Question

If we have an inverse trigonometric expression as ${{\sin }^{-1}}a+{{\sin }^{-1}}b+{{\sin }^{-1}}c=\dfrac{3\pi }{2}$ and $f\left( 2 \right)=2,f\left( x+y \right)=f\left( x \right)f\left( y \right)\forall x,y\in R$ then ${{a}^{f\left( 2 \right)}}+{{b}^{f\left( 4 \right)}}+{{c}^{f\left( 6 \right)}}-\dfrac{3{{a}^{f\left( 2 \right)}}{{b}^{f\left( 4 \right)}}{{c}^{f\left( 6 \right)}}}{{{a}^{f\left( 2 \right)}}+{{b}^{f\left( 4 \right)}}+{{c}^{f\left( 6 \right)}}}$ equals to
(A) 2
(B) 4
(C) 6
(D) 8

Answer 1

So here in the question we have to find the values of ${{a}^{f\left( 2 \right)}}+{{b}^{f\left( 4 \right)}}+{{c}^{f\left( 6 \right)}}-\dfrac{3{{a}^{f\left( 2 \right)}}{{b}^{f\left( 4 \right)}}{{c}^{f\left( 6 \right)}}}{{{a}^{f\left( 2 \right)}}+{{b}^{f\left( 4 \right)}}+{{c}^{f\left( 6 \right)}}}$ from given equation. Find the function and then use that function in the given expression. Then solve this in this question we have to find the value of a, b, c and then with the help of this formula $f\left( x \right)={{a}^{kx}}$ we will find the values of $f\left( 2 \right),f\left( 4 \right),f\left( 6 \right)$ we will put these in this ${{a}^{f\left( 2 \right)}}+{{b}^{f\left( 4 \right)}}+{{c}^{f\left( 6 \right)}}-\dfrac{3{{a}^{f\left( 2 \right)}}{{b}^{f\left( 4 \right)}}{{c}^{f\left( 6 \right)}}}{{{a}^{f\left( 2 \right)}}+{{b}^{f\left( 4 \right)}}+{{c}^{f\left( 6 \right)}}}$ and then we will get the final answer.

Complete step-by-step solution:
So here we will start the solution by taking the given equation,
${{\sin }^{-1}}a+{{\sin }^{-1}}b+{{\sin }^{-1}}c=\dfrac{3\pi }{2}$
As we know the range of x${{\sin }^{-1}}\theta$is $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$. Maximum value of ${{\sin }^{-1}}\theta$ will be $\dfrac{\pi }{2}$. And the minimum value of the ${{\sin }^{-1}}\theta$ will be $-\dfrac{\pi }{2}$ Then only the above equation can reach $\dfrac{3\pi }{2}$.
So, ${{\sin }^{-1}}a={{\sin }^{-1}}b={{\sin }^{-1}}c=\dfrac{\pi }{2}$
Then only we will reach the value of $\dfrac{3\pi }{2}$.

& {{\sin }^{-1}}a=\dfrac{\pi }{2} \\\ &\Rightarrow \sin \left( {{\sin }^{-1}}a \right)=\sin \left( \dfrac{\pi }{2} \right) \\\ &\Rightarrow a=1 \\\ \end{aligned}$$ So here the value of a, b, c will be as we have given $${{\sin }^{-1}}a+{{\sin }^{-1}}b+{{\sin }^{-1}}c=\dfrac{3\pi }{2}$$’ $$a=b=c=1$$ Also, we have Given that $$f\left( 2 \right)=2,f\left( x+y \right)=f\left( x \right)f\left( y \right)\forall x,y\in R$$ $$\Rightarrow f\left( x+y \right)=f\left( x \right)f\left( y \right)$$ This type of operation only happens in exponential functions. So, put $$f\left( x \right)={{a}^{kx}}$$ $$\begin{aligned} & \Rightarrow f\left( x \right)={{a}^{kx}} \\\ & \Rightarrow f\left( y \right)={{a}^{ky}} \\\ & \Rightarrow f\left( x \right)f\left( y \right)={{a}^{k\left( x+y \right)}} \\\ & \Rightarrow f\left( x+y \right)={{a}^{k\left( x+y \right)}} \\\ & \Rightarrow f\left( x+y \right)=f\left( x \right)f\left( y \right)={{a}^{k\left( x+y \right)}} \\\ \end{aligned}$$ And we have Given that $$f\left( 2 \right)=2$$,so put this in function $$f\left( x \right)={{a}^{kx}}$$ then we will get. $$\begin{aligned} & \Rightarrow f\left( 2 \right)={{a}^{2k}} \\\ & \Rightarrow {{a}^{2k}}={{2}^{1}} \\\ \end{aligned}$$ Base and powers should be equal. $$\begin{aligned} &\Rightarrow a=2 \\\ &\Rightarrow 2k=1 \\\ &\Rightarrow k=\dfrac{1}{2} \\\ \end{aligned}$$ So, the function is $$f\left( x \right)={{2}^{\dfrac{x}{2}}}$$ We have the Give expression is $$\begin{aligned} & \Rightarrow {{a}^{f\left( 2 \right)}}+{{b}^{f\left( 4 \right)}}+{{c}^{f\left( 6 \right)}}-\dfrac{3{{a}^{f\left( 2 \right)}}{{b}^{f\left( 4 \right)}}{{c}^{f\left( 6 \right)}}}{{{a}^{f\left( 2 \right)}}+{{b}^{f\left( 4 \right)}}+{{c}^{f\left( 6 \right)}}} \\\ & a=b=c=1 \\\ &\Rightarrow f\left( x \right)={{2}^{\dfrac{x}{2}}} \\\ &\Rightarrow f\left( 2 \right)={{2}^{\dfrac{2}{2}}}=2 \\\ &\Rightarrow f\left( 4 \right)={{2}^{\dfrac{4}{2}}}=4 \\\ &\Rightarrow f\left( 6 \right)={{2}^{\dfrac{6}{2}}}=8 \\\ \end{aligned}$$ Put these values in the given expression $$\begin{aligned} & \Rightarrow {{a}^{f\left( 2 \right)}}+{{b}^{f\left( 4 \right)}}+{{c}^{f\left( 6 \right)}}-\dfrac{3{{a}^{f\left( 2 \right)}}{{b}^{f\left( 4 \right)}}{{c}^{f\left( 6 \right)}}}{{{a}^{f\left( 2 \right)}}+{{b}^{f\left( 4 \right)}}+{{c}^{f\left( 6 \right)}}} \\\ & \Rightarrow {{1}^{2}}+{{1}^{4}}+{{1}^{8}}-\dfrac{{{3.1}^{2}}{{.1}^{2}}{{.1}^{8}}}{{{1}^{2}}+{{1}^{2}}+{{1}^{8}}} \\\ & \Rightarrow 3-\dfrac{3}{3} \\\ & \Rightarrow 3-1 \\\ & \Rightarrow 2 \\\ \end{aligned}$$ **Hence the correct option is option A).** **Note:** We should remember range and domain for trigonometric and inverse trigonometric functions. If you don’t remember then you won’t be able to solve this type of question. Students make sure when you have given the equations like $${{\sin }^{-1}}a=\dfrac{\pi }{2}$$ here they are asking for the value of a then we need to multiply it by the $$\sin \left( {{\sin }^{-1}}a \right)=\sin \left( \dfrac{\pi }{2} \right)$$ then we will get the value of $$a=1$$.