Question
Question: The maximum value of f(x) = sin x + cos x is: \[\left( a \right)1\] \[\left( b \right)2\] \[\l...
The maximum value of f(x) = sin x + cos x is:
(a)1
(b)2
(c)21
(d) None of these
Solution
First, differentiate the given function f(x) to find f’(x). Substitute f’(x) = 0 and find the value of x from the first quadrant. Now, again differentiate f’(x) to find f’’(x) and substitute the value of x, found in the first step, in f’’(x). If it is negative, then x will be a point of maxima. Then substitute this value of x in f(x) to get the correct option. If f’’(x) at x turns out to be positive, then x will be a point of minima, then we will have to consider x from another quadrant and then again do the same process.
Complete step by step answer:
We have been provided with the function f(x) = sin x + cos x. To find its maxima, we need to differentiate it. So, on differentiating f(x), we get,
f′(x)=dxd(sinx)+dxd(cosx)
⇒f′(x)=cosx−sinx
Substituting f’(x) = 0, we get,
⇒cosx−sinx=0
⇒cosx=sinx
Dividing both the sides with cos x, we get,
⇒1=tanx
⇒tanx=1
⇒x=4π
(Considered from the first quadrant)
Now, again differentiating f’(x), we get,
⇒f′′(x)=dxd(cosx)−dxd(sinx)
⇒f′′(x)=−sinx−cosx
Substituting x=4π in f’’(x), we get,
f′′(4π)=2−1−21
f′′(4π)=−2
Clearly, we can see that f’’(x) is negative at x=4π. Therefore x=4π is a point of maxima. Therefore, the maxima value of f(x) will be obtained by substituting x=4π in that.
Therefore, maxima of f(x) will be
=f(4π)
=sin(4π)+cos(4π)
=21+21
=2
So, the correct answer is “Option d”.
Note: One may note that when f’(x) is substituted as 0, then we get tan x = 1. Here, we know that there are many values of x for which tan x = 1, but we have considered x=4π from the first quadrant. This is because in the first quadrant both sin x and cos x are positive. So, we have a chance of getting the maxima. If the value of f’’(x) at x=4π would have been positive then it would have been a point of minima. Then we would have considered another value of x for which tan x = 1 and carried out the same process.