f(x) = \(\frac{sin(x)}{x}\), How many points exist such that f'(c) = – McqOptions

MCQOPTIONS

1.	f(x) = \(\frac{sin(x)}{x}\), How many points exist such that f'(c) = 0 in the interval [0, 18π].
A.	18
B.	17
C.	= 0 in the interval [0, 18π].a) 18b) 17c) 8
D.	9view answer
Answer» B. 17

Discussion

No Comment Found

Related MCQs

If the domain of a function can be broken into infinite number of disjoint subsets such that every subset has a Rolles point then the function cannot be in a polynomial structure.
For any second degree polynomial with two real unequal roots. The relation between Rolles point r1 and the two roots r2 is
For second degree polynomial it is seen that the roots are equal. Then what is the relation between the Rolles point c and the root x?
For all second degree polynomials with y = ax2 + bx + k, it is seen that the Rolles’ point is at c = 0. Also the value of k is zero. Then what is the value of b?
Let f(x) = x + sin(x) Every point on the graph is rotated by 45 degree with respect to the origin along the radius equal to the radius vector at that point. How many c that belong to [0, 11π] exist Such that f'(c) = 0.
f(x) = \(\frac{sin(x)}{x}\), How many points exist such that f'(c) = 0 in the interval [0, 18π].
For the function f(x) = \(\frac{sin(x)}{x^2}\) How many points exist in the interval [0, 7π] Such that f'(c) = 0.
For y = -x2 + 2x there exist a c in the interval [- 19765, 19767] Such that f'(c) = 0.