Let `f(x)=cos(a_1+x)+1/2cos(a_2+x)+1/(2^2)cos(a_1+x)++1/(2^(n-1))cos(a_n+x)` where `a)1,a_2 a_n in Rdot` If `f(x_1)=f(x_2)=0,t h e n|x_2-x_1|` may be equal to `pi` (b) `2pi` (c) `3pi` (d) `pi/2`

Let f(x)=cos(a_1+x)+1/2cos(a_2+x)+1/(2^2)cos(a_3+x)+ ........+ 1/(2^(n-1))cos(a_n+x) where a)1,a_2 a_n in Rdot If f(x_1)=f(x_2)=0,t h e n|x_2-x_1| may be equal to (a) pi (b) 2pi (c) 3pi (d) pi/2

Let f(x)=cos(a_1+x)+1/2cos(a_2+x)+1/(2^2)cos(a_3+x)+ ........+ 1/(2^(n-1))cos(a_n+x) where a)1,a_2 a_n in Rdot If f(x_1)=f(x_2)=0,t h e n|x_2-x_1| may be equal to (a) pi (b) 2pi (c) 3pi (d) pi/2

Let f(x)=cos(a_1+x)+1/2cos(a_2+x)+1/(2^2)cos(a_3+x)++1/(2^(n-1))cos(a_n+x) where a)1,a_2 a_n in Rdot If f(x_1)=f(x_2)=0,t h e n|x_2-x_1| may be equal to pi (b) 2pi (c) 3pi (d) pi/2

Let f(x)=cos(a_(1)+x)+(1)/(2)cos(a_(2)+x)+(1)/(2^(2))cos(a_(1)+x)+...+(1)/(2^(n-1))cos(a_(n)+x) where a)1,a_(2)...a_(n)in R. If f(x_(1))=f(x_(2))=0, then |x_(2)-x_(1)| may be equal to pi(b)2 pi(c)3 pi(d)(pi)/(2)

If f(x) = cos(x)cos(2x)cos(2^(2)x)…cos(2^(n-1)x) and n gt1 , then f'((pi)/(2)) is

If f(x) = cos x cos 2x cos 2^2 x cos^(2^3) x .....cos 2^(n-1) x and n gt 1 then f^(1)(pi/2) is

Let f(x)=(2cos x-1)(2cos2x-1)(2cos2^(2)x-1)...(2cos2^(n)x-1) (where n>=1). Then prove that f((2 pi k)/(2^(n)+-1))=1AA k in I

If f(x) = cos x\ cos 2x\ cos 2^2\ x\ cos 2^3 x\ ....cos2^(n-1) x and n gt 1, then f'(pi/2) is

Let a_n=int_0^(pi/2) (1-cos2nxdx)/(1-cos2x) , then a1, a2, a3 is in