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Find the values of p and q so that f(x)={x2+3x+p,x≤1qx+2,x>1 is differentiable at x=1 .

0 5 Views | Posted 3 months ago
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    Answered by

    alok kumar singh | Contributor-Level 10

    3 months ago

    This is a Long Answer Type Questions as classified in NCERT Exemplar G i v e n     t h a t           f ( x ) = { x 2 + 3 x + p , x ≤ 1 q x + 2 , x > 1     a t     x = 1 . L . H . L .     f ' ( c ) = l i m x → 1 − f ( x ) − f ( c ) x − c                                   f ' ( 1 ) = l i m x → 1 − f ( x ) − f ( 1 ) x − 1                                                             = l i m x → 1 − ( x 2 + 3 x + p ) − ( 1 + 3 + p ) x − 1                                                             = l i m h → 0 [ ( 1 − h ) 2 + 3 ( 1 − h ) + p ] − ( 1 + 3 + p ) 1 − h − 1                                                             = l i m h → 0 [ 1 + h 2 − 2 h + 3 − 3 h + p ] − ( 4 + p ) − h                                                             = l i m h → 0 [ h 2 − 5 h + 4 + p ] − [ 4 + p ] − h                                                             = l i m h → 0 h 2 − 5 h + 4 + p − 4 − p − h                                                             = l i m h → 0 h 2 − 5 h − h = l i m h → 0 h [ h − 5 ] − h = 5 R . H . L .     f ' ( 1 ) = l i m x → 1 + f ( x ) − f ( 1 ) x − 1                                                           = l i m x → 1 + ( q x + 2 ) − ( 1 + 3 + p ) x − 1                                                           = l i m h → 0 [ q ( 1 + h ) + 2 ] − [ 4 + p ] 1 + h − 1                                                           = l i m h → 0 q + q h + 2 − 4 − p h = l i m h → 0 q + q h − 2 − p h For  existing  the  limit q − 2 − p = 0             ⇒ q − p = 0                                                       … ( 1 ) ⇒ l i m h → 0 q h − 0 h = q I f     L . H . L .     f ' ( 1 ) = R . H . L .     f ' ( 1 )     t h e n     q = 5 . N o w     p u t t

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