Question

The equation ${{\sin }^{6}}x+{{\cos }^{6}}x={{a}^{2}}$ has a real solution if,
(a) $a\in \left[ -1,1 \right]$
(b) $a\in \left[ -1,-\dfrac{1}{2} \right]$
(c) $a\in \left[ \dfrac{1}{2},1 \right]\cup \left[ -1,-\dfrac{1}{2} \right]$
(d) $a\in \left[ \dfrac{1}{2},1 \right]$

Answer 1

Write ${{\sin }^{6}}x+{{\cos }^{6}}x$ in the form of ${{a}^{3}}+{{b}^{3}}$ and apply the identity, ${{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)$. Use the trigonometric identity, ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ to simplify the expression. Now, use the formula, $2\sin \theta .\cos \theta =\sin 2\theta$ to convert L.H.S to a function containing a single trigonometric function. Now use the information, $0\le {{\sin }^{2}}\theta \le 1$ to find the range of ‘a’.

Complete step-by-step solution:
We have been provided with the equation, ${{\sin }^{6}}x+{{\cos }^{6}}x={{a}^{2}}$. Let us consider the L.H.S. = ${{\sin }^{6}}x+{{\cos }^{6}}x={{a}^{2}}$. This can be written as,
$\Rightarrow L.H.S={{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}}$
We know that, ${{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)$, therefore, we have,
$\Rightarrow L.H.S={{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{3}}-3{{\sin }^{2}}x{{\cos }^{2}}x\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)$
Applying the identity, ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, we get,
$\begin{aligned} & \Rightarrow L.H.S=1-3{{\sin }^{2}}x{{\cos }^{2}}x\times 1 \\\ & \Rightarrow L.H.S=1-3{{\sin }^{2}}x{{\cos }^{2}}x \\\ \end{aligned}$
The above relation can be written as,
$\begin{aligned} & \Rightarrow L.H.S=1-\dfrac{3}{4}\times 4{{\sin }^{2}}x{{\cos }^{2}}x \\\ & \Rightarrow L.H.S=1-\dfrac{3}{4}\times {{\left( 2\sin x\cos x \right)}^{2}} \\\ \end{aligned}$
We know that $2\sin x\cos x=\sin 2x$, therefore we have,
$\Rightarrow L.H.S=1-\dfrac{3}{4}{{\sin }^{2}}2x$
So, we can write the original equation as,
$1-\dfrac{3}{4}{{\sin }^{2}}2x={{a}^{2}}$
To find the range of a, we must find the range of $1-\dfrac{3}{4}{{\sin }^{2}}2x$. We know that,
$0\le {{\sin }^{2}}2x\le 1$, because the range of sine function is [-1, 1]. Multiplying $\dfrac{3}{4}$, we get,
$\Rightarrow 0\le \dfrac{3}{4}{{\sin }^{2}}2x\le \dfrac{3}{4}$
Now, multiplying -1 and reversing the direction of inequality, we get,
$\begin{aligned} & \Rightarrow 0\ge -\dfrac{3}{4}{{\sin }^{2}}2x\ge -\dfrac{3}{4} \\\ & \Rightarrow -\dfrac{3}{4}\le -\dfrac{3}{4}{{\sin }^{2}}2x\le 0 \\\ \end{aligned}$
Now, adding 1, we get,
$\begin{aligned} & \Rightarrow 1-\dfrac{3}{4}\le 1-\dfrac{3}{4}{{\sin }^{2}}2x\le 1+0 \\\ & \Rightarrow \dfrac{1}{4}\le {{a}^{2}}\le 1 \\\ \end{aligned}$
Case (i) : When ${{a}^{2}}\ge \dfrac{1}{4}$,
$\begin{aligned} & \Rightarrow {{a}^{2}}\ge {{\left( \dfrac{1}{2} \right)}^{2}} \\\ & \Rightarrow a\in \left[ -\infty .-\dfrac{1}{2} \right]\cup \left[ \dfrac{1}{2},\infty \right] \\\ \end{aligned}$
Case (ii) : When ${{a}^{2}}\le 1$,
$\begin{aligned} & \Rightarrow {{a}^{2}}\le {{1}^{2}} \\\ & \Rightarrow -1\le a\le +1 \\\ & \Rightarrow a\in \left[ -1,1 \right] \\\ \end{aligned}$
Since, we have to satisfy both the cases, therefore, to find the values of ‘a’, we need to take the intersection of the above sets of values of ‘a’. Therefore, taking intersection of the set, we get,
$\begin{aligned} & a\in \left[ -1,1 \right]\cap \left[ -\infty ,\dfrac{1}{2} \right]\cup \left[ \dfrac{1}{2},\infty \right] \\\ & \Rightarrow a\in \left[ -1,-\dfrac{1}{2} \right]\cup \left[ \dfrac{1}{2},1 \right] \\\ \end{aligned}$
Hence, option (c) is the correct answer.

Note: One may note that when we multiply (-1) with $\dfrac{3}{4}{{\sin }^{2}}2x$, then the direction of inequality gets reversed. This is because when any negative number is multiplied to the inequality, its direction changes. This also happens when we take the reciprocal. You must note that we have to satisfy both cases (i) and case (ii) and therefore, we need to take the intersection of the sets and not the union.