Give that f(x) =\(\frac {1-cos4x}{x^2}\) if x < 0 ,f(x) = a if x = 0 , f(x) =\(\frac {\sqrt {x}}{\sqrt {16 + \sqrt {x} }- 4}\) if x > 0, is continuous at x = 0, then a will be
Give that f(x) =\(\frac {1-cos4x}{x^2}\) if x < 0 ,f(x) = a if x = 0 , f(x) =\(\frac {\sqrt {x}}{\sqrt {16 + \sqrt {x} }- 4}\) if x > 0, is continuous at x = 0, then a will be
- 16
- 2
- 4
- 8
The Correct Option is D
Solution and Explanation
As we know that function is continuous at x = 0, then:
lim (x→0-) f(x)= lim (x→0+) f(x) = f(a)
lim (x→0-) f(x) = lim (x→0-) \(\frac {1-cos4x}{x^2}\)
lim (x→0-) f(x) = lim (x→0-) \(\frac {(2 sin^2 2x)}{(2x)^2}\ .4\) = 8
lim (x→0+) f(x) = lim (x→0+)\(\frac {\sqrt {x}}{\sqrt {16 + \sqrt {x} }- 4}\)
lim (x→0+) = \({\sqrt {16 + \sqrt {x} } + 4}\)= 8
and f(0)= 8
so, a = 8
Therefore the correct option is (D) 8.
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Concepts Used:
Continuity
A function is said to be continuous at a point x = a, if
limx→a
f(x) Exists, and
limx→a
f(x) = f(a)
It implies that if the left hand limit (L.H.L), right hand limit (R.H.L) and the value of the function at x=a exists and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is undefined or does not exist, then we say that the function is discontinuous.
Conditions for continuity of a function: For any function to be continuous, it must meet the following conditions:
- The function f(x) specified at x = a, is continuous only if f(a) belongs to real number.
- The limit of the function as x approaches a, exists.
- The limit of the function as x approaches a, must be equal to the function value at x = a.