1.

दिया है - ` f( x ) = {{:( ( 1 - cos 4 x ) /(x ^(2))",",, "यदि " x lt 0), ( a",",, "यदि " x = 0 ), ( (sqrt x)/(sqrt ( 16 + sqrt x ) - 4 )",",,"यदि " x gt 0 ):}` a का मान ज्ञात कीजिये यदि ` x = ( pi ) /(2) ` पर यह सतत है |

Answer» ` f ( x ) ` की परिभाषानुसार,
` f( 0 ) = a `
और ` f ( 0 + 0 ) = lim _ ( h to 0 ) f ( 0 + h ) `
` = lim _ ( hto 0 ) { ( sqrth) /( sqrt ( 16 + sqrth ) - 4 ) } `
` = lim _ ( hto 0 ) { ( sqrt h ) /( sqrt ( 16 + sqrth ) - 4 )} xx ( sqrt ( 16 + sqrth ) + 4 ) /( sqrt ( 16 + sqrt h ) + 4) `
`
= lim _ ( hto 0) ( sqrth ( sqrt ( 16 + sqrt h ) + 4 ) ) /( ( 16 + sqrt h ) - 16 ) `
` = lim _ ( hto 0 ) ( sqrth * { sqrt (16 + sqrth) +4 })/( sqrth ) `
` = lim _ ( hto 0 ) [ sqrt ( 16+sqrt h ) + 4] = 4 + 4 = 8 `
और ` f ( 0 - 0 ) = lim _ ( hto 0 ) f ( 0 - h ) `
` = lim _ ( hto 0 ) ( [ 1 - cos 4 ( -h ) ] ) /( [ - h ]^(2)) = lim _ ( hto 0 ) ( [ 1 - cos ( - 4h ) ] ) /( h ^(2)) `
` = lim _ ( hto 0) (( 1 - cos 4h ))/(h ^(2))= lim _ ( hto 0 ) ( 2 sin ^2 2h ) /( h ^(2)) `
` = 2 xx 4 *lim _ ( hto 0 ) ( sin ^2 2h ) /( ( 2h )^(2)) = 8 lim _ ( hto 0) (( sin 2h )/(2h))^(2) `
` = 8 xx1 ^(2) = 8 `
` f( 0 + 0 ) = f ( 0 - 0 ) `
` rArr lim _ ( x to 0 ) f ( x) = 8 `
परन्तु, बिंदु ` x = 0 ` पर ` f ( x ) ` की सततता के अनुसार,
` lim _ ( x to 0 ) f ( x) = f ( 0 ) rArr a = 8 `
अतः ` a = 8 `


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