39. Match the following
(i) Two rays in the first quadrant $x + y = | a |$ and ax $- y = 1$ intersects each other in the interval $a \in \left( a _ { 0 } , \infty \right)$, the value of $a _ { 0 }$ is
(A) 2
(ii) Point ( $\alpha , \beta , \gamma$ ) lies on the plane $\mathrm { x } + \mathrm { y } + \mathrm { z } = 2$. Let $\overrightarrow { \mathrm { a } } = \alpha \hat { \mathrm { i } } + \beta \hat { \mathrm { j } } + \gamma \hat { \mathrm { k } } , \hat { \mathrm { k } } \times ( \hat { \mathrm { k } } \times \overrightarrow { \mathrm { a } } ) = 0$, then $\gamma =$.
(B) $4 / 3$
(iii) $\left| \int _ { 0 } ^ { 1 } \left( 1 - y ^ { 2 } \right) d y \right| + \left| \int _ { 1 } ^ { 0 } \left( y ^ { 2 } - 1 \right) d y \right|$
(C) $\left| \int _ { 0 } ^ { 1 } \sqrt { 1 - \mathrm { x } } \mathrm { dx } \right| + \left| \int _ { - 1 } ^ { 0 } \sqrt { 1 + \mathrm { x } } \mathrm { dx } \right|$
(iv) If $\sin \mathrm { A } \sin \mathrm { B } \sin \mathrm { C } + \cos \mathrm { A } \cos \mathrm { B } = 1$, then the value of $\sin \mathrm { C } =$
(D) 1
Sol. (i) Solving the two equations of ray i.e. $x + y = | a |$ and $a x - y = 1$ we get $x = \frac { | a | + 1 } { a + 1 } > 0$ and $y = \frac { | a | - 1 } { a + 1 } > 0$ when $\mathrm { a } + 1 > 0$; we get $\mathrm { a } > 1 \quad \therefore \mathrm { a } _ { 0 } = 1$.
(ii) We have $\overrightarrow { \mathrm { a } } = \alpha \hat { \mathrm { i } } + \beta \hat { \mathrm { j } } + \gamma \hat { \mathrm { k } } \Rightarrow \overrightarrow { \mathrm { a } } \cdot \hat { \mathrm { k } } = \gamma$
Now; $\hat { \mathrm { k } } \times ( \hat { \mathrm { k } } \times \hat { \mathrm { a } } ) = ( \hat { \mathrm { k } } \cdot \overrightarrow { \mathrm { a } } ) \hat { \mathrm { k } } - ( \hat { \mathrm { k } } \cdot \hat { \mathrm { k } } ) \overrightarrow { \mathrm { a } }$ $= \gamma \hat { \mathrm { k } } - ( \alpha \hat { \mathrm { i } } + \beta \hat { \mathrm { j } } + \gamma \hat { \mathrm { k } } )$ $= \alpha \hat { \mathrm { i } } + \beta \hat { \mathrm { j } } = \overrightarrow { 0 } \quad \Rightarrow \alpha = \beta = 0$ As $\alpha + \beta + \gamma = 2 \Rightarrow \gamma = 2$.
(iii) $\quad \left| \int _ { 0 } ^ { 1 } \left( 1 - y ^ { 2 } \right) d y \right| + \left| \int _ { 1 } ^ { 0 } \left( y ^ { 2 } - 1 \right) d y \right|$ $= 2 \int _ { 0 } ^ { 1 } \left( 1 - y ^ { 2 } \right) d y = \frac { 4 } { 3 }$ $\left| \int _ { 0 } ^ { 1 } \sqrt { 1 - \mathrm { x } } \mathrm { dx } \right| + \left| \int _ { - 1 } ^ { 0 } \sqrt { 1 + \mathrm { x } } \mathrm { dx } \right| = 2 \int _ { 0 } ^ { 1 } \sqrt { 1 - \mathrm { x } } \mathrm { dx }$ $= 2 \int _ { 0 } ^ { 1 } \sqrt { \mathrm { x } } \mathrm { d } x = \left. 2 \cdot \frac { 2 } { 3 } \cdot \mathrm { x } ^ { 3 / 2 } \right| _ { 0 } ^ { 1 } = \frac { 4 } { 3 }$.
(iv) $\quad \sin \mathrm { A } \sin \mathrm { B } \sin \mathrm { C } + \cos \mathrm { A } \cos \mathrm { B } \leq \sin \mathrm { A } \sin \mathrm { B } + \cos \mathrm { A } \cos \mathrm { B } = \cos ( \mathrm { A } - \mathrm { B } )$
$$\Rightarrow \quad \cos ( \mathrm { A } - \mathrm { B } ) \geq 1 \Rightarrow \cos ( \mathrm {~A} - \mathrm { B } ) = 1 \Rightarrow \sin \mathrm { C } = 1$$
- Match the following
(i) $\sum _ { i = 1 } ^ { \infty } \tan ^ { - 1 } \left( \frac { 1 } { 2 i ^ { 2 } } \right) = t$, then $\tan t =$
(A) 0
(ii) Sides $\mathrm { a } , \mathrm { b } , \mathrm { c }$ of a triangle ABC are in AP and
$$\cos \theta _ { 1 } = \frac { \mathrm { a } } { \mathrm {~b} + \mathrm { c } } , \cos \theta _ { 2 } = \frac { \mathrm { b } } { \mathrm { a } + \mathrm { c } } , \cos \theta _ { 3 } = \frac { \mathrm { c } } { \mathrm { a } + \mathrm { b } } , \text { then } \tan ^ { 2 } \left( \frac { \theta _ { 1 } } { 2 } \right) + \tan ^ { 2 } \left( \frac { \theta _ { 3 } } { 2 } \right) = \quad \text { (B) } 1$$
(iii) A line is perpendicular to $x + 2 y + 2 z = 0$ and passes through $( 0,1,0 )$.
(C) $\frac { \sqrt { 5 } } { 3 }$
The perpendicular distance of this line from the origin is
(D) $2 / 3$
(iv) Data could not be retrieved.
Sol. (i) $\quad \sum _ { \mathrm { i } = 1 } ^ { \infty } \tan ^ { - 1 } \left[ \frac { 1 } { 2 \mathrm { i } ^ { 2 } } \right] = \mathrm { t }$
$$\begin{aligned}
\text { Now } & ; \sum _ { \mathrm { i } = 1 } ^ { \infty } \tan ^ { - 1 } \left[ \frac { 2 } { 4 \mathrm { i } ^ { 2 } - 1 + 1 } \right] \\
= & \sum _ { \mathrm { i } = 1 } ^ { \infty } \left[ \tan ^ { - 1 } ( 2 \mathrm { i } + 1 ) - \tan ^ { - 1 } ( 2 \mathrm { i } - 1 ) \right] \\
= & { \left[ \left( \tan ^ { - 1 } 3 - \tan ^ { - 1 } 1 \right) + \left( \tan ^ { - 1 } 5 - \tan ^ { - 1 } 3 \right) + \cdots + \tan ^ { - 1 } ( 2 \mathrm { n } + 1 ) - \tan ^ { - 1 } ( 2 \mathrm { n } - 1 ) \ldots . . \infty \right] } \\
& \mathrm { t } = \tan ^ { - 1 } ( 2 \mathrm { n } + 1 ) - \tan ^ { - 1 } 1 = \lim _ { \mathrm { n } \rightarrow \infty } \tan ^ { - 1 } \frac { 2 \mathrm { n } } { 1 + ( 2 \mathrm { n } + 1 ) } \\
\Rightarrow \quad & \tan \mathrm { t } = \lim _ { \mathrm { n } \rightarrow \infty } \frac { \mathrm { n } } { \mathrm { n } + 1 } \Rightarrow \mathrm { t } = \frac { \pi } { 4 }
\end{aligned}$$
(ii) We have $\cos \theta _ { 1 } = \frac { 1 - \tan ^ { 2 } \frac { \theta _ { 1 } } { 2 } } { 1 + \tan ^ { 2 } \frac { \theta _ { 1 } } { 2 } } = \frac { \mathrm { a } } { \mathrm { b } + \mathrm { c } } \Rightarrow \tan ^ { 2 } \frac { \theta _ { 1 } } { 2 } = \frac { \mathrm { b } + \mathrm { c } - \mathrm { a } } { \mathrm { b } + \mathrm { c } + \mathrm { a } }$
Also, $\cos \theta _ { 3 } = \frac { 1 - \tan ^ { 2 } \frac { \theta _ { 3 } } { 2 } } { 1 + \tan ^ { 2 } \frac { \theta _ { 3 } } { 2 } } = \frac { \mathrm { c } } { \mathrm { a } + \mathrm { b } } \Rightarrow \tan ^ { 2 } \frac { \theta _ { 3 } } { 2 } = \frac { \mathrm { a } + \mathrm { b } - \mathrm { c } } { \mathrm { a } + \mathrm { b } + \mathrm { c } }$ $\therefore \quad \tan ^ { 2 } \frac { \theta _ { 1 } } { 2 } + \tan ^ { 2 } \frac { \theta _ { 3 } } { 2 } = \frac { 2 \mathrm {~b} } { 3 \mathrm {~b} } = \frac { 2 } { 3 }$
(iii) Line through $( 0,1,0 )$ and perpendicular to plane $x + 2 y + 2 z = 0$ is given by $\frac { x - 0 } { 1 } = \frac { y - 1 } { 2 } = \frac { z - 1 } { 2 } = r$.
Let $\mathrm { P } ( \mathrm { r } , 2 \mathrm { r } + 1,2 \mathrm { r } )$ be the foot of perpendicular on the straight line then
$$r \times 1 + ( 2 r + 1 ) 2 + 2 \times 2 r = 0 \Rightarrow r = - \frac { 2 } { 9 }$$
$\therefore \quad$ Point is given by $\left( - \frac { 2 } { 9 } , \frac { 5 } { 9 } , - \frac { 4 } { 9 } \right)$ $\therefore \quad$ Required perpendicular distance $= \sqrt { \frac { 4 + 25 + 16 } { 81 } } = \frac { \sqrt { 5 } } { 3 }$ units.
(iv) Data could not be retrieved.