Question

Let the value of f(x) be \[{{f}_{k}}x=\dfrac{1}{k}\left( \text{si}{{\text{n}}^{\text{k}}}\text{x+co}{{\text{s}}^{\text{k}}}\text{x} \right)\] where $x\in R\text{ and R}\ge \text{1}$ then the value of ${{f}_{4}}(x)-{{f}_{6}}(x)$ equals to
\[\begin{align}
  & A.\dfrac{1}{6} \\
 & B.\dfrac{1}{3} \\
 & C.\dfrac{1}{4} \\
 & D.\dfrac{1}{12} \\
\end{align}\]

Answer 1

Hint: To solve this question, we will first calculate the value of ${{f}_{4}}(x)$ and value of ${{f}_{6}}(x)$ then subtract them. In between we will use the identity as \[\left( {{a}^{2}}+{{b}^{2}} \right)={{\left( a+b \right)}^{2}}-2ab\text{ and }{{\text{a}}^{\text{3}}}\text{+}{{\text{b}}^{\text{3}}}\text{=}{{\left( a+b \right)}^{3}}-3ab\left( a+b \right)\]

Complete step by step answer:
Given that, \[{{f}_{k}}x=\dfrac{1}{k}\left( \text{si}{{\text{n}}^{\text{k}}}\text{x+co}{{\text{s}}^{\text{k}}}\text{x} \right)\]
Where $x\in R\text{ and R}\ge \text{1}$
Consider k = 4 in above equation, we get:
\[{{f}_{4}}x=\dfrac{1}{4}\left( \text{si}{{\text{n}}^{4}}\text{x+co}{{\text{s}}^{4}}\text{x} \right)\]
We have a identity given as \[\left( {{a}^{2}}+{{b}^{2}} \right)={{\left( a+b \right)}^{2}}-2ab\]
Let \[a={{\sin }^{2}}x\text{ and b=co}{{\text{s}}^{\text{2}}}\text{x}\] and using identity stated above, we get:
\[{{f}_{4}}x=\dfrac{1}{4}\left\{ {{\left( \text{si}{{\text{n}}^{2}}\text{x+co}{{\text{s}}^{2}}\text{x} \right)}^{2}}-2\text{si}{{\text{n}}^{2}}\text{xco}{{\text{s}}^{2}}\text{x} \right\}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}\]
Now, we have trigonometric identity given as:
\[\text{si}{{\text{n}}^{2}}\text{x+co}{{\text{s}}^{2}}\text{x}=1\]
Using this above, we get:
\[\begin{align}
  & {{f}_{4}}x=\dfrac{1}{4}\left\{ 1-2\text{si}{{\text{n}}^{2}}\text{xco}{{\text{s}}^{2}}\text{x} \right\} \\
 & \Rightarrow {{f}_{4}}x=\dfrac{1}{4}-\dfrac{1}{2}\text{si}{{\text{n}}^{2}}\text{xco}{{\text{s}}^{2}}\text{x} \\
\end{align}\]
Similarly, we will now calculate the value of ${{f}_{6}}(x)$
\[{{f}_{6}}x=\dfrac{1}{6}\left( \text{si}{{\text{n}}^{6}}\text{x+co}{{\text{s}}^{6}}\text{x} \right)\]
Now, we have an identity as \[{{\text{a}}^{\text{3}}}\text{+}{{\text{b}}^{\text{3}}}\text{=}{{\left( a+b \right)}^{3}}-3ab\left( a+b \right)\]
Let, \[a={{\sin }^{2}}x\text{ and b=co}{{\text{s}}^{\text{2}}}\text{x}\]
Using this in above, we get:
\[{{f}_{6}}x=\dfrac{1}{6}\left[ {{\left( \text{si}{{\text{n}}^{2}}\text{x+co}{{\text{s}}^{2}}\text{x} \right)}^{3}}-3\text{co}{{\text{s}}^{2}}\text{xsi}{{\text{n}}^{2}}\text{x}\left( \text{si}{{\text{n}}^{2}}\text{x+co}{{\text{s}}^{2}}\text{x} \right) \right]\]
We have value of \[\text{si}{{\text{n}}^{2}}\text{x+co}{{\text{s}}^{2}}\text{x}=1\]
Using this in above, we get:
\[\begin{align}
  & {{f}_{6}}x=\dfrac{1}{6}\left[ 1-3\text{si}{{\text{n}}^{2}}\text{xco}{{\text{s}}^{2}}\text{x} \right] \\
 & \Rightarrow {{f}_{6}}x=\dfrac{1}{6}-\dfrac{1}{2}\text{si}{{\text{n}}^{2}}\text{xco}{{\text{s}}^{2}}\text{x }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)} \\
\end{align}\]
Now, from equation (i) and (ii) we get:
\[{{f}_{4}}x-{{f}_{6}}x=\dfrac{1}{4}-\dfrac{1}{2}\text{si}{{\text{n}}^{2}}\text{xco}{{\text{s}}^{2}}\text{x}-\dfrac{1}{6}+\dfrac{1}{2}\text{si}{{\text{n}}^{2}}\text{xco}{{\text{s}}^{2}}\text{x}\]
Cancelling common term, we have:
\[\begin{align}
  & {{f}_{4}}x-{{f}_{6}}x=\dfrac{1}{4}-\dfrac{1}{6} \\
 & \Rightarrow {{f}_{4}}x-{{f}_{6}}x=\dfrac{6-4}{24}=\dfrac{2}{24}=\dfrac{1}{12} \\
\end{align}\]

So, the correct answer is “Option D”.

Note: The biggest possibility of mistake in this question is at the point where $a={{\sin }^{2}}x$ b is assumed to be $b={{\cos }^{2}}x$ do not go for assuming $a={{\sin }^{4}}x\text{ and b=co}{{\text{s}}^{\text{4}}}\text{x}$ this way, you would not be able to use the formula
\[\begin{align}
  & \left( {{a}^{2}}+{{b}^{2}} \right)={{\left( a+b \right)}^{2}}-2ab\text{ and }{{\text{f}}_{\text{4}}}\text{x=}{{\sin }^{4}}x+\text{co}{{\text{s}}^{\text{4}}}\text{x} \\
 & \Rightarrow {{\text{f}}_{\text{4}}}\text{x=}{{\left( \sin x+\cos x \right)}^{2}}\text{-2}{{\sin }^{4}}x\text{co}{{\text{s}}^{\text{4}}}\text{x} \\
\end{align}\]
This would not give any solution and hence, we would stick. So, go for assuming $a={{\sin }^{2}}x\text{ and b=co}{{\text{s}}^{2}}\text{x}$