3 Marks Question

Question 13 Marks

Three vertices of a parallelogram ABCD are A(3, -1, 2), B(1, 2, -4) and C(-1, 1, 2). Find the coordinates of the fourth vertex.

Answer

Let D (x, y, z) be the fourth vertex of parallelogram ABCD.
We know that diagonals of a parallelogram bisect each other. So the mid points of AC and BD coincide.

$\therefore$ Coordinates of mid point of $AC \left(\frac{3-1}{2}, \frac{-1+1}{2}, \frac{2+2}{2}\right)$
= (1, 0, 2)
Also coordinates of mid point of $BD \left(\frac{x+1}{2}, \frac{y+2}{2}, \frac{z-4}{2}\right)$
$\therefore \frac{x+1}{2}=1 \Rightarrow x+1=2 \Rightarrow x=1$
$\begin{array}{l}\frac{y+2}{2}=0 \Rightarrow y+2=0 \Rightarrow y=-2 \\ \frac{z-4}{2}=2 \Rightarrow z-4=4 \Rightarrow z=8\end{array}$
Thus the coordinates of point D are (1, -2, 8)

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Question 23 Marks

If ${ }^{22} P_{r+1}:{ }^{20} P_{r+2}=11: 52$. find r .

Answer

Here ${ }^{22} P_{r+1}:{ }^{20} P_{r+2}=11: 52$
$\begin{array}{l}\Rightarrow \frac{22!}{(21-r)!} \times \frac{(18-r)!}{20!}=\frac{11}{52} \\ \Rightarrow=\frac{22 \times 21 \times 20!}{(21-r)(20-r)(19-r)(18-r)!} \times \frac{(18-r)!}{20!}=\frac{11}{52} \\ \Rightarrow \frac{22 \times 21}{(21-r)(20-r)(19-r)}=\frac{11}{52} \\ \Rightarrow(21-r)(20-r)(19-r)=2 \times 21 \times 52 \\ \Rightarrow(21-r)(20-r)(19-r)=14 \times 13 \times 12 \\ \Rightarrow(21-r)(20-r)(19-r)=(21-7)(20-7)(19-7) \\ \Rightarrow r=7\end{array}$

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Question 33 Marks

In a class, 18 students took Physics, 23 students took Chemistry and 24 students took
Mathematics of these 13 took both Chemistry and Mathematics, 12 took both Physics and Chemistry and 11 took both Physics an Mathematics. If 6 students offered all the three subjects, find:
i. The total number of students.
ii. How many took Maths but not Chemistry.
iii. How many took exactly one of the three subjects.

Answer

$\begin{array}{l}\text { Given, } n(p)=18, n(C)=23, n(M)=24, n(C \cap M)=13 \\ n(P \cap C)=12, n(P \cap M)=11 \text { and } n(P \cap C \cap M)=6\end{array}$
i. Total no. of students in the class
$\begin{array}{l}=n(P \cup C \cup M) \\ =n(P)+n(C)+n(M)-n(P \cap C)-n(P \cap M)-n(C \cap M)+n(P \cap C \cap M) \\ =18+23+24-12-11-13+6=35\end{array}$
ii. No. of students who took Mathematics but not Chemistry
$\begin{array}{l}=n(M-C) \\ =n(M)-n(M \cap C) \\ =24-13=11\end{array}$
iii. No. of students who took exactly one of the three subjects
$\begin{array}{l}=n(P)+n(C)+n(M)-2 n(M \cap P)-2 n(P \cap C)-2 n(M \cap C)+3 n(P \cap C \cap M) \\ =18+23+24-2 \times 11-2 \times 12-2 \times 13+3 \times 6 \\ =65-22-24-26+18 \\ =83-72=11\end{array}$

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Question 43 Marks

Evaluate: $\sum_{k=1}^{11}\left(2+3^k\right)$

Answer

Given: $\sum_{k=1}^{11}\left(2+3^k\right)$
$=\left(2+3^1\right)+\left(2+3^2\right)+\left(2+3^3\right)+\left(2+3^{11}\right)$
$=(2+2+2+\ldots \ldots \ldots 11$ times $)+\left(3+3^2+3^3+\ldots \ldots . .+3^{11}\right)$
$=22+\left(3+3^2+3^3+\ldots \ldots+3^{11}\right) \ldots \ldots \ldots$ (i)
Here $3,3^2, 3^3 \ldots \ldots . ., 3^{11}$ is in G.P.
$\therefore a=3$ and $r=\frac{3^2}{3}=3$
$S _n=\frac{3\left(3^{11}-1\right)}{3-1}=\frac{3}{2}\left(3^{11}-1\right)$
Putting the value of $S _{ n }$ in eq. (i), we get $\sum_{k=1}^{11}\left(2+3^k\right)=22+\frac{3}{2}\left(3^{11}-1\right)$

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Question 53 Marks

If the AM and GM of two positive numbers a and b are in the ratio $m : n$, show that $a : b =\left(m+\sqrt{m^2-n^2}\right):\left(m-\sqrt{m^2-n^2}\right)$

Answer

$\begin{array}{l}\frac{a+b}{2}=\frac{m}{n} \\ \frac{a+b}{2 \sqrt{a b}}=\frac{m}{n}\end{array}$
By C and D
$\begin{array}{l}\frac{a+b+2 \sqrt{b}}{a+b-2 \sqrt{b}}=\frac{m+n}{m-n} \\ \frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2}=\frac{m+n}{m-n} \\ \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{m+n}}{\sqrt{m-n}}\end{array}$
By C and D
$\frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}$
Squaring both side
$\begin{array}{l}\frac{a}{b}=\frac{m+n+m-n+2 \sqrt{m^2-n^2}}{m+n+m-n-2 \sqrt{m^2-n^2}} \\ \frac{a}{b}=\frac{m+\sqrt{m^2-n^2}}{m-\sqrt{m^2-n^2}}\end{array}$

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Question 63 Marks

Find the derivative of $\frac{(x-1)(x-2)}{(x-3)(x-4)}$.

Answer

Let $y=\frac{(x-1)(x-2)}{(x-3)(x-4)}$
On differentiating both sides w.r.t. x, we get
$\begin{array}{l}\frac{d y}{d x}=\frac{\left[\begin{array}{r}{[x-3)(x-4) \frac{d}{d x}[(x-1)(x-2)]-(x-1)} \\ (x-2) \frac{d}{d x}[(x-3)(x-4)]\end{array}\right]}{[(x-3)(x-4)]^2} \\ {\left[\because \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^2}\right]}\end{array}$
$\begin{array}{l}=\frac{(x-3)(x-4)\left[(x-1) \frac{d}{d x}(x-2)+(x-2) \frac{d}{d x}(x-1)\right]-(x-1)(x-2)\left[(x-3) \frac{d}{d x}(x-4)+(x-4) \frac{d}{d x}(x-3)\right]}{(x-3)^2(x-4)^2} \\ =\frac{(x-3)(x-4)[(x-1) \cdot 1+(x-2) \cdot 1]-(x-1)(x-2)[(x-3) \cdot 1+(x-4) \cdot 1)]}{(x-3)^2(x-4)^2} \\ =\frac{(x-3)(x-4)[2 x-3]-(x-1)(x-2)[2 x-7]}{(x-3)^2(x-4)^2} \\ =\frac{\left(x^2-7 x+12\right)(2 x-3)-\left(x^2-3 x+2\right)(2 x-7)}{(x-3)^2(x-4)^2} \\ =\frac{2 x^3-14 x^2+24 x-3 x^2+21 x-36-2 x^3+6 x^2-4 x+7 x^2-21 x+14}{\left.(x-3)^{2( } x-4\right)^2} \\ =\frac{-4 x^2+20 x-22}{(x-3)^2(x-4)^2}\end{array}$

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Question 73 Marks

Find the derivative of $x^{-4}\left(3-4 x^{-5}\right)$

Answer

Here $f(x)=x^{-4}\left(3-4 x^{-5}\right)$
$f^{\prime}(x)=\frac{d}{d x}\left[x^{-4}\left(3-4 x^{-5}\right)\right]$
$=x^{-4} \frac{d}{d x}\left(3-4 x^{-5}\right)+\left(3-4 x^{-5}\right) \frac{d}{d x}\left(x^{-4}\right)$
$\begin{array}{l}=x^{-4}\left(20 x^{-6}\right)+\left(3-4 x^{-5}\right)\left(-4 x^{-5}\right) \\ =20 x^{-10}-12 x^{-5}+16 x^{-10} \\ =36 x^{-10}-12 x^{-5}=\frac{36}{x^{10}}-\frac{12}{x^5}\end{array}$

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Question 83 Marks

Show that $2^{4 n+4}-15 n-16$ where $n \in N$ is divisible by 225

Answer

From the given equation we have $2^{4 n+4}-15 n-16=2^{4(n+1)}-15 n-16$
$\begin{array}{l}=16^{n+1}-15 n-16 \\ =(1+15)^{n+1}-15 n-16\end{array}$
Using binomial expression we have
$\begin{array}{l}={ }^{n+1} C _0 15^0+{ }^{n+1} C _1 15^1+{ }^{n+1} C _2 15^2+{ }^{n+1} C _3 15^3 \\ +\ldots+x+[C],(15)^{n+1}-15 n-16 \\ =1+(n+1) 15+{ }^{n+1} C _2 15^2+{ }^{n+1} C _3 15^3\end{array}$
$\begin{array}{l}+\ldots+n+1 C_{n+1}(15)^{n+1}-15 n-16 \\ =1+15 n+15+{ }^{n+1} C _2 15^2+{ }^{n+1} C _3 15 \\ +\ldots+{ }^{n+1} C _{n+1}(15)^{n+1}-15 n-16 \\ =15^2\left[{ }^{n+1} C _2+{ }^{n+1} C _3 15+\ldots \text { so on }\right]\end{array}$
Thus, $2^{4 n+4}-15 n-16$ is divisible 225 .

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Question 93 Marks

Using binomial theorem, expand: $\left(x^2-\frac{2}{x}\right)^7$.

Answer

To find: Expansion of $\left(x^2-\frac{3 x}{7}\right)^7$
Formula used: ${ }^n C_r=\frac{n!}{(n-r)!(r)!}$
We know that, $(a+b)^n={ }^n C_0 a^n+{ }^n C_1 a^{n-1} b+{ }^n C_2 a^{n-2} b^2+\ldots \ldots .+{ }^n C_{n-1} a^{n-1}+{ }^n C_n b^n$
Here We have, $\left(x^2-\frac{3 x}{7}\right)^7$
$\begin{array}{l}\Rightarrow\left[{ }^7 C _0\left( x ^2\right)^{7-0}\right]+\left[7 C _1\left( x ^2\right)^{7-1}\left(-\frac{3 x }{7}\right)^1\right]+\left[7 c_2\left(x^2\right)^{7-2}\left(-\frac{3 x}{7}\right)^2\right]+\left[7 C _3\left( x ^2\right)^{7-3}\left(-\frac{3 x }{7}\right)^3\right]+\left[7 C _4\left( x ^2\right)^{7-4}\left(-\frac{3 x }{7}\right)^4\right] \\ +\left[7 C _5\left( x ^2\right)^{7-5}\left(-\frac{3 x }{7}\right)^5\right]+\left[7 C _6\left( x ^2\right)^{7-6}\left(-\frac{3 x }{7}\right)^6\right]+\left[7 C _7\left(-\frac{3 x }{7}\right)^7\right] \\ \Rightarrow\left[\frac{7!}{0!(7-0)!}\left(x^2\right)^7\right]-\left[\frac{7!}{1!(7-1)!}\left(x^2\right)^6\left(\frac{3 x}{7}\right)\right]+\left[\frac{7!}{2!(7-2)!}\left(x^2\right)^5\left(\frac{9 x^2}{49}\right)\right]-\left[\frac{7!}{3!(7-3)!}\left(x^2\right)^4\left(\frac{27 x^3}{343}\right)\right] \\ +\left[\frac{7!}{4!(7-4)!}\left(x^2\right)^3\left(\frac{81 x^4}{2401}\right)\right]-\left[\frac{7!}{5!(7-5)!}\left(x^2\right)^2\left(\frac{243 x^5}{16807}\right)\right]+\left[\frac{7!}{6!(7-6)!}\left(x^2\right)^1\left(\frac{729 x^6}{117649}\right)\right]-\left[\frac{7!}{7!(7-7)!}\left(\frac{2187 x^7}{823543}\right)\right] \\ -\left[\frac{7!}{7!(7-7)!}\left(\frac{2187 x^7}{823543}\right)\right]+\left[21\left(x^{10}\right)\left(\frac{9 x^2}{49}\right)\right]-\left[35\left(x^8\right)\left(\frac{27 x^3}{343}\right)\right] \\ +\left[35\left(x^6\right)\left(\frac{81 x^4}{2401}\right)\right]-\left[21\left(x^4\right)\left(\frac{243 x^5}{16807}\right)\right]+\left[7\left(x^2\right)\left(\frac{729 x^6}{117649}\right)\right]-\left[1\left(\frac{2187 x^7}{823543}\right)\right] \\ \Rightarrow x^{24}-3 x^{13}+\left(\frac{27}{7}\right) x^{12}-\left(\frac{135}{49}\right) x^{11}+\left(\frac{405}{343}\right) x^{10}-\left(\frac{729}{2401}\right) x^9+\left(\frac{729}{16807}\right) x^8-\left(\frac{2187}{823543}\right) x^7 \\ x^{14}-3 x^{13}+\left(\frac{27}{7}\right) x^{12}-\left(\frac{135}{49}\right) x^{11}+\left(\frac{405}{343}\right) x^{10}-\left(\frac{729}{2901}\right) x^9+\left(\frac{729}{12807}\right) x^8-\left(\frac{2187}{825353}\right) x^7\end{array}$

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MATHS 3 Marks Question

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