JEE Advance - Mathematics (2010 - Paper 2 Offline - No. 1)
$$\,\,\,\,$$ $$\,\,\,\,$$ $$\,\,\,\,$$ Column-$$I$$
(A)$$\,\,\,\,$$ A line from the origin meets the lines $$\,{{x - 2} \over 1} = {{y - 1} \over { - 2}} = {{z + 1} \over 1}$$
and $${{x - {8 \over 3}} \over 2} = {{y + 3} \over { - 1}} = {{z - 1} \over 1}$$ at $$P$$ and $$Q$$ respectively. If length $$PQ=d,$$ then $${d^2}$$ is
(B)$$\,\,\,\,$$ The values of $$x$$ satisfying $${\tan ^{ - 1}}\left( {x + 3} \right) - {\tan ^{ - 1}}\left( {x - 3} \right) = {\sin ^{ - 1}}\left( {{3 \over 5}} \right)$$ are
(C)$$\,\,\,\,$$ Non-zero vectors $$\overrightarrow a ,\overrightarrow b $$ and $$\overrightarrow c \,\,$$ satisfy $$\overrightarrow a \,.\,\overrightarrow b \, = 0.$$
$$\left( {\overrightarrow b - \overrightarrow a } \right).\left( {\overrightarrow b + \overrightarrow c } \right) = 0$$ and $$2\left| {\overrightarrow b + \overrightarrow c } \right| = \left| {\overrightarrow b - \overrightarrow a } \right|.$$
If $$\overrightarrow a = \mu \overrightarrow b + 4\overrightarrow c \,\,,$$ then the possible values of $$\mu $$ are
(D)$$\,\,\,\,$$ Let $$f$$ be the function on $$\left[ { - \pi ,\pi } \right]$$ given by $$f(0)=9$$
and $$f\left( x \right) = \sin \left( {{{9x} \over 2}} \right)/\sin \left( {{x \over 2}} \right)$$ for $$x \ne 0$$
The value of $${2 \over \pi }\int_{ - \pi }^\pi {f\left( x \right)dx} $$ is
$$\,\,\,\,$$ $$\,\,\,\,$$ $$\,\,\,\,$$Column-$$II$$
(p)$$\,\,\,\,$$ $$-4$$
(q)$$\,\,\,\,$$ $$0$$
(r)$$\,\,\,\,$$ $$4$$
(s)$$\,\,\,\,$$ $$5$$
(t)$$\,\,\,\,$$ $$6$$
Explanation
(A) Equation of the line passing through origin is
$${x \over a} = {y \over b} = {z \over c}$$
$$\therefore$$ $$\left| {\matrix{ 2 & 1 & { - 1} \cr 1 & { - 2} & 1 \cr a & b & c \cr } } \right| = 0$$
$$ \Rightarrow a( - 1) - b(3) + c( - 5) = 0$$
$$ \Rightarrow - a - 3b - 5c = 0$$
$$ \Rightarrow a + 3b + 5c = 0$$ ....... (i)
Also, $$\left| {\matrix{ {{8 \over 3}} & { - 3} & 1 \cr 2 & { - 1} & 1 \cr a & b & c \cr } } \right| = 0$$
$$\therefore$$ $$a( - 2) - b\left( {{2 \over 3}} \right) + c\left( {{{10} \over 3}} \right) = 0$$
$$ \Rightarrow 2a + {{2b} \over 3} - {{10c} \over 3} = 0$$
$$3a + b - 5c = 0$$ ...... (ii)
From Eqs. (i) and (ii),
$${a \over { - 20}} = {b \over {20}} = {c \over { - 8}}$$
$${a \over 5} = {b \over { - 5}} = {c \over 4}$$
Equation of line is
$${x \over 5} = {y \over { - 5}} = {z \over 4} = \lambda $$ (say) ....... (iii)
Also, $${{x - 2} \over 1} = {{y - 1} \over 2} = {{z + 1} \over 1} = {k_1}$$ (say) ...... (iv)
Now, $${{x - {8 \over 3}} \over 2} = {{y + 3} \over { - 1}} = {{z - 1} \over 1} = {k_2}$$ (say) ...... (v)
Point on (iii) is $$(5\lambda ,\, - 5\lambda ,\, + 4\lambda )$$
Point on (iv) is $$(2 + {k_1},\,1 - 2{k_1},\, - 1 + {k_1})$$
Point on (v) is $$\left( {{8 \over 3} + 2{k_2},\, - 3 - {k_2},\,1 + {k_2}} \right)$$
On solving, $$2 + {k_1} + 1 - 2{k_1} = 0$$
$$ - {k_1} + 3 = 0$$
$${k_1} = 3$$
$$P \equiv (5,\, - 5,\,2)$$
Again, for Q
$${8 \over 3} + 2{k_2} - 3 - {k_2} = 0$$
$${k_2} - {1 \over 3} = 0$$
$${k_2} = {1 \over 3}$$
$$Q \equiv \left( {{{10} \over 3},{{ - 10} \over 3},{4 \over 3}} \right)$$
Now, $$PQ = \sqrt {{{\left( {{5 \over 3}} \right)}^2} + {{\left( {{5 \over 3}} \right)}^2} + {{\left( {{2 \over 3}} \right)}^2}} $$
$$ = {{\sqrt {54} } \over 3}$$
$$P{Q^2} = {d^2} = {{54} \over 9} = 6$$
(B) $${\tan ^{ - 1}}\left( {{{x + 3 - x + 3} \over {1 + ({x^2} - 9)}}} \right) = {\tan ^{ - 1}}\left( {{3 \over 4}} \right)$$
$$ \Rightarrow {6 \over {{x^2} - 8}} = {3 \over 4}$$
$$ \Rightarrow 3{x^2} = 48$$
$$ \Rightarrow x = \, \pm \,4$$
(C) $$\left( {\overrightarrow b - \overrightarrow a } \right)\,.\,\left( {\overrightarrow b + {{\overrightarrow a - \mu \overrightarrow b } \over 4}} \right) = 0$$
$$ \Rightarrow \left( {\overrightarrow b - \overrightarrow a } \right)\,.\,\left( {4\overrightarrow b + \overrightarrow a - \mu \overrightarrow b } \right) = 0$$
$$\left( {4 - \mu } \right){\overrightarrow b ^2} - {\overrightarrow a ^2} = 0$$ ...... (i)
Also, $$2\left| {\overrightarrow b + {{\overrightarrow a - \mu \overrightarrow b } \over 4}} \right| = \left| {\overrightarrow b - \overrightarrow a } \right|$$
$$ \Rightarrow 2\left| {{{(4 - \mu )\overrightarrow b + \overrightarrow a } \over 4}} \right| = \left| {\overrightarrow b - \overrightarrow a } \right|$$
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