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Find the number of solutions of 2cos x = | sin x | , 0 ≤ x ≤ 4π.
Given is the graph of y = 2 cosx and y = | sin x |.
We can see that graphs intersects at 4 points. Hence the equation has 4 solutions
If the range of the function y = does not contain any values belonging to the interval then find the the integral value (s) of ‘a’.
Range f = { - 1, 0, 1 }
If F (n + 1) = n = 1, 2,.........and F (1) = 1 then find the value of F (2009).
then find the range of f (x).
If f (x) is an even function and satisfies the relation x2 f (x) - 2f = g (x) where g (x) is an odd function ,then find the value of f(5).
... g (x) and x2 are odd and even function respectively . So, f (x) is an odd function . But f (x) is given even. ... f (x) = 0 ∀x , Hence f (5) = 0
If [ 2sin x ] + [ cos x ] = - 3 ,then find the range of the function
[ 2sin x ] + [ cos x ] = - 3 only if [ 2sin x ] = - 2 and [ cos x ] = - 1 ... - 2≤ 2 sin x < - 1 and - 1 ≤ cosx < 0 ⇒ - 1≤ sin x < - 1/2 and - 1 ≤ cos x < 0
What will be the number of solutions of the equation 2cosx = | sin x |, when x ∈ [ 0, 2π ].
See the graph y = 2cosx and y = | sin x |. Two curves meets at four points for x ∈ [ 0, 2π ]
So, the equation 2cosx = | sin x | has four solutions.
Let A = {1, 2, 3, 4, 5}. If ‘f’ be a bijective function from A to A, then the number of such functions for which f (k) ≠ k, k = 1, 2, 3, 4, 5 is
The problem is equivalent to derangement . The required number of functions
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