if
Question: If $A=\left[\begin{array}{rr}2 3 \\ -1 0\end{array}\right]$, show that $A^{2}-2 A+3 / 2=0$ Solution: Given : $A=\left[\begin{array}{cc}2 3 \\ -1 0\end{array}\right]$ Now, $A^{2}=A A$ $\Rightarrow A^{2}=\left[\begin{array}{cc}2 3 \\ -1 0\end{array}\right]\left[\begin{array}{cc}2 3 \\ -1 0\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{cc}4-3 6+0 \\ -2+0 -3+0\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{cc}1 6 \\ -2 -3\end{array}\right]$ $A^{2}-2 A+3 I_{2}$ $\Righ...
Read More →The probability of an impossible event is
Question: The probability of an impossible event is (a) 0(b) 1(c) 1/2(d) non-existent Solution: GIVEN: 4 options of probability of some events TO FIND: Which of the given options is the probability of impossible event? We know that, probability of an impossible event is 0. Hence the correct answer is option...
Read More →The probability of a certain event is
Question: The probability of a certain event is (a) 0(b) 1(c) 1/2(d) no existent Solution: GIVEN: 4 options of probability of some events TO FIND: Which of the given options is the probability of sure event? We know that, probability of a certain event is 1. Hence the correct answer is option...
Read More →Solve the following
Question: IfSpdenotes the sum of the series 1 +rp+r2p+ ... to andspthe sum of the series 1 rp+r2p ... to , prove thatSp+sp= 2 .S2p. Solution: We have: $\mathrm{S}_{\mathrm{p}}=1+\mathrm{r}^{\mathrm{p}}+\mathrm{r}^{2 \mathrm{p}}+\ldots \infty$ $\therefore \mathrm{S}_{\mathrm{p}}=\frac{1}{1-\mathrm{r}^{\mathrm{p}}}$ Similarly, $\mathrm{s}_{\mathrm{p}}=1-\mathrm{r}^{\mathrm{p}}+\mathrm{r}^{2 \mathrm{p}}-\ldots \infty$ $\therefore \mathrm{s}_{\mathrm{p}}=\frac{1}{1-\left(-\mathrm{r}^{\mathrm{p}}\rig...
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Question: If $A=\left[\begin{array}{rr}3 1 \\ -1 2\end{array}\right]$, show that $A^{2}-5 A+7 / 2=0$ Solution: Given : $A=\left[\begin{array}{cc}3 1 \\ -1 2\end{array}\right]$ Now, $A^{2}=A A$c $\Rightarrow A^{2}=\left[\begin{array}{cc}3 1 \\ -1 2\end{array}\right]\left[\begin{array}{cc}3 1 \\ -1 2\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{cc}9-1 3+2 \\ -3-2 -1+4\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{cc}8 5 \\ -5 3\end{array}\right]$ $A^{2}-5 A+7 I_{2}$ $\Righ...
Read More →Which of the following cannot be the probability of occurence of an event?
Question: Which of the following cannot be the probability of occurence of an event? (a) 0.2(b) 0.4(c) 0.8(d) 1.6 Solution: GIVEN: 4 options of probability of some events TO FIND: Which of the given options cannot be the probability of an event? We know that As the probability of an event cannot be more than 1 Hence the correct answer is option...
Read More →If P(E) = 0.05, then P(not E) =
Question: If P(E) = 0.05, then P(not E) = (a) 0.05(b) 0.5(c) 0.9(d) 0.95 Solution: Given: $P(E)=0.05$ TO FIND: $P(\bar{E})$ CALCULATION:We know that sum of probability of occurrence of an event and probability of non occurrence of an event is 1. Therefore $P(E)+P(\bar{E})=1$ $0.05+P(\bar{E})=1$ $P(\bar{E})=1-0.05$ $P(\bar{E})=0.95$ Hence the correct answer is option $(d)$...
Read More →Which of the following is a false statement?
Question: Which of the following is a false statement?(a) If the diagonals of a rhombus are 18 cm and 14 cm, then its area is 126 cm2. (b) Area of a $\| \mathrm{gm}=\frac{1}{2} \times$ base $\times$ corresp $o$ nding height. (c) A parallelogram and a rectangle on the same base and between the same parallels are equal in area.(d) If the area of a || gm with one side 24 cm and corresponding heighthcm is 192 cm2, thenh= 8 cm. Solution: (b) Area of a $\| g m=\frac{1}{2} \times$ base $\times$ corresp...
Read More →Prove that:
Question: Prove that: (21/4. 41/8. 81/16. 161/32... ) = 2. Solution: $\mathrm{LHS}=2^{\frac{1}{4}} \cdot 4^{\frac{2}{8}} \cdot 8^{\frac{3}{16}} \cdot 16^{\frac{4}{32}} \ldots \infty$ $=2^{\left(\frac{1}{4}+\frac{2}{8}+\frac{3}{16} \frac{3}{16} \frac{4}{32} \cdot \infty\right)}$ $=2^{\left(\frac{1}{2^{2}}+\frac{2}{2^{3}}+\frac{3}{2^{4}}+\frac{4}{2^{5}}+\ldots \infty\right)}$ $=2^{\frac{1}{2^{2}}\left\{1+\frac{2}{2}+\frac{3}{2^{2}}+\frac{4}{2^{3}} \ldots \infty\right\}}$ $=2^{\frac{1}{2^{2}}}\left...
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Question: If $A=\left[\begin{array}{rr}3 1 \\ -1 2\end{array}\right]$ and $I=\left[\begin{array}{ll}1 0 \\ 0 1\end{array}\right]$, then find $\lambda$ so that $A^{2}=5 A+N$. Solution: Given : $A=\left[\begin{array}{cc}3 1 \\ -1 2\end{array}\right]$ Now, $A^{2}=A A$ $\Rightarrow A^{2}=\left[\begin{array}{cc}3 1 \\ -1 2\end{array}\right]\left[\begin{array}{cc}3 1 \\ -1 2\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{cc}9-1 3+2 \\ -3-2 -1+4\end{array}\right]$ $\Rightarrow A^{2}=\left[\b...
Read More →Which of the following cannot be the probability of an event?
Question: Which of the following cannot be the probability of an event? (a) $\frac{2}{3}$ (b) $-1.5$ (c) $15 \%$ (d) $0.7$ Solution: GIVEN: 4 options of probability of some events TO FIND: Which of the given options cannot be the probability of an event? We know that . As the probability of an event cannot be negative In option (b)P=-1.5 Hence the correct answer is option...
Read More →Prove that:
Question: Prove that: (91/3. 91/9. 91/27... ) = 3. Solution: $\mathrm{LHS}=9^{\frac{1}{3}} \cdot 9^{\frac{1}{9}} \cdot 9^{\frac{1}{27} \ldots \infty}$ $=9\left(\frac{1}{3}+\frac{1}{9} \frac{1}{27} .\right)$ $=9\left\{\frac{\left(\frac{1}{3}\right)}{\left(1-\frac{1}{3}\right)}\right\}$ $=9^{\frac{\left(\frac{1}{3}\right)}{\left(1-\frac{1}{3}\right)}}$ $=\sqrt{9}$ $=3=\mathrm{RHS}$...
Read More →In a single throw of a pair of dice, the probability of getting the sum a perfect square is
Question: In a single throw of a pair of dice, the probability of getting the sum a perfect square is (a) $\frac{1}{18}$ (b) $\frac{7}{36}$ (c) $\frac{1}{6}$ (d) $\frac{2}{9}$ Solution: GIVEN: A pair of dice is thrown TO FIND: Probability of getting the sum a perfect square Let us first write the all possible events that can occur (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6),...
Read More →if
Question: If $A=\left[\begin{array}{ll}3 -2 \\ 4 -2\end{array}\right]$ and $I=\left[\begin{array}{ll}1 0 \\ 0 1\end{array}\right]$, then prove that $A^{2}-A+2 l=0$ Solution: Given : $A=\left[\begin{array}{ll}3 -2 \\ 4 -2\end{array}\right]$ Now, $A^{2}=A A$ $\Rightarrow A^{2}=\left[\begin{array}{ll}3 -2 \\ 4 -2\end{array}\right]\left[\begin{array}{ll}3 -2 \\ 4 -2\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{cc}9-8 -6+4 \\ 12-8 -8+4\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{a...
Read More →A number is selected from numbers 1 to 25. The probability that it is prime is
Question: A number is selected from numbers 1 to 25. The probability that it is prime is (a) $\frac{2}{3}$ (b) $\frac{1}{6}$ (c) $\frac{1}{3}$ (d) $\frac{5}{6}$ Solution: GIVEN: A number is selected from 1 to 25 TO FIND: Probability that, the number is a prime Total number is 25 Numbers from 1 to 25 that are primes are 2, 3, 5, 7, 11, 13, 17, 19 and 23 Total numbers that are primes from 1 to 25 is 9 We know that PROBABILITY = Hence probability of getting a prime number from 1 to 25 is equal to H...
Read More →The probability that a non-leap year has 53 sundays, is
Question: The probability that a non-leap year has 53 sundays, is (a) $\frac{2}{7}$ (b) $\frac{5}{7}$ (C) $\frac{6}{7}$ (d) $\frac{1}{7}$ Solution: GIVEN: A non leap year TO FIND: Probability that a non leap year has 53 Sundays. Total number of days in non leap year is 365days Hence number of weeks in a non leap year is In a non leap year we have 52 complete weeks and 1 day which can be any day of the week e.g. Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, and Saturday To make 53 Sundays...
Read More →A card is accidently dropped from a pack of 52 playing cards.
Question: A card is accidently dropped from a pack of 52 playing cards. The probability that it is an ace is (a) $\frac{1}{4}$ (b) $\frac{1}{13}$ (c) $\frac{1}{52}$ (d) $\frac{12}{13}$ Solution: GIVEN: One card is drawn from a well shuffled deck of 52 playing cards TO FIND: Probability of getting an Ace Total number of cards is 52 Cards which are Ace are 1 from each suit Total number of Ace cards is We know that PROBABILITY = Hence probability of getting an Ace is equal to Hence the correct opti...
Read More →Find the sum of the following series to infinity:
Question: Find the sum of the following series to infinity: (i) $1-\frac{1}{3}+\frac{1}{3^{2}}-\frac{1}{3^{3}}+\frac{1}{3^{4}}+\ldots \infty$ (ii) $8+4 \sqrt{2}+4+\ldots \infty$ (iii) $2 / 5+3 / 5^{2}+2 / 5^{3}+3 / 5^{4}+\ldots^{\infty}$. (iv) $10-9+8.1-7.29+\ldots \infty$ (v) $\frac{1}{3}+\frac{1}{5^{2}}+\frac{1}{3^{3}}+\frac{1}{5^{4}}+\frac{1}{3^{5}}+\frac{1}{56}+\ldots \infty$ Solution: (i) In the given G.P., first term, $a=1$ and common ratio, $r=-\frac{1}{3}$ Hence, the sum $S$ to infinity ...
Read More →Find the sum of 2n terms of the series whose every even term is 'a'
Question: Find the sum of 2nterms of the series whose every even term is 'a' times the term before it and every odd term is 'c' times the term before it, the first term being unity. Solution: Let the given series be $a_{1}+a_{2}+a_{3}+a_{4}+\ldots+a_{2 n}$. Now, it is given that $a_{1}=1, a_{2}=a a_{1}, a_{3}=c a_{2}, a_{4}=a a_{3}, a_{5}=c a_{4}$ and so on. $\because a_{1}=1$ $\Rightarrow a_{1}=1, a_{2}=a, a_{3}=a c, a_{4}=a^{2} c, a_{5}=a^{2} c^{2,} a_{6}=a^{3} c^{2}, \ldots$ $\therefore$ Sum ...
Read More →Solve the following
Question: Let $a_{n}$ be the $n$th term of the G.P. of positive numbers. Let $\sum_{n=1}^{100} a_{2 n}=\alpha$ and $\sum_{n=1}^{100} a_{2 n-1}=\beta$, such that $\alpha \neq \beta$. Prove that the common ratio of the G.P. is $\alpha / \beta$. Solution: Letabe the first term andrbe the common ratio of the G.P. $\therefore \sum_{n=1}^{100} a_{2 n}=\alpha$ and $\sum_{n=1}^{100} a_{2 n-1}=\beta$ $\therefore a_{2}+a_{4}+\ldots+a_{200}=\alpha$ and $a_{1}+a_{3}+\ldots+a_{199}=\beta$ $\Rightarrow a r+a ...
Read More →A G.P. consists of an even number of terms.
Question: A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of the terms occupying the odd places. Find the common ratio of the G.P. Solution: Let there be 2nterms in the given G.P. with the first term beingaand the common ratio beingr. According to the question Sum of all the terms = 5 (Sum of the terms occupying the odd places) $\Rightarrow a_{1}+a_{2}+\ldots+a_{2 n}=5\left(a_{1}+a_{3}+a_{5}+\ldots+a_{2 n-1}\right)$ $\Rightarrow a+a r+\ldots+a r^{2 n-1}...
Read More →Solve the following
Question: IfS1,S2, ...,Snare the sums ofnterms ofnG.P.'s whose first term is 1 in each and common ratios are 1, 2, 3, ...,nrespectively, then prove thatS1+S2+ 2S3+ 3S4+ ... (n 1)Sn= 1n+ 2n+ 3n+ ... +nn. Solution: Given: $S_{1}, S_{2}, \ldots, S_{n}$ are the sum of $n$ terms of an G.P. whose first term is 1 in each case and the common ratios are $1,2,3, \ldots, n$. $\therefore S_{1}=1+1+1+\ldots n$ terms $=n$ ...(1) $S_{2}=\frac{1\left(2^{n}-1\right)}{2-1}=2^{n}-1$ ...(2) $S_{3}=\frac{1\left(3^{n...
Read More →A person has 2 parents, 4 grandparents, 8 great grandparents, and so on.
Question: A person has 2 parents, 4 grandparents, 8 great grandparents, and so on. Find the number of his ancestors during the ten generations preceding his own. Solution: Here, the ancestors of the person form the G.P. 2, 4, 8, 16, ........ Now, first term,a= 2 And,r= 2 Number of his ancestors during the ten generations preceding his own, $S_{10}=2\left(\frac{2^{10}-1}{2-1}\right)=2(1024-1)=2046$...
Read More →How many terms of the G.P. 3,
Question: How many terms of the G.P. $3, \frac{3}{2}, \frac{3}{4} \ldots . .$ are needed to give the sum $\frac{3069}{512} ?$ Solution: Here, $a=3$ and Common ratio, $r=\frac{1}{2}$ And, $S_{n}=\frac{3069}{512}$ $\therefore S_{n}=3\left\{\frac{1-\left(\frac{1}{2}\right)^{n}}{1-\frac{1}{2}}\right\}$ $\Rightarrow \frac{3069}{512}=3\left\{\frac{1-\frac{1}{2^{n}}}{1-\frac{1}{2}}\right\}$ $\Rightarrow \frac{3069}{512}=6\left\{1-\frac{1}{2^{n}}\right\}$ $\Rightarrow \frac{3069}{3072}=1-\frac{1}{2^{n}}...
Read More →If a and b are the roots of
Question: Ifaandbare the roots ofx2 3x+p= 0 andc,dare the rootsx2 12x+q= 0, wherea,b,c,dform a G.P. Prove that (q+p) : (qp) = 17 : 15. Solution: We have, a+b= 3,ab=p,c+d =12 andcd = q a, b, c and d form a G.P. First term =a,b=ar,c=ar2andd =ar3 Then, we have a + b= 3 andc + d= 12 $\Rightarrow a+a r=3$ $\Rightarrow a(1+r)=3$ ...(1) Similarly, $a r^{2}(1+r)=12$ ...(2) $\Rightarrow \frac{a r^{2}(1+r)}{a(1+r)}=\frac{12}{3}$ $\Rightarrow r^{2}=4$ $\Rightarrow r=2$ $\therefore a(1+r)=3$ $\Rightarrow a=...
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