Maximum value of (sinx + cosx) is
(a) 1
(b) 2
(c)\[\sqrt{2}\]
(d) \[\dfrac{1}{\sqrt{2}}\]
Answer
593.4k+ views
Hint: For maximizing the function f(x)=sinx + cosx, we will first multiply and divide the equation with \[\sqrt{2}\] that is \[\dfrac{\sqrt{2}(\sin x+\cos x)}{\sqrt{2}}\] then , as we know that \[\dfrac{1}{\sqrt{2}}\] can be written as \[\sin \dfrac{\pi }{4}\ and\ \cos \dfrac{\pi }{4}\] , therefore, the expression can be written as \[\sqrt{2}\left( \cos \dfrac{\pi }{4}\sin x+\sin \dfrac{\pi }{4}\cos x \right)\] then we can apply the formula for the sum of angles in the sin function. Also another important formula that would be used in the solution would be as follows \[sin\left( x+y \right)=sinx\cdot cosy+siny\cdot cosx\]
Complete step-by-step answer:
Now, in this question, we will first multiply and divide the function with \[\sqrt{2}\] . Then, we will write the value of \[\dfrac{1}{\sqrt{2}}\] as \[\sin \dfrac{\pi }{4}\ and\ \cos \dfrac{\pi }{4}\] .
Now, we will use the following formula which is \[sin\left( x+y \right)=sinx\cdot cosy+siny\cdot cosx\] to get a single sine function whose value, we can then maximize.
As mentioned in the question, we need to maximize the given expression and for that we would follow the exact procedure which is mentioned in the hint that is as follows
We will first multiply and divide the equation with \[\sqrt{2}\] .
\[=\dfrac{\sqrt{2}(\sin x+\cos x)}{\sqrt{2}}\]
Then, as we know that \[\dfrac{1}{\sqrt{2}}\] can be written as \[\sin \dfrac{\pi }{4}\ and\ \cos \dfrac{\pi }{4}\] , therefore, the expression can be written as
\[=\sqrt{2}\left( \cos \dfrac{\pi }{4}\sin x+\sin \dfrac{\pi }{4}\cos x \right)\]
Now, using the formula mentioned in the hint, we get
f(x)
\[\begin{align}
& =\sqrt{2}\left( \sin x\cdot \cos \dfrac{\pi }{4}+\sin x\cdot \cos \dfrac{\pi }{4} \right) \\
& =\sqrt{2}\left( \sin \left( x+\dfrac{\pi }{4} \right) \right) \\
\end{align}\]
Now, we know that over the entire domain of sin function, the maximum value that it can attain is 1, hence, the maximum value of the above mentioned expression is as follows
\[\begin{align}
& =\sqrt{2}\left( \sin \left( x+\dfrac{\pi }{4} \right) \right) \\
& =\sqrt{2} \\
\end{align}\]
so, the maximum value of the ( sinx + cosx ) is equal to \[\sqrt{2}\].
Hence, option ( c ) is correct.
So, the correct answer is “Option (c)”.
Note: The students can make an error in evaluating the maximum value of the given expression if they don’t have any idea about the procedure that is mentioned in the hint and also the students can get confused in finding the maximum value of the sin function at the end of the simplification.
Another method of solving this question is by taking derivative of the function that is
f(x)=sinx + cosx and then equating it to zero which will give the value of x at which the function will attain the maximum value.
Complete step-by-step answer:
Now, in this question, we will first multiply and divide the function with \[\sqrt{2}\] . Then, we will write the value of \[\dfrac{1}{\sqrt{2}}\] as \[\sin \dfrac{\pi }{4}\ and\ \cos \dfrac{\pi }{4}\] .
Now, we will use the following formula which is \[sin\left( x+y \right)=sinx\cdot cosy+siny\cdot cosx\] to get a single sine function whose value, we can then maximize.
As mentioned in the question, we need to maximize the given expression and for that we would follow the exact procedure which is mentioned in the hint that is as follows
We will first multiply and divide the equation with \[\sqrt{2}\] .
\[=\dfrac{\sqrt{2}(\sin x+\cos x)}{\sqrt{2}}\]
Then, as we know that \[\dfrac{1}{\sqrt{2}}\] can be written as \[\sin \dfrac{\pi }{4}\ and\ \cos \dfrac{\pi }{4}\] , therefore, the expression can be written as
\[=\sqrt{2}\left( \cos \dfrac{\pi }{4}\sin x+\sin \dfrac{\pi }{4}\cos x \right)\]
Now, using the formula mentioned in the hint, we get
f(x)
\[\begin{align}
& =\sqrt{2}\left( \sin x\cdot \cos \dfrac{\pi }{4}+\sin x\cdot \cos \dfrac{\pi }{4} \right) \\
& =\sqrt{2}\left( \sin \left( x+\dfrac{\pi }{4} \right) \right) \\
\end{align}\]
Now, we know that over the entire domain of sin function, the maximum value that it can attain is 1, hence, the maximum value of the above mentioned expression is as follows
\[\begin{align}
& =\sqrt{2}\left( \sin \left( x+\dfrac{\pi }{4} \right) \right) \\
& =\sqrt{2} \\
\end{align}\]
so, the maximum value of the ( sinx + cosx ) is equal to \[\sqrt{2}\].
Hence, option ( c ) is correct.
So, the correct answer is “Option (c)”.
Note: The students can make an error in evaluating the maximum value of the given expression if they don’t have any idea about the procedure that is mentioned in the hint and also the students can get confused in finding the maximum value of the sin function at the end of the simplification.
Another method of solving this question is by taking derivative of the function that is
f(x)=sinx + cosx and then equating it to zero which will give the value of x at which the function will attain the maximum value.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

Which gland is known as mixed gland class 11 biology CBSE

