यदि ` lim _( x to 0) [ 1 + log ( 1+ b^(2)) ]^(1//x) = 2b sin^(2) theta , b gt 0 ` तथा ` theta in [ - pi , pi ] ` तब ` theta` का मान है :

Options

A. ` pm pi/4`
B. ` pm pi/3`
C. ` pm pi/6`
D. ` pm pi/2`

Answer

B

Explanation

` :' lim _( x to 0 ) [ 1 + xlog ( 1+ b^(2) )]^(1//x)= 2b sin^(2) theta `
` :. lim _( x to 0 )e ^((x log (1+b^(2))/x))= 2b sin^(2) theta `
` rArr 1 + b^(2) = 2b sin^(2) theta `
` rArr sin^(2) theta = (1+b^(2))/(2b) = 1/(2b) + (b^(2))/(2b)`
` rArr sin^(2) theta = 1/2 (1/b+b)`
` :' 1/b + b ge 2 `
` :. sin^(2) theta ge 1 "लेकिन " sin^(2) theta le 1 `
तथा ` sin^(2) theta = 1 `
अतः ` theta = pm pi/2`