1

IIT-JEE 2010 Paper 1 Offline

Numerical
The maximum value of the expression $${1 \over {{{\sin }^2}\theta + 3\sin \theta \cos \theta + 5{{\cos }^2}\theta }}$$ is
Your Input ________

Answer

Correct Answer is 2

Explanation

Let

$$f(\theta ) = {1 \over {{{\sin }^2}\theta + 3\sin \theta \cos \theta + 5{{\cos }^2}\theta }}$$

Again let

$$g(\theta ) = {\sin ^2}\theta + 3\sin \theta \cos \theta + 5{\cos ^2}\theta $$

$$ = {{1 - \cos 2\theta } \over 2} + 5\left( {{{1 + \cos 2\theta } \over 2}} \right) + {3 \over 2}\sin 2\theta $$

$$ = 3 + 2\cos 2\theta + {3 \over 2}\sin 2\theta $$

$$\therefore$$ $$g{(\theta )_{\min }} = 3 - \sqrt {4 + {9 \over 4}} = 3 - {5 \over 2} = {1 \over 2}$$

$$\therefore$$ $$f(\theta ) = {1 \over {g{{(\theta )}_{\min }}}} = 2$$

2

IIT-JEE 2010 Paper 1 Offline

Numerical
The number of values of $$\theta $$ in the interval, $$\left( { - {\pi \over 2},\,{\pi \over 2}} \right)$$ such that$$\,\theta \ne {{n\pi } \over 5}$$ for $$n = 0,\, \pm 1,\, \pm 2$$ and $$\tan \,\theta = \cot \,5\theta \,$$ as well as $$\sin \,2\theta = \cos \,4 \theta $$ is
Your Input ________

Answer

Correct Answer is 3

Explanation

Given, $$\tan \theta = \cot 5\theta $$

$$ \Rightarrow \tan \theta = \tan \left( {{\pi \over 2} - 5\theta } \right)$$

$$ \Rightarrow {\pi \over 2} - 5\theta = n\pi + \theta $$

$$ \Rightarrow 6\theta = {\pi \over 2} - {{n\pi } \over 6}$$

$$ \Rightarrow \theta = {\pi \over {12}} - {{n\pi } \over 6}$$

Also $$\cos 4\theta = \sin 2\theta = \cos \left( {{\pi \over 2} - 2\theta } \right)$$

$$ \Rightarrow 4\theta = 2n\pi \, \pm \,\left( {{\pi \over 2} - 2\theta } \right)$$

Taking positive

$$6\theta = 2n\pi + {\pi \over 2} \Rightarrow \theta = {{n\pi } \over 3} + {\pi \over {12}}$$

Taking negative

$$2\theta = 2n\pi - {\pi \over 2} \Rightarrow \theta = n\pi - {\pi \over 4}$$

Above values of $$\theta$$ suggests that there are only 3 common solutions.

3

IIT-JEE 2010 Paper 1 Offline

Numerical
The number of all possible values of $$\theta $$ where $$0 < \theta < \pi ,$$ for which the system of equations $$$\left( {y + z} \right)\cos {\mkern 1mu} 3\theta = \left( {xyz} \right){\mkern 1mu} \sin 3\theta $$$ $$$x\sin 3\theta = {{2\cos 3\theta } \over y} + {{2\sin 3\theta } \over z}$$$ $$$\left( {xyz} \right){\mkern 1mu} \sin 3\theta = \left( {y + 2z} \right){\mkern 1mu} \cos 3\theta + y{\mkern 1mu} sin3\theta $$$

have a solution $$\left( {{x_0},{y_0},{z_0}} \right)$$ with $${y_0}{z_0}{\mkern 1mu} \ne {\mkern 1mu} 0,$$ is

Your Input ________

Answer

Correct Answer is 3

Explanation

View the equation in xyz, y and t.

We have,

$$(xyz)\sin 3\theta - y\cos 3\theta - z\cos 3\theta = 0$$

$$(xyz)\sin 3\theta - 2y\sin 3\theta - 2z\cos 3\theta = 0$$

$$(xyz)\sin 3\theta - y(\cos 3\theta + \sin 3\theta ) - 2z\cos 3\theta = 0$$

$$xyz \ne 0$$

Hence, the equation has non-trivial solution which gives

$$\left| {\matrix{ {\sin 3\theta } & { - \cos 3\theta } & { - \cos 3\theta } \cr {\sin 3\theta } & { - 2\sin 3\theta } & { - 2\cos 3\theta } \cr {\sin 3\theta } & { - (\cos 3\theta + \sin 3\theta )} & { - 2\cos 3\theta } \cr } } \right| = 0$$

$$ \Rightarrow \sin 3\theta \cos 3\theta (\sin 3\theta - \cos 3\theta ) = 0$$

$$ \Rightarrow \sin 3\theta = 0$$ then $$xyz = 0$$ (not possible)

$$\cos 3\theta = 0$$ not possible

$$\sin 3\theta = \cos 3\theta \Rightarrow \tan 3\theta = 1$$

$$3\theta = n\pi + {\pi \over 4},n \in z$$

$$\theta = {{n\pi } \over 3} + {\pi \over {12}}$$ ; $$\theta = {\pi \over {12}},{{5\pi } \over {12}},{{9\pi } \over {12}}$$

Thus there are 3 solutions.

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