If f(x)={2mxn1−cos(1−2cosx)1x=0,x=0 and f(0)=1 is continuous at x=0 then the value of m+n is a. 2 b. 3 c. -3 d. 7

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If {f{{({x})}}}={{{{1}-{{{({1}-{{{{x}}}}{{2}})}}}}}{{{2}^{{m}}

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If $f (x) = {\frac{1 - cos ( 1 - \frac{c o s x}{2} )}{2 ^{m} x ^{n}} 1 x = 0, x \neq = 0$ and $f (0) = 1$ is continuous at $x = 0$ then the value of $m + n$ is a. 2 b. 3 c. -3 d. 7

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Lucas Discussed

f (x) = {\frac{1 - cos ( 1 - \frac{c o s x}{2} )}{2 ^{m} x ^{n}} 1 x = 0, x \neq = 0

and

f (0) = 1

is continuous at

x = 0

then the value of

m + n

is a. 2 b. 3 c. -3 d. 7

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Question Text

f (x) = {\frac{1 - cos ( 1 - \frac{c o s x}{2} )}{2 ^{m} x ^{n}} 1 x = 0, x \neq = 0

and

f (0) = 1

is continuous at

x = 0

then the value of

m + n

is a. 2 b. 3 c. -3 d. 7