If `("lim")_(xvec0)[1+x1n(1+b^2)]^(1/x)=2bsin^2theta,b >0,s mftheta in (-pi,pi],` then the value of `theta` is `+-pi/4` (b) `+-pi/3` (c) `+-pi/6` (d) `+-pi/2`

Options

A. `+-(pi)/(4)`
B. `+-(pi)/(3)`
C. `+-(pi)/(6)`
D. `+-(pi)/(2)`

Answer

D

Explanation

We have,
`lim_(xto0) {1+x In (1+b^2)}^(1//x)=2bsin^2theta`
`rArr e^(lim_(xto0)xIn (1+b^2)xx(1)/(x))=2bsin^2theta `
`rArr ^In (1+b^2=2bsin^2theta`
`rArr 1+b^2=2bsin^2theta`
`rArr sin^2theta (1+b^2)/(2b)`
`rArr sin^2theta=(1)/(2)(b+(1)/(b))[because b+(1)/(b)ge 2therefore(1)/(2))ge 1]`
`rArr sin^2theta=1`
`rArr theta=+-(pi)/(2)`.