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

& A.\dfrac{1}{6} \\\ & B.\dfrac{1}{3} \\\ & C.\dfrac{1}{4} \\\ & D.\dfrac{1}{12} \\\ \end{aligned}$$

Answer 1

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:

& {{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{aligned}$$ Similarly, we will now calculate the value of ${{f}_{6}}(x)$ $${{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{aligned} & {{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{aligned}$$ Now, from equation (i) and (ii) we get: $${{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{aligned} & {{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{aligned}$$ **So, the correct answer is “Option D”.** **Note:** The biggest possibility of mistake in this question is at the point where $a={{\sin }^{2}}x$ b is assumed to be $b={{\cos }^{2}}x$ do not go for assuming $a={{\sin }^{4}}x\text{ and b=co}{{\text{s}}^{\text{4}}}\text{x}$ this way, you would not be able to use the formula $$\begin{aligned} & \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{aligned}$$ This would not give any solution and hence, we would stick. So, go for assuming $a={{\sin }^{2}}x\text{ and b=co}{{\text{s}}^{2}}\text{x}$