MHT CET20242 May 2024Evening ShiftMathematicsContinuity and DifferentiabilityActual

The value of k , for which the function f(x)= cases ( 4 5 )^ 4 x 5 x & , 0 x 2 k + 2 5 & , x= 2 cases is continuous at x= 2 , is

Options

A17 20
B3 5
C- 2 5
D2 5

Correct answer

B. 3 5

Step-by-step solution

Since f (x) is continuous at x= 2 . aligned & f ( 2 )= _ x 2 ( 4 5 )^ 4 x 5 x & k + 2 5 = _ x 2 ( 4 5 )^ _ x 2 ( 4 x 5 x) & k + 2 5 = ( 4 5 )^0=1 & k = 3 5 aligned

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