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Question : 76 of 120
Marks:
+1,
-0
Solution:
The given function is
f(x)= for
x<0 and
f(x)= for
x>0. Let's calculate the left-hand
(x→0−) and the right-hand
(x→0+) limits of the function:
f(x)= which is an indeterminate from
. Let's rationalize by multiplying with the conjugate:
= ×| √1+kx+√1−kx |
| √1+kx+√1−kx |
=×| (1+kx)−(1−kx) |
| √1+kx+√1−kx |
=| lim |
| x→2k√1+kx+√1−kx |
=k And,
f(x)==1 Since, f(x) is given to be continuous at x = 0, both the left-hand and the right-hand side limits must be equal.
∴ k = 1.
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