Question

If $f(x)={{\sin }^{6}}x+{{\cos }^{6}}x$ then the range of f(x) is:

a)$\left[ \dfrac{1}{4},1 \right]$
b)$\left[ \dfrac{1}{4},\dfrac{3}{4} \right]$
c)$\left[ \dfrac{3}{4},1 \right]$
d)None of these

Answer 1

Hint: Here, first rewrite f(x) as $f(x)={{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}}$. Then, we have to apply the formula, ${{(a+b)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$. Then do the simplification and obtain the function in terms of ${{\sin }^{2}}2x$. Now, with the help of this find the range of the function.

Complete step-by-step answer:


Here, we are given that $f(x)={{\sin }^{6}}x+{{\cos }^{6}}x$.

Now, we have to find the range of f(x).

First, consider the function:

$f(x)={{\sin }^{6}}x+{{\cos }^{6}}x$

Now, we can rewrite the above function as:

$f(x)={{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}}$

We know the expansion of ${{(a+b)}^{3}}$, which is :

$\begin{align}

  & {{(a+b)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}} \\

 & \Rightarrow {{(a+b)}^{3}}={{a}^{3}}+3ab(a+b)+{{b}^{3}} \\

\end{align}$

Now, from the above equation we can find ${{a}^{3}}+{{b}^{3}}$ by taking 3ab (a + b) to the left side,

$\Rightarrow {{(a+b)}^{3}}-3ab(a+b)={{a}^{3}}+{{b}^{3}}$

Next, by applying this expansion we can write:

$\begin{align}

  & {{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{3}}={{\left( {{\sin }^{2}}x \right)}^{3}}+3{{\sin }^{2}}x{{\cos }^{2}}x({{\sin }^{2}}x+{{\cos }^{2}}x)+{{\left( {{\cos }^{2}}x \right)}^{3}} \\

 & \Rightarrow {{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}}={{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{3}}-3{{\sin }^{2}}x{{\cos }^{2}}x({{\sin }^{2}}x+{{\cos }^{2}}x) \\

\end{align}$

We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. Hence our above equation becomes:

$\begin{align}

  & \Rightarrow {{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}}={{1}^{3}}-3{{\sin }^{2}}x{{\cos }^{2}}x\times 1 \\

 & \Rightarrow {{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}}=1-3{{\sin }^{2}}x{{\cos }^{2}}x \\

\end{align}$

Now, by multiplying and dividing 4 in $3{{\sin }^{2}}x{{\cos }^{2}}x$ we get:

${{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}}=1-\dfrac{3}{4}\times 4{{\sin }^{2}}x{{\cos }^{2}}x$

We know that:

$4{{\sin }^{2}}x{{\cos }^{2}}x={{(2\sin x\cos x)}^{2}}$

Hence, our equation becomes:

${{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}}=1-\dfrac{3}{4}{{(2\sin x\cos x)}^{2}}$

We also have an identity that:

$2\sin x\cos x=\sin 2x$

Thus, we obtain:

$\begin{align}

  & \Rightarrow {{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}}=1-\dfrac{3}{4}{{(\sin 2x)}^{2}} \\

 & \Rightarrow {{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}}=1-\dfrac{3}{4}{{\sin }^{2}}2x \\

\end{align}$

We have that the range of $\sin x\in [-1,1]$

Then the range of ${{\sin }^{2}}2x\in [0,1]$

Now, when ${{\sin }^{2}}2x=0$ we have,

$\begin{align}

  & \Rightarrow {{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}}=1-\dfrac{3}{4}\times 0 \\

 & \Rightarrow {{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}}=1-0 \\

 & \Rightarrow {{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}}=1 \\

 & \Rightarrow f(x)=1 \\

\end{align}$

Next, when ${{\sin }^{2}}2x=1$ we have,

$\begin{align}

  & \Rightarrow {{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}}=1-\dfrac{3}{4}\times 1 \\

 & \Rightarrow {{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}}=1-\dfrac{3}{4} \\

\end{align}$

Next, by taking LCM,

$\begin{align}

  & \Rightarrow {{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}}=\dfrac{4-3}{4} \\

 & \Rightarrow {{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}}=\dfrac{1}{4} \\

 & \Rightarrow f(x)=\dfrac{1}{4} \\

\end{align}$

Therefore, we can say that the range of f(x) will be $\left[ \dfrac{1}{4},1 \right]$.

Hence, the correct answer for this question is option (a).


Note: We have the range of $\sin x\in [-1,1]$. Then the range of ${{\sin }^{2}}x\in [0,1]$. Similarly, we can say that the range of ${{\sin }^{2}}2x\in [0,1]$ as it is a square function the range can never be negative. With the help of the range of ${{\sin }^{2}}2x$, we can determine the range of the function f(x).