Mathematics Test - 32

\(\operatorname{Let} A=\sqrt{\frac{a(a+b+c)}{b c}}\)
\(A B C=\frac{a+b+c}{a b c}\)
\(A+B+C=\sqrt{\frac{a+b+c}{a b c}}(a+b+c)\)
\(A B=\frac{a+b+c}{c}\)
\(B C=\frac{a+b+c}{a}\)
\(\rightarrow \tan ^{-1} A+\tan ^{-1} B+\tan ^{-1} C\)
\(=\tan ^{-1}\left(\frac{A+B}{1-A B}\right)+\tan ^{-1} C\)
\(=\tan ^{-1}\left(\frac{A+B}{1-A B}\right)+\tan ^{-1} C\)
\(\tan ^{-1} \frac{\left(\frac{A+B}{1-A B}+C\right)}{1-\left(\frac{A C+B C}{1-A B}\right)}\)
\(=\tan ^{-1}\left(\frac{A+B+C-A B C}{1-A B-B C-A C}\right)\)
\(=\tan ^{-1} \frac{\left(\frac{a+b+c}{a b c}-(a+b+c) \sqrt{\frac{a+b+c}{a b c}}\right)}{1-(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\)

Principle value of \(\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
\(=\cos ^{-1}\left(\cos \left(\pi-\frac{\pi}{4}\right)\right)\)
\(=\pi-\frac{\pi}{4}\)
\(=\frac{3 \pi}{4}\)
Hence, the correct option is (A)

We have \(\tan ^{-1}\left(\cot \frac{3 \pi}{4}\right)\)
\(=\tan ^{-1}\left(\cot \left(\frac{\pi}{2}+\frac{\pi}{4}\right)\right)\)
\(=\tan ^{-1}\left(-\tan \frac{\pi}{4}\right)\)
\(\tan ^{-1}(-1)=-\frac{\pi}{4}\)
Hence, the correct option is (C)

Rearrange the expression as, \(\left(\frac{1+\cos ^{4} x}{\cos ^{2} x}\right)\left(2+\cot ^{2} y\right)(4+\sin 4 z) \leq 12\)
\(\therefore\left(\cos ^{2} x+\sec ^{2} x\right)\left(2+\cot ^{2} y\right)(4+\sin 4 z) \leq 12 \ldots \ldots \ldots \ldots \ldots .(1)\)
\(\left(\cos ^{2} x+\sec ^{2} x\right) \geq 2 \ldots \ldots \ldots \ldots \ldots \ldots .(A . M \geq G . M)\)
\(\left(2+\cot ^{2} y\right) \geq 2 \ldots \ldots \ldots \ldots \ldots \ldots .\left(\cot ^{2} y \geq 0\right)\)
\((4+\sin 4 z) \geq 3 \ldots \ldots \ldots \ldots \ldots \ldots \ldots(\sin 4 z \geq-1)\)
\(\therefore\left(\cos ^{2} x+\sec ^{2} x\right)\left(2+\cot ^{2} y\right)(4+\sin 4 z) \geq 12 \ldots \ldots \ldots \ldots \ldots \ldots .(2)\)
From (1) and (2) it implies that only limiting case is possible. Therefore, \(\cos ^{2} x=\sec ^{2} x=1 \rightarrow x=0\) or \(x=\pi\) or \(x=2 \pi\) gives 3 values.
and \(\cot ^{2} y=0 \rightarrow x=\frac{\pi}{2}\) or \(x=\frac{3 \pi}{2}\) gives 2 values.
and \(\sin 4 z=-1 \rightarrow z=\frac{3 \pi}{8}\) or \(z=\frac{7 \pi}{8}\) or \(z=\frac{11 \pi}{8}\) or \(z=\frac{15 \pi}{8}\) gives 4 values.
\(\therefore\) Total number of triplets is \(3 \times 2 \times 4\) Hence, answer is 24
Hence, the correct option is (D)

Substituting \(x=\tan A\) \(\tan ^{-1}\left(\frac{\sec A-1}{\tan A}\right)\)
\(=\tan ^{-1}\left(\frac{1-\cos A}{\sin A}\right)\)
\(=\tan ^{-1}\left(\tan \left(\frac{A}{2}\right)\right)\)
\(4^{0}=\frac{A}{2}\)
\(\Rightarrow A=8^{0}\)
\(\tan ^{-1}(x)=8^{0}\)
\(x=\tan \left(8^{0}\right)\)
Hence, the correct option is (D)

Case \(\mathrm{I}: \frac{x^{2}-1}{2 x}>0\)
\(\Rightarrow x \in(-1,0) \cup(1, \infty)\)
Given \(\Rightarrow 2 \tan ^{-1}\left(\frac{2 x}{x^{2}-1}\right)=\frac{2 \pi}{3}\)
\(\frac{2 x}{x^{2}-1}=\sqrt{3} \Rightarrow x=-\frac{1}{\sqrt{3}}, \sqrt{3}\)
Case II: \(\frac{x^{2}-1}{2 x}<0\)
\(\Rightarrow x \in(-\infty,-1) \cup(0,1)\)
Given \(\Rightarrow \pi+2 t a n^{-1}\left(\frac{2 x}{x^{2}-1}\right)=-\frac{1}{\sqrt{3}} \Rightarrow x=\pm 2-\sqrt{3}\)
\(\therefore\) four solutions
Hence, the correct option is (A)

Consider \(p=\frac{1}{2}\)
Therefore the above expression reduces to \(\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)+\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)+\cos ^{-1}(\sqrt{1-q})\)
\(=\frac{\pi}{4}+\frac{\pi}{4}+\cos ^{-1}(\sqrt{1-q})\)
\(=\frac{2 \pi}{4}+\cos ^{-1}(\sqrt{1-q})\)
\(=\frac{3 \pi}{4}\)
Therefore \(\cos ^{-1}(\sqrt{1-q})=\frac{\pi}{4}\)
\(\Rightarrow \sqrt{1-q}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow 2-2 q=1\)
\(\Rightarrow q=\frac{1}{2}\)
Hence, the correct option is (C)

Substituting \(\boldsymbol{x}=\tan A\)
Hence \(\cos ^{-1}\left(\frac{1-\tan ^{2} A}{1+\tan ^{2} A}\right)+\sin ^{-1}\left(\frac{2 \tan A}{1+\tan ^{2} A}\right)\)
\(=\cos ^{-1}(\cos 2 A)+\sin ^{-1}(\sin 2 A)\)
Now, \(x \in[-1,0]\)
Hence \(A \in\left[\frac{-\pi}{4}, 0\right]\)
Therefore A lies in the fourth quadrant. Hence \(\cos ^{-1}(\cos 2 A)\)
\(=2 A\)
And \(\sin ^{-1}(\sin 2 A)\)
\(=-2 A\)
Therefore \(K=2 A-2 A\)
\(=0\)
Hence, the correct option is (C)

\(\lim _{x \rightarrow 1}\left[\sec \left(\frac{\pi x}{2}\right) \log x\right]\)
\(=\lim _{x \rightarrow 1} \frac{\log x}{\cos \left(\frac{\pi x}{2}\right)}\)
It is of the form \(\frac{0}{0},\) So applying L-Hospital's rule, \(=\lim _{x \rightarrow 1} \frac{\frac{1}{x}}{-\frac{\pi}{2} \sin \left(\frac{\pi x}{2}\right)}\)
\(=-\frac{2}{\pi}\)
Hence, the correct option is (A)