Question

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

• A. 16
• B. 2
• C. 4
• D. 8

Answer 1

Answer: (D)

8

Answer 2

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.