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Ex 5.7, 16 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at May 29, 2018 by Teachoo
Transcript
Ex 5.7, 16 (Method 1) If (x+1)= 1, show that 2 2 = 2 We need to show that 2 2 = 2 (x+1)= 1 Differentiating . . . x+1 = (1) x + 1 = 0 ( ) . (x+1) + (x + 1) . = 0 ( ) (x+1) + ( ) + (1) . = 0 ( ) (x+1) + 1+0 . = 0 (x+1) + = 0 (x+1) = = ( + 1) = 1 ( + 1) Again Differentiating . . . = 1 +1 2 2 = (1) . +1 +1 . 1 ( +1) 2 = 0 . +1 +1 . 1 ( +1) 2 = 0 1 + 0 . 1 ( +1) 2 = 1 ( +1) 2 = 1 ( +1) 2 Hence 2 2 = 1 ( +1) 2 = 1 +1 2 = 2 Hence proved Ex 5.7, 16 (Method 2) If = (x+1)= 1, show that 2 2 = 2 If = (x+1)= 1 We need to show that 2 2 = 2 ( +1)= 1 Differentiating . . . x+1 = (1) x + 1 = 0 ( ) . (x+1) + (x + 1) . = 0 ( ) (x+1) + ( ) + (1) . = 0 ( ) (x+1) + 1+0 . = 0 (x+1) + = 0 (x+1) = = ( + 1) = 1 ( + 1) Given, x + 1 = 1 = + Putting (2) in (1) = 1 ( + 1) = Again Differentiating . . . = ( ) 2 2 = ( ) 2 2 = ( ) 2 2 = ( ) 2 2 = 2 2 = 2 2 = 2 2 = 2 Hence proved .
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