InterviewSolution
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InterviewSolution
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यदि ` f ( x ) = {{:( (sin (a + 1) x + 2 sin x ) /( x ),",", xlt 0 ), ( 2, ",", x = 0 ), ( (sqrt( 1 +bx)- 1 )/( x ),",", x gt 0 ):}` ` x = 0 ` पर सतत है, तब a व b के मान ज्ञात कीजिये | |
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Answer» दिया है, ` f ( x ) = {{:( (sin (a + 1) x + 2 sin x ) /( x ),",", xlt 0 ), ( 2, ",", x = 0 ), ( (sqrt( 1 +bx)- 1 )/( x ),",", x gt 0 ):}` दिया है कि फलन ` f ( x ) , x = 0 ` पर सतत है | ` therefore ` (बायाँ पक्ष ) `""_(x = 0 ) ` = ( दायाँ पक्ष) `""_(x =0) ` = ` f ( 0 ) ` ...(1) अब, बायाँ पक्ष = ` lim_ ( x to 0^-) f (x) ` ` = lim _ ( hto 0) f ( 0 - h ) ` ` [ x = 0 - h ` रखने पर, जब ` x to 0 ` , तब ` h to 0 ] ` ` = lim _ ( hto 0) ( sin ( a + 1 ) (0 - h ) + 2 sin ( 0-h ) ) /( ( 0 - h)) ` ` = lim _ ( hto 0) ( - sin ( a + 1 ) h - 2 sin h )/( - h) ` `" " [ because sin ( -theta) = -sin theta ] ` ` = lim _ ( h to 0) ( sin ( a + 1 ) h+ 2 sin h ) /( h) ` ` = lim _ ( hto 0 ) ( sin ( a + 1 ) h ) /( h ) + lim _ ( hto 0) ( 2 sin h ) /( h ) ` ` = lim _ ( hto 0 ) ( sin ( a+ 1 ) h ) /( ( a + 1 ) h) xx ( a + 1 ) + 2 lim_ ( hto0 ) ( sin h ) /( h) ` ` = 1xx ( a + 1 ) + 2 xx 1 " " [ because lim _ ( x to 0) ( sin x ) /(x) = 1 ] ` ` = a + 1 + 2 = a + 3 ` तथा दायाँ पक्ष ` = lim _ ( xto 0^+) f (x) ` ` = lim _ ( hto 0) f ( 0 + h) ` ` [ x = 0 + h` रखने पर, यदि ` x to 0 ` तब ` h to 0 ] ` ` = lim _ ( hto 0) ( sqrt ( 1 + b ( 0 + h) ) - 1 ) /( 0 + h ) = lim_ ( hto 0 ) ( sqrt ( 1 + bh) - 1) /( h ) ` ` = lim _ ( hto 0) (sqrt ( 1 + bh ) - 1 ) /( h ) xx ( sqrt ( 1 + bh ) + 1 ) /( sqrt ( 1 + bh ) + 1 ) ` [अंश व हर में ` sqrt ( 1 + bh ) + 1 ` से गुणा करने पर ] ` = lim _ ( hto 0 ) (( 1 + bh ) - 1 ) /( h ( sqrt ( 1 + bh ) + 1 )) = lim _ ( hto 0) ( bh ) /(hsqrt( 1 +bh ) + 1 ) ` ` = lim _ (h to 0) ( b) /(( sqrt ( 1 + bh ) + 1 )) = (b) /(sqrt ( 1+ 0 ) + 1 ) = (b) /(2) ` समीकरण (1 ) से, ` a + 3 = ( b) /(2) = 2 ` ` rArr a + 3 = 2 ` तथा `(b)/(2) = 2 ` ` rArr a = - 1 ` तथा ` b = 4 ` |
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