+h)sin((pi)/(2)+h))`
`=underset(hto0)lim(-((pi)/(2)+h)sin h)/((pi)/(2)(1-cos h)-hcos h)`
`=underset(hto0)lim(-((pi)/(2)+h)((sin h)/(h)))/((pi)/(2)((1-cos h))/(h)-cos h)`(Divide `N^(r)` and `D^(r)` by `h`)
`=(-((pi)/(2)+0)1)/(0-1)=(pi)/(2)`