Question

Let the function ${{f}_{k}}\left( x \right)=\dfrac{1}{k}\left( {{\sin }^{k}}x+{{\cos }^{k}}x \right)$ for $k=1,2,3,....$ Then for all $x\in R$, the value of ${{f}_{4}}\left( x \right)-{{f}_{6}}\left( x \right)$ is equal to:
(A) $\dfrac{5}{12}$
(B) $\dfrac{-1}{12}$
(C) $\dfrac{1}{4}$
(D) $\dfrac{1}{12}$

Answer 1

We solve this question by first considering ${{f}_{4}}\left( x \right)$ and simplify it using the formulas ${{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab$ and ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. Then we consider the function ${{f}_{6}}\left( x \right)$ and find its value by simplifying the value using the formulas $\left( {{a}^{3}}+{{b}^{3}} \right)=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$ and ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. Then we use these values and find the value of ${{f}_{4}}\left( x \right)-{{f}_{6}}\left( x \right)$.

Complete step-by-step solution:
We are given that the function, ${{f}_{k}}\left( x \right)=\dfrac{1}{k}\left( {{\sin }^{k}}x+{{\cos }^{k}}x \right)$.
We need to find the value of ${{f}_{4}}\left( x \right)-{{f}_{6}}\left( x \right)$.
Now let us consider the ${{f}_{4}}\left( x \right)$. So,
$\begin{aligned} & \Rightarrow {{f}_{4}}\left( x \right)=\dfrac{1}{4}\left( {{\sin }^{4}}x+{{\cos }^{4}}x \right) \\\ & \Rightarrow {{f}_{4}}\left( x \right)=\dfrac{1}{4}\left( {{\left( {{\sin }^{2}}x \right)}^{2}}+{{\left( {{\cos }^{2}}x \right)}^{2}} \right) \\\ \end{aligned}$
Now let us consider the formula
$\begin{aligned} & \Rightarrow {{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}} \\\ & \Rightarrow {{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab \\\ \end{aligned}$
Using this formula, we can write ${{\sin }^{4}}x+{{\cos }^{4}}x$ as,
${{\sin }^{4}}x+{{\cos }^{4}}x={{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{2}}-2{{\sin }^{2}}x{{\cos }^{2}}x$
Now let us use the formula
${{\sin }^{2}}x+{{\cos }^{2}}x=1$
Using that we can write,
$\begin{aligned} & {{\sin }^{4}}x+{{\cos }^{4}}x={{\left( 1 \right)}^{2}}-2{{\sin }^{2}}x{{\cos }^{2}}x \\\ & {{\sin }^{4}}x+{{\cos }^{4}}x=1-2{{\sin }^{2}}x{{\cos }^{2}}x..........\left( 1 \right) \\\ \end{aligned}$
Using equation (1) we can write ${{f}_{4}}\left( x \right)$ as,
$\begin{aligned} & \Rightarrow {{f}_{4}}\left( x \right)=\dfrac{1}{4}\left( {{\sin }^{4}}x+{{\cos }^{4}}x \right) \\\ & \Rightarrow {{f}_{4}}\left( x \right)=\dfrac{1}{4}\left( 1-2{{\sin }^{2}}x{{\cos }^{2}}x \right) \\\ & \Rightarrow {{f}_{4}}\left( x \right)=\dfrac{1}{4}-\dfrac{1}{2}{{\sin }^{2}}x{{\cos }^{2}}x...........\left( 2 \right) \\\ \end{aligned}$
Now let us consider the ${{f}_{6}}\left( x \right)$. So,
$\begin{aligned} & \Rightarrow {{f}_{6}}\left( x \right)=\dfrac{1}{6}\left( {{\sin }^{6}}x+{{\cos }^{6}}x \right) \\\ & \Rightarrow {{f}_{6}}\left( x \right)=\dfrac{1}{6}\left( {{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}} \right) \\\ \end{aligned}$
Now let us consider the formula,
$\left( {{a}^{3}}+{{b}^{3}} \right)=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$
Using this formula, we can write the above equation as
$\Rightarrow {{f}_{6}}\left( x \right)=\dfrac{1}{6}\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)\left( {{\sin }^{4}}x+{{\cos }^{4}}x-{{\sin }^{2}}x{{\cos }^{2}}x \right)$
Now let us consider the formula,
${{\sin }^{2}}x+{{\cos }^{2}}x=1$
Using this formula, we can write the above equation as
$\begin{aligned} & \Rightarrow {{f}_{6}}\left( x \right)=\dfrac{1}{6}\left( 1 \right)\left( {{\sin }^{4}}x+{{\cos }^{4}}x-{{\sin }^{2}}x{{\cos }^{2}}x \right) \\\ & \Rightarrow {{f}_{6}}\left( x \right)=\dfrac{1}{6}\left( {{\sin }^{4}}x+{{\cos }^{4}}x-{{\sin }^{2}}x{{\cos }^{2}}x \right) \\\ \end{aligned}$
Now let use the value in equation (1) and we write it as,
$\begin{aligned} & \Rightarrow {{f}_{6}}\left( x \right)=\dfrac{1}{6}\left( 1-2{{\sin }^{2}}x{{\cos }^{2}}x-{{\sin }^{2}}x{{\cos }^{2}}x \right) \\\ & \Rightarrow {{f}_{6}}\left( x \right)=\dfrac{1}{6}\left( 1-3{{\sin }^{2}}x{{\cos }^{2}}x \right) \\\ & \Rightarrow {{f}_{6}}\left( x \right)=\dfrac{1}{6}-\dfrac{1}{2}{{\sin }^{2}}x{{\cos }^{2}}x............\left( 3 \right) \\\ \end{aligned}$
Now using equations (2) and (3) we can find the value of ${{f}_{4}}\left( x \right)-{{f}_{6}}\left( x \right)$ as,
$\begin{aligned} & \Rightarrow {{f}_{4}}\left( x \right)-{{f}_{6}}\left( x \right)=\dfrac{1}{4}-\dfrac{1}{2}{{\sin }^{2}}x{{\cos }^{2}}x-\left( \dfrac{1}{6}-\dfrac{1}{2}{{\sin }^{2}}x{{\cos }^{2}}x \right) \\\ & \Rightarrow {{f}_{4}}\left( x \right)-{{f}_{6}}\left( x \right)=\left( \dfrac{1}{4}-\dfrac{1}{6} \right)-\dfrac{1}{2}{{\sin }^{2}}x{{\cos }^{2}}x+\dfrac{1}{2}{{\sin }^{2}}x{{\cos }^{2}}x \\\ & \Rightarrow {{f}_{4}}\left( x \right)-{{f}_{6}}\left( x \right)=\dfrac{1}{12} \\\ \end{aligned}$
So, we get the value of ${{f}_{4}}\left( x \right)-{{f}_{6}}\left( x \right)$ as $\dfrac{1}{12}$.
Hence answer is Option D.

Note: We can also solve this question in another process.
First let us consider the ${{f}_{6}}\left( x \right)$. So,
$\begin{aligned} & \Rightarrow {{f}_{6}}\left( x \right)=\dfrac{1}{6}\left( {{\sin }^{6}}x+{{\cos }^{6}}x \right) \\\ & \Rightarrow {{f}_{6}}\left( x \right)=\dfrac{1}{6}\left( {{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}} \right) \\\ \end{aligned}$
Now let us consider the formula,
$\left( {{a}^{3}}+{{b}^{3}} \right)=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$
Using this formula, we can write the above equation as
$\Rightarrow {{f}_{6}}\left( x \right)=\dfrac{1}{6}\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)\left( {{\sin }^{4}}x+{{\cos }^{4}}x-{{\sin }^{2}}x{{\cos }^{2}}x \right)$
Now let us consider the formula,
${{\sin }^{2}}x+{{\cos }^{2}}x=1$
Using this formula, we can write the above equation as
$\begin{aligned} & \Rightarrow {{f}_{6}}\left( x \right)=\dfrac{1}{6}\left( 1 \right)\left( {{\sin }^{4}}x+{{\cos }^{4}}x-{{\sin }^{2}}x{{\cos }^{2}}x \right) \\\ & \Rightarrow 6{{f}_{6}}\left( x \right)={{\sin }^{4}}x+{{\cos }^{4}}x-{{\sin }^{2}}x{{\cos }^{2}}x...........\left( 1 \right) \\\ \end{aligned}$
As we have
$\begin{aligned} & \Rightarrow {{f}_{4}}\left( x \right)=\dfrac{1}{4}\left( {{\sin }^{4}}x+{{\cos }^{4}}x \right) \\\ & \Rightarrow 4{{f}_{4}}\left( x \right)={{\sin }^{4}}x+{{\cos }^{4}}x \\\ \end{aligned}$
Substituting this value in equation (1), we get,

& \Rightarrow 6{{f}_{6}}\left( x \right)=4{{f}_{4}}\left( x \right)-{{\sin }^{2}}x{{\cos }^{2}}x \\\ & \Rightarrow 4{{f}_{4}}\left( x \right)-6{{f}_{6}}\left( x \right)={{\sin }^{2}}x{{\cos }^{2}}x \\\ \end{aligned}$$ Let us add $2{{f}_{4}}\left( x \right)$ on both sides. Then we get, $$\Rightarrow 6{{f}_{4}}\left( x \right)-6{{f}_{6}}\left( x \right)=2{{f}_{4}}\left( x \right)+{{\sin }^{2}}x{{\cos }^{2}}x.......\left( 2 \right)$$ As we know, ${{f}_{4}}\left( x \right)=\dfrac{1}{4}\left( {{\sin }^{4}}x+{{\cos }^{4}}x \right)$ we get, $\Rightarrow 2{{f}_{4}}\left( x \right)=\dfrac{1}{2}\left( {{\sin }^{4}}x+{{\cos }^{4}}x \right)$ Substituting this value in the above equation (2) we get $$\begin{aligned} & \Rightarrow 6{{f}_{4}}\left( x \right)-6{{f}_{6}}\left( x \right)=\dfrac{1}{2}\left( {{\sin }^{4}}x+{{\cos }^{4}}x \right)+{{\sin }^{2}}x{{\cos }^{2}}x \\\ & \Rightarrow 6\left( {{f}_{4}}\left( x \right)-{{f}_{6}}\left( x \right) \right)=\dfrac{1}{2}\left( {{\sin }^{4}}x+{{\cos }^{4}}x+2{{\sin }^{2}}x{{\cos }^{2}}x \right) \\\ & \Rightarrow {{f}_{4}}\left( x \right)-{{f}_{6}}\left( x \right)=\dfrac{1}{12}{{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{2}} \\\ \end{aligned}$$ Using the formula ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, we can write the above equation as $$\Rightarrow {{f}_{4}}\left( x \right)-{{f}_{6}}\left( x \right)=\dfrac{1}{12}{{\left( 1 \right)}^{2}}=\dfrac{1}{12}$$ Hence answer is Option D.