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# If we are given a trigonometric expression as $f\left( x \right)={{\cos }^{-1}}\left( \dfrac{2\cos x-3\sin x}{\sqrt{13}} \right)$ then, $\left[ f'{{\left( x \right)}^{2}} \right]$ is equal to (a) $\sqrt{1+x}$ (b) 1 + 2x(c) 2(d) 1(e) 0

Last updated date: 22nd Mar 2024
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Hint: We will apply here the formula of basic identity in trigonometry which is given by ${{\sin }^{2}}\left( p \right)+{{\cos }^{2}}\left( p \right)=1$ or ${{\sin }^{2}}\left( p \right)=1-{{\cos }^{2}}\left( p \right)$. Also we will use the formula $\cos \left( p \right)\cos \left( x \right)-\sin \left( p \right)\sin \left( x \right)=\cos \left( p+x \right)$ in order to solve the question further.

We will consider $f\left( x \right)={{\cos }^{-1}}\left( \dfrac{2\cos x-3\sin x}{\sqrt{13}} \right)$...(i).
We can also write this equation as $f\left( x \right)={{\cos }^{-1}}\left( \dfrac{2\cos x}{\sqrt{13}}-\dfrac{3\sin x}{\sqrt{13}} \right)$ or $f\left( x \right)={{\cos }^{-1}}\left( \dfrac{2}{\sqrt{13}}\cos x-\dfrac{3}{\sqrt{13}}\sin x \right)$.
Now, we will substitute the value $\cos \left( p \right)=\dfrac{2}{\sqrt{13}}$. This results in $p={{\cos }^{-1}}\left( \dfrac{2}{\sqrt{13}} \right)$.
As we know that $\cos \left( p \right)\cos \left( x \right)-\sin \left( p \right)\sin \left( x \right)=\cos \left( p+x \right)$ therefore, to convert the equation $f\left( x \right)={{\cos }^{-1}}\left( \dfrac{2}{\sqrt{13}}\cos x-\dfrac{3}{\sqrt{13}}\sin x \right)$ into $\cos \left( p \right)\cos \left( x \right)-\sin \left( p \right)\sin \left( x \right)=\cos \left( p+x \right)$ form we will need to find the value of $\sin \left( p \right)$.
For this we will apply the formula of basic identity in trigonometry. The formula is given by ${{\sin }^{2}}\left( p \right)+{{\cos }^{2}}\left( p \right)=1$ or ${{\sin }^{2}}\left( p \right)=1-{{\cos }^{2}}\left( p \right)$. Now we will substitute the value of $\cos \left( p \right)=\dfrac{2}{\sqrt{13}}$ into this formula. Therefore, we get \begin{align} & {{\sin }^{2}}\left( p \right)=1-{{\cos }^{2}}\left( p \right) \\ & \Rightarrow {{\sin }^{2}}\left( p \right)=1-{{\left( \dfrac{2}{\sqrt{13}} \right)}^{2}} \\ & \Rightarrow {{\sin }^{2}}\left( p \right)=1-\dfrac{4}{13} \\ & \Rightarrow {{\sin }^{2}}\left( p \right)=\dfrac{13-4}{13} \\ & \Rightarrow {{\sin }^{2}}\left( p \right)=\dfrac{9}{13} \\ & \Rightarrow \sin \left( p \right)=\sqrt{\dfrac{9}{13}} \\ & \Rightarrow \sin \left( p \right)=\pm \dfrac{3}{\sqrt{13}} \\ \end{align}
Now, we have two values of sine given by $\sin \left( p \right)=\dfrac{3}{\sqrt{13}}$ and $\sin \left( p \right)=-\dfrac{3}{\sqrt{13}}$.
We will reject the negative value here as it will not satisfy the formula ${{\sin }^{2}}\left( p \right)+{{\cos }^{2}}\left( p \right)=1$. Therefore, we now have $\cos \left( p \right)=\dfrac{2}{\sqrt{13}}$ and $\sin \left( p \right)=\dfrac{3}{\sqrt{13}}$.
Now we will substitute these values in equation $f\left( x \right)={{\cos }^{-1}}\left( \dfrac{2}{\sqrt{13}}\cos x-\dfrac{3}{\sqrt{13}}\sin x \right)$. Thus, we get
$f\left( x \right)={{\cos }^{-1}}\left( \cos p\cos x-\sin p\sin x \right)$
At this step we will use the formula given by $\cos \left( p \right)\cos \left( x \right)-\sin \left( p \right)\sin \left( x \right)=\cos \left( p+x \right)$. This results into $f\left( x \right)={{\cos }^{-1}}\left( \cos \left( p+x \right) \right)$. As we know that ${{\cos }^{-1}}\left( \cos p \right)=p$ therefore, we get $f\left( x \right)=p+x$.
Now, we will differentiate this equation with respect to x. So, we now get \begin{align} & \dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{d}{dx}\left( p+x \right) \\ & \Rightarrow \dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{d}{dx}p+\dfrac{d}{dx}x \\ \end{align}
As we know that $\dfrac{dx}{dx}=1$ therefore we get $\dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{d}{dx}p+1$. Also, as we know that p works as a constant here because we are differentiating with respect to x only therefore, by the fact that differentiating a constant with respect to x, we will use the formula $\dfrac{dp}{dx}=0$ in the equation $\dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{d}{dx}p+1$.
Therefore, we get $\dfrac{d}{dx}\left( f\left( x \right) \right)=1$. As we know that $\dfrac{d}{dx}\left( f\left( x \right) \right)=f'\left( x \right)$ therefore, we have $f'\left( x \right)=1$.
Now, we will apply the square on both sides of the equation $f'\left( x \right)=1$. Thus we get ${{\left[ f'\left( x \right) \right]}^{2}}={{\left( 1 \right)}^{2}}$ or $\left[ f'{{\left( x \right)}^{2}} \right]=1$.
Therefore, the correct option is (d).

Note: As we have substituted $\cos \left( p \right)=\dfrac{2}{\sqrt{13}}$ in equation $f\left( x \right)={{\cos }^{-1}}\left( \dfrac{2}{\sqrt{13}}\cos x-\dfrac{3}{\sqrt{13}}\sin x \right)$ Instead of this we can also substitute $\sin \left( p \right)=\dfrac{3}{\sqrt{13}}$ and get $\cos \left( p \right)=\dfrac{2}{\sqrt{13}}$ after solving. We can also chose the form of the $\sin \left( p \right)\cos \left( x \right)-\cos \left( p \right)\sin \left( x \right)=\sin \left( p-x \right)$ instead of $\cos \left( p \right)\cos \left( x \right)-\sin \left( p \right)\sin \left( x \right)=\cos \left( p+x \right)$ for the conversion of equation (i). But as we can clearly see that there is an inverse cosine term on the outside of the expression also we know that ${{\cos }^{-1}}\left( \cos p \right)=p$ therefore we have chosen $\cos \left( p \right)\cos \left( x \right)-\sin \left( p \right)\sin \left( x \right)=\cos \left( p+x \right)$ as for the conversion in this question.