Question

The maximum value of f(x) = sin x + cos x is:
\[\left( a \right)1\]
\[\left( b \right)2\]
\[\left( c \right)\dfrac{1}{\sqrt{2}}\]
(d) None of these

Answer 1

Hint: 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,
\[{{f}^{'}}\left( x \right)=\dfrac{d\left( \sin x \right)}{dx}+\dfrac{d\left( \cos x \right)}{dx}\]
\[\Rightarrow {{f}^{'}}\left( x \right)=\cos x-\sin x\]
Substituting f’(x) = 0, we get,
\[\Rightarrow \cos x-\sin x=0\]
\[\Rightarrow \cos x=\sin x\]
Dividing both the sides with cos x, we get,
\[\Rightarrow 1=\tan x\]
\[\Rightarrow \tan x=1\]
\[\Rightarrow x=\dfrac{\pi }{4}\]
(Considered from the first quadrant)
Now, again differentiating f’(x), we get,
\[\Rightarrow {{f}^{''}}\left( x \right)=\dfrac{d\left( \cos x \right)}{dx}-\dfrac{d\left( \sin x \right)}{dx}\]
\[\Rightarrow {{f}^{''}}\left( x \right)=-\sin x-\cos x\]
Substituting \[x=\dfrac{\pi }{4}\] in f’’(x), we get,
\[{{f}^{''}}\left( \dfrac{\pi }{4} \right)=\dfrac{-1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}\]
\[{{f}^{''}}\left( \dfrac{\pi }{4} \right)=-\sqrt{2}\]
Clearly, we can see that f’’(x) is negative at \[x=\dfrac{\pi }{4}.\] Therefore \[x=\dfrac{\pi }{4}\] is a point of maxima. Therefore, the maxima value of f(x) will be obtained by substituting \[x=\dfrac{\pi }{4}\] in that.
Therefore, maxima of f(x) will be
\[=f\left( \dfrac{\pi }{4} \right)\]
\[=\sin \left( \dfrac{\pi }{4} \right)+\cos \left( \dfrac{\pi }{4} \right)\]
\[=\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}\]
\[=\sqrt{2}\]

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 \[x=\dfrac{\pi }{4}\] 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 \[x=\dfrac{\pi }{4}\] 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.