Question

The minimum and maximum values of ${{\sin }^{4}}x+{{\cos }^{4}}x$ are
A. $\dfrac{1}{2},\dfrac{3}{2}$
B. $\dfrac{1}{2},1$
C. $1,\dfrac{3}{2}$
D. $1,2$

Answer 1

Hint: Here we have been given a trigonometric function and we have to find the minimum and maximum value of it. Firstly we will simplify the function given by using basic algebraic formulas. Then we will simplify the values by using square relation and double angle formulas. Finally we will see for which value of the trigonometric function we get the maximum and minimum value of the function and our desired answer.

Complete step-by-step solution:
We have to find the maximum and minimum value of,
${{\sin }^{4}}x+{{\cos }^{4}}x$
So let the above value as,
$f\left( x \right)={{\sin }^{4}}x+{{\cos }^{4}}x$
We can write the above value as,
$f\left( x \right)={{\left( {{\sin }^{2}}x \right)}^{2}}+{{\left( {{\cos }^{2}}x \right)}^{2}}$
Now we will add and subtract $2{{\cos }^{2}}x{{\sin }^{2}}x$ above and get,
$f\left( x \right)={{\left( {{\sin }^{2}}x \right)}^{2}}+{{\left( {{\cos }^{2}}x \right)}^{2}}+2{{\cos }^{2}}x{{\sin }^{2}}x-2{{\cos }^{2}}x{{\sin }^{2}}x$
Now as we can see that the first three terms are satisfying the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ so,
$f\left( x \right)={{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{2}}-2{{\cos }^{2}}x{{\sin }^{2}}x$
Now by Square relation ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ we get,
$f\left( x \right)={{\left( 1 \right)}^{2}}-2{{\cos }^{2}}x{{\sin }^{2}}x$
Multiply and divide the second term by $2$ as follows,
$f\left( x \right)=1-\dfrac{2{{\cos }^{2}}x{{\sin }^{2}}x}{2}\times 2$
Using Double Angle formula in second term we get,
$f\left( x \right)=1-\dfrac{{{\left( \sin 2x \right)}^{2}}}{2}$
We cannot simplify the above value further so now,
For minimum value of function $\sin 2x=1$ or $-1$ in both case ${{\left( \sin 2x \right)}^{2}}=1$ so,
$f\left( x \right)=1-\dfrac{1}{2}$
$\Rightarrow f\left( x \right)=\dfrac{1}{2}$
For maximum value of the function we should have $\sin 2x=0$
$f\left( x \right)=1-0$
$\Rightarrow f\left( x \right)=1$
So we got the minimum and maximum value as $\dfrac{1}{2},1$ respectively.
Hence the correct option is (B).

Note: In such a type of question it is necessary that we have only one trigonometric function with an unknown variable so that we can take the highest or lowest value of it according to our requirement of minimum or maximum value. The basic relation like the square relation or the double angle formulas are very important and useful in simplifying such types of questions.