Question 14 Marks
The area of the region bounded by the curve $y=\max \{|x|, x|x-2|\}$, then $x$-axis and the lines $x=-2$ and $x=4$ is equal to __________.
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SECTION - B [MATHS - NUMERIC]
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Question 24 Marks
If the number of seven-digit numbers, such that the sum of their digits is even, is $m \cdot n \cdot 10^{ n }$; $m, n \in\{1,2,3, \ldots, 9\}$, then $m+n$ is equal to __________.
Answer
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Total 7 digit nos. $=9000000$
7 digit nos. having sum of digits
$
\begin{array}{l}
\text { Even }=4500000 \\
=9.5 \cdot 10^5 \\
m=9, n=5 \\
m+n=14
\end{array}
$
Total 7 digit nos. $=9000000$
7 digit nos. having sum of digits
$
\begin{array}{l}
\text { Even }=4500000 \\
=9.5 \cdot 10^5 \\
m=9, n=5 \\
m+n=14
\end{array}
$
Question 34 Marks
Let $\overrightarrow{ a }=\hat{ i }+\hat{ j }+\hat{ k }, \overrightarrow{ b }=3 \hat{ i }+2 \hat{ j }-\hat{ k }, \overrightarrow{ c }=\lambda \hat{ j }+\mu \hat{ k }$ and $\hat{ d }$ be a unit vector such that $\overrightarrow{ a } \times \hat{ d }=\overrightarrow{ b } \times \hat{ d }$ and $\overrightarrow{ c } . \hat{ d }=1$, If $\vec{c}$ is perpendicular to $\vec{a}$, then $|3 \lambda \hat{d}+\mu \overrightarrow{ c }|^2$ is equal to __________.
Answer
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$\begin{array}{l}\overrightarrow{ a } \times \overrightarrow{ d }-\overrightarrow{ b } \times \overrightarrow{ d }=0 \\ ( a -\overrightarrow{ b }) \times \overrightarrow{ d }=0 \\ \overrightarrow{d}= t (\overrightarrow{ a }-\overrightarrow{ b }) \\ \overrightarrow{ d }= t (-2 \hat{ i }-\hat{ j }+2 \hat{ k }) \\ |\overrightarrow{ d }|=1 \\ | t |=\frac{1}{3}\end{array}$
$\begin{array}{l}\overrightarrow{ c } \cdot \overrightarrow{ a }=0 \\ \lambda+\mu=0 \\ \mu=-\lambda \\ \overrightarrow{ c }=\lambda(\hat{ j }-\hat{ k }),|\overrightarrow{ c }|^2=2 \lambda^2 \\ \overrightarrow{ c } \cdot \hat{ d }=1 \\ t (-2,-1,2) \cdot \lambda(0,1,-1)=1 \\ \lambda t =\frac{-1}{3} \Rightarrow \lambda^2=1\end{array}$
$\begin{array}{l}|3 \lambda \hat{d}+\mu c |^2=9 \lambda^2|\hat{d}|^2+\mu^2|\overrightarrow{ c }|^2+6 \lambda \mu(\hat{d} \cdot \overrightarrow{ c }) \\ =3 \lambda^2+2 \lambda^4 \\ =5\end{array}$
$\begin{array}{l}\overrightarrow{ a } \times \overrightarrow{ d }-\overrightarrow{ b } \times \overrightarrow{ d }=0 \\ ( a -\overrightarrow{ b }) \times \overrightarrow{ d }=0 \\ \overrightarrow{d}= t (\overrightarrow{ a }-\overrightarrow{ b }) \\ \overrightarrow{ d }= t (-2 \hat{ i }-\hat{ j }+2 \hat{ k }) \\ |\overrightarrow{ d }|=1 \\ | t |=\frac{1}{3}\end{array}$
$\begin{array}{l}\overrightarrow{ c } \cdot \overrightarrow{ a }=0 \\ \lambda+\mu=0 \\ \mu=-\lambda \\ \overrightarrow{ c }=\lambda(\hat{ j }-\hat{ k }),|\overrightarrow{ c }|^2=2 \lambda^2 \\ \overrightarrow{ c } \cdot \hat{ d }=1 \\ t (-2,-1,2) \cdot \lambda(0,1,-1)=1 \\ \lambda t =\frac{-1}{3} \Rightarrow \lambda^2=1\end{array}$
$\begin{array}{l}|3 \lambda \hat{d}+\mu c |^2=9 \lambda^2|\hat{d}|^2+\mu^2|\overrightarrow{ c }|^2+6 \lambda \mu(\hat{d} \cdot \overrightarrow{ c }) \\ =3 \lambda^2+2 \lambda^4 \\ =5\end{array}$
Question 44 Marks
Let the product of the focal distances of the point $P (4,2 \sqrt{3})$ on the hyperbola $H : \frac{ x ^2}{ a ^2}-\frac{ y ^2}{b^2}=1$ be 32. Let the length of the conjugate axis of H be p and the length of its latus rectum be $q$. Then $p^2+q^2$ is equal to __________.
Answer
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$\begin{array}{l}\frac{ x ^2}{ a ^2}-\frac{ y ^2}{b^2}=1\quad \quad \ldots \ldots(1)\ \\ P (4,2 \sqrt{3}) \\ PS _1 \cdot PS _2=32 \\ \left| PS _1- PS _2\right|=2 a \\ P (4,2 \sqrt{3}) \text { lies on } H \\ \therefore \frac{16}{ a ^2}-\frac{12}{b^2}=1\end{array}$
$
\begin{array}{l}
16 b^2-12 a^2=a^2 b^2\quad \quad \ldots \ldots(2) \\
\left|PS_1-PS_2\right|^2=4 a^2 \\
PS_1{ }^2+PS_2{ }^2-2 PS_1 \cdot PS_2=4 a^2 \\
(ae-4)^2+12+(ae+4)^2+12-64=4 a^2 \\
2 a^2 e^2-8=4 a^2 \\
a^2+b^2-4=2 a^2 \\
b^2-a^2=4
\end{array}
$
$\begin{array}{l}\text { (2) } \&(3) \Rightarrow 16\left(a^2+4\right)-12 a^2=a^2\left(a^2+4\right) \\ \Rightarrow 16 a ^2+64-12 a ^2= a ^4+4 a ^2 \\ \Rightarrow a ^4=64 \\ \Rightarrow a ^2=8 \\ \therefore b^2=12 \\ p ^2+ q ^2=4 b^2+\frac{4 b^4}{ a ^2} \\ =120\end{array}$
$\begin{array}{l}\frac{ x ^2}{ a ^2}-\frac{ y ^2}{b^2}=1\quad \quad \ldots \ldots(1)\ \\ P (4,2 \sqrt{3}) \\ PS _1 \cdot PS _2=32 \\ \left| PS _1- PS _2\right|=2 a \\ P (4,2 \sqrt{3}) \text { lies on } H \\ \therefore \frac{16}{ a ^2}-\frac{12}{b^2}=1\end{array}$
$
\begin{array}{l}
16 b^2-12 a^2=a^2 b^2\quad \quad \ldots \ldots(2) \\
\left|PS_1-PS_2\right|^2=4 a^2 \\
PS_1{ }^2+PS_2{ }^2-2 PS_1 \cdot PS_2=4 a^2 \\
(ae-4)^2+12+(ae+4)^2+12-64=4 a^2 \\
2 a^2 e^2-8=4 a^2 \\
a^2+b^2-4=2 a^2 \\
b^2-a^2=4
\end{array}
$
$\begin{array}{l}\text { (2) } \&(3) \Rightarrow 16\left(a^2+4\right)-12 a^2=a^2\left(a^2+4\right) \\ \Rightarrow 16 a ^2+64-12 a ^2= a ^4+4 a ^2 \\ \Rightarrow a ^4=64 \\ \Rightarrow a ^2=8 \\ \therefore b^2=12 \\ p ^2+ q ^2=4 b^2+\frac{4 b^4}{ a ^2} \\ =120\end{array}$
Question 54 Marks
All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers. Let the word at serial number $n$ be denoted by $W _{ n }$. Let the probability $P \left( W _{ n }\right)$ of choosing the word $W _{ n }$ satisfy $P \left( W _{ n }\right)=2 P \left( W _{ n -1}\right), n >1$.
If $P ( CDBEA )=\frac{2^\alpha}{2^\beta-1}, \alpha, \beta \in N$, then $\alpha+\beta$ is equal to : __________.
View full question & answer→If $P ( CDBEA )=\frac{2^\alpha}{2^\beta-1}, \alpha, \beta \in N$, then $\alpha+\beta$ is equal to : __________.
