Question

$\begin{array}{cccccc}& \text{Column 1}& & \text{Column 2}& & \text{Column 3}\\ & & & & & f\left(x\right)=3x^3-13x^2+14x-2\text{has}\\ \text{(I)}& \cos \left({\sin }^{-1}\left(\sin \frac{5\pi }{6}\right)\right)& \left(i\right)& 1& \text{P}& \text{three roots}\alpha ,\beta ,\gamma ,\text{then value of}\\ & & & & & |\sin ({\tan }^{-1}\alpha +{\tan }^{-1}\beta +{\tan }^{-1}\gamma )|=\\ \text{(II)}& \cos \left({\cos }^{-1}(-\sin \frac{7\pi }{6})\right)=& \text{(ii)}& \frac{1}{\sqrt{2}}& \text{Q}& \frac{(\sqrt{3}\csc {20}^{\circ }-\sec {20}^{\circ })}{8}=\\ \text{(III)}& \left|\cos \left\{{\tan }^{-1}(-\tan \frac{15\pi }{4})\right\}\right|=& \text{(iii)}& \frac{1}{2}& \text{R}& 4\sqrt{2}\left(\sin {12}^{\circ }\right)\left(\sin {48}^{\circ }\right)\left(\sin {54}^{\circ }\right)=\\ \text{(IV)}& \sin ({({\cos }^{-1}\left\{\frac{1}{\sqrt{2}}(\cos \frac{9\pi }{10}-\sin \frac{9\pi }{10})\right\}+\frac{7\pi }{20})}^{-1}\times {\pi }^2)& \text{(iv)}& \frac{\sqrt{3}}{2}& \text{S}& \text{If the angles A, B and C of a}\\ & & & & & \text{triangle are in an arithmetic}\\ & & & & & \text{progression and if a, b and c denote}\\ & & & & & \text{the lengths of sides opposite to A, B}\\ & & & & & \text{and C respectively, then the value}\\ & & & & & \text{of the expression}\\ & & & & & \frac{1}{2}(\frac{a}{c}\sin 2C+\frac{c}{a}\sin 2A)\text{is}\end{array}$

Which of the following options is only INCORRECT combination ?

• A. (III), (ii), (P)
• B. (IV), (iii), (Q)
• C. (III), (ii), (R)
• D. (IV), (iii), (P)

Answer 1

The correct option is D (IV), (iii), (P)

$\text{(I)}\cos \left({\sin }^{-1}\left(\sin \frac{5\pi }{6}\right)\right)=\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}\text{(II)}\cos \left({\cos }^{-1}(-\sin \frac{7\pi }{6})\right)=\cos \left(\frac{\pi }{3}\right)=\frac{1}{2}\text{(III)}\left|\cos \left({\tan }^{-1}(-\tan \frac{15\pi }{4})\right)\right|=\left|\cos (-\frac{\pi }{4})\right|=\frac{1}{\sqrt{2}}\text{(IV)}\sin ({({\cos }^{-1}\left\{\frac{1}{\sqrt{2}}(\cos \frac{9\pi }{10}-\sin \frac{9\pi }{10})\right\}+\frac{7\pi }{20})}^{-1}\times {\pi }^2)=\sin ({({\cos }^{-1}\left(\cos \frac{23\pi }{20}\right)+\frac{7\pi }{20})}^{-1}\times {\pi }^2)=\sin ({(2\pi -\frac{23\pi }{20}+\frac{7\pi }{20})}^{-1}\times {\pi }^2)=\sin ({\left(\frac{24\pi }{20}\right)}^{-1}\times {\pi }^2)=\sin \left(\frac{5\pi }{6}\right)=\frac{1}{2}$
$\text{(P)}f\left(x\right)=3x^3-13x^2+14x-2\alpha ,\beta ,\gamma \text{are the roots of}f\left(x\right)\therefore \alpha +\beta +\gamma =\frac{13}{3}\alpha \beta +\beta \gamma +\gamma \alpha =\frac{14}{3}\alpha \beta \gamma =\frac{2}{3}$

$\therefore \left|\sin \right({\tan }^{-1}\alpha +{\tan }^{-1}\beta +{\tan }^{-1}\gamma \left)\right|=\left|\sin \left({\tan }^{-1}\left(\frac{\frac{13}{3}-\frac{2}{3}}{1-\frac{14}{3}}\right)\right)\right|=\left|\sin \right({\tan }^{-1}(-1)\left)\right|=\frac{1}{\sqrt{2}}$

$\text{(Q)}\frac{\sqrt{3}\csc {20}^{\circ }-\sec {20}^{\circ }}{8}=\frac{\frac{\sqrt{3}}{2}\cos {20}^{\circ }-\frac{1}{2}\sin {20}^{\circ }}{4\cos {20}^{\circ }\sin {20}^{\circ }}=\frac{\sin {60}^{\circ }\cos {20}^{\circ }-\cos {60}^{\circ }\sin {20}^{\circ }}{2\left(2\sin {20}^{\circ }\cos {20}^{\circ }\right)}=\frac{\sin {40}^{\circ }}{2\sin {40}^{\circ }}=\frac{1}{2}$

$\text{(R)}4\sqrt{2}\left(\sin {12}^{\circ }\right)\left(\sin {48}^{\circ }\right)\left(\sin {54}^{\circ }\right)=2\sqrt{2}\left(2\sin {12}^{\circ }\sin {48}^{\circ }\right)\sin {54}^{\circ }=2\sqrt{2}[\cos {36}^{\circ }-\cos {60}^{\circ }]\sin {54}^{\circ }=2\sqrt{2}[\cos {36}^{\circ }-\frac{1}{2}]\sin {54}^{\circ }=2\sqrt{2}[{\left(\frac{\sqrt{5}+1}{4}\right)}^2-\frac{1}{2}\left(\frac{\sqrt{5}+1}{4}\right)]=\frac{1}{\sqrt{2}}$


$\text{(S)}\text{Since}a,b,c\text{are in A.P.}\therefore \angle B={60}^{\circ }$
$\therefore \frac{1}{2}(\frac{a}{c}\sin 2C+\frac{c}{a}(\sin 2A)=\frac{1}{2}\left(\frac{a}{c}\right(2\sin C\cos C)+\frac{c}{a}(2\sin A\cos A\left)\right)=\frac{\sin C}{c}\left(a\cos C\right)+\frac{\sin A}{a}\left(c\cos A\right)=\frac{1}{2R}(a\cos C+c\cos A)[\because \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R]=\frac{b}{2R}=\sin B=\frac{\sqrt{3}}{2}$