Question

The equation ${{\sin }^{4}}x+{{\cos }^{4}}x=a$ has a real solution, if
A) $0 < a \le 1$
B) $\dfrac{1}{2}\le a\le 1$
C) $\dfrac{1}{4}\le a\le \dfrac{1}{2}$
D) $-1\le a\le 1$

Answer 1

Hint:
Here we have to find the range of a. Here the value of $a$ is given which is${{\sin }^{4}}x+{{\cos }^{4}}x$. We will first expand the terms by putting the formula of ${{\sin }^{2}}x\And {{\cos }^{2}}x$ in terms of cos2x in the value of $a$. We will convert the value of $a$in terms of cos2x. We will use the range of cos2x which is between -1 and 1.
From there we will get the required range of $a$.

Complete step by step solution:
It is given:-
$a={{\sin }^{4}}x+{{\cos }^{4}}x$…………… $\left( 1 \right)$
We will expand the terms now.
We know the formula of ${{\sin }^{2}}x$ in terms of $\cos 2x$ which is
$\dfrac{1-\cos 2x}{2}$
Similarly, we will put the value of ${{\cos }^{2}}x$in terms of $\cos 2x$ which is
$\dfrac{1+\cos 2x}{2}$
Now, we will put the value of ${{\sin }^{2}}x$and ${{\cos }^{2}}x$ in equation $\left( 1 \right)$
$a={{\left( \dfrac{1-\cos 2x}{2} \right)}^{2}}+{{\left( \dfrac{1+\cos 2x}{2} \right)}^{2}}$
Now, we will expand the terms by applying exponents on the base.
$a=\dfrac{1+{{\cos }^{2}}2x-2\cos 2x}{4}+\dfrac{1+{{\cos }^{2}}2x+2\cos 2x}{4}$
On further simplifying the terms, we get
$a=\dfrac{2\left( 1+{{\cos }^{2}}2x \right)}{4}$
It can be further simplified as
$a=\dfrac{\left( 1+{{\cos }^{2}}2x \right)}{2}$…………………. $\left( 2 \right)$
We have to find the range of. We know the range of;
$-1\le \cos x\le 1$
The range of $\cos 2x$ will be the same.
 $-1\le \cos 2x\le 1$
We will find the range of ${{\cos }^{2}}2x$now;
$0\le {{\cos }^{2}}2x\le 1$
The range of $1+{{\cos }^{2}}2x$ will be:-
$0+1\le 1+{{\cos }^{2}}2x\le 1+1$
On addition of 1 to each term, we get
$1\le 1+{{\cos }^{2}}2x\le 2$
Now, we will divide the terms by 2.
$\dfrac{1}{2}\le \dfrac{1+{{\cos }^{2}}2x}{2}\le \dfrac{2}{2}$
After simplifying the terms, we get
$\dfrac{1}{2}\le \dfrac{1+{{\cos }^{2}}2x}{2}\le 1$…………….$\left( 3 \right)$
From equation$\left( 2 \right)$, we know;
$a=\dfrac{\left( 1+{{\cos }^{2}}2x \right)}{2}$
Putting value of in equation$\left( 3 \right)$, we get
$\dfrac{1}{2}\le a\le 1$

Hence the correct option is B.

Note:
We have used half angle formulas. We have used the formula of ${{\sin }^{2}}x \And {{\cos }^{2}}x$. These have been derived from the half angle formulas.
We know the formula, $\sin x=\pm \sqrt{\dfrac{\left( 1-\cos 2x \right)}{2}}$, squaring both sides, we get ${{\sin }^{2}}x=\dfrac{\left( 1-\cos 2x \right)}{2}$.
This is the formula of ${{\sin }^{2}}x$ which we have derived from the half angle formula. Similarly for ${{\cos }^{2}}x$, formula would be ${{\cos }^{2}}x=\dfrac{1+\cos 2x}{2}$