InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
p व q के किस मान के लिए फलन ` f ( x) = {{:( ( 1 - sin ^(3) x ) /( 3 cos ^(2) x )",",, "यदि"x lt pi//2), (p",",, "यदि" x = pi//2), ( (q ( 1 -sin x )) /((pi_ 2x ) ^(2))",",, "यदि" x gt pi//2):}` बिंदु ` x = pi//2 ` पर सतत है ? |
|
Answer» दिया है, ` f ( x) = {{:( ( 1 - sin ^(3) x ) /( 3 cos ^(2) x )",",, "यदि"x lt pi//2), (p",",, "यदि" x = pi//2), ( (q ( 1 -sin x )) /((pi_ 2x ) ^(2))",",, "यदि" x gt pi//2):}` ` x = pi //2 ` पर सतत है | ` therefore ` बायाँ पक्ष = दायाँ पक्ष = ` f ( pi//2)`... (1) अब, बायाँ पक्ष = ` lim _ ( xto pi//2) f ( x ) = lim _ ( hto 0 ) f ( (pi)/(2) - h ) ` ` [ x = (pi)/(2) - h ` रखने पर, यदि ` x to (pi)/(2) ` तब ` h to 0 ] ` ` = lim _ ( hto 0) ( 1 - sin ^3 ((pi)/(2) - h))/( 3 cos ^2 ((pi)/(2) - h)) = lim _ ( hto0) ( 1 - cos ^3 h ) /( 3sin ^2 h ) ` `[ because cos ((pi)/(2) - theta ) = sin theta, sin ((pi)/(2) - theta ) = cos theta ] ` ` = lim _ ( h to 0 ) (( 1 - cos h ) ( 1 ^2 +cos ^2 h + 1 xx cos h ) )/( 3 ( 1 - cos ^2h)) ` ` = lim _ ( hto 0) ((1- cos h ) ( 1 + cos ^2 h + cos h ) )/( 3( 1 - cos h ) ( 1 + cos h)) ` ` = ( 1 + cos ^2 0 + cos 0 )/( 3 (1 + cos 0)) = ( 1+ 1 +1)/( 3 ( 1 + 1 )) = (3) /( 3 xx 2) = (1)/(2) ` तथा daayan पक्ष ` = lim _ ( x to pi//2) [ f ( x) ] = lim _ ( hto 0 ) f (( pi ) /(2) + h ) ` ` [ x = (pi)/(2) + h`, रखने पर, यदि ` x to (pi)/(2) ` तब ` h to 0 ] ` ` = lim _ ( hto 0 ) ( q [ 1 - sin ( ( pi ) /(2) + h)] ) /([ pi - 2 ((pi) /(2) +h)]^2) ` ` = lim _ ( hto 0 ) ( q ( 1 - cos h ) ) /( (pi - pi - 2h )^2) = lim _ ( hto 0 ) ( q ( 1 - cos h ) ) /( 4h^2) ` ` = lim_ ( hto 0 ) ( q ( 2 sin ^2 h//2))/(4h ^2)" " [ because cos x = 1 - 2 sin ^2"" (x)/(2)] ` ` = lim _ ( hto 0) ( q sin ^2 h//2)/(2) = (q)/(8) lim _ ( hto 0) [ ( sin (h//2))/( h//2)]^2 ` ` = (q) /(8) xx 1 = (q)/8" " [ because lim _ ( hto 0) ( sin h ) /(h) = 1] ` समीकरण (1 ) से, ` ( 1 ) /(2) = (q) /(8) = p ` ` ( 1 ) /(2) = (q) /(8) ` तथा ` ( 1 ) /(2) = p ` ` q = 4 ` तथा ` p = ( 1 ) /(2) ` |
|