Question: $1^{2}+3^{2}+5^{2}+\ldots+(2 n-1)^{2}=\frac{1}{3} n\left(4 n^{2}-1\right)$ Solution: LetP(n) be the given statement. Now, $P(n)=1^{2}+3^{2}+5^{2}+\ldots+(2 n-1)^{2}=\frac{1}{3} n\left(4 n^{2}-1\right)$ Step 1: $P(1)=1^{2}=1=\frac{1}{3} \times 1 \times(4-1)$ Hence, $P(1)$ is true. Step 2 : Let $P(m)$ be true. Then, $1^{2}+3^{2}+\ldots+(2 m-1)^{2}=\frac{1}{3} m\left(4 m^{2}-1\right)$ To prove : $P(m+1)$ is true whenever $P(m)$ is true. That $i s$ $1^{2}+3^{2}=\ldots+(2 m+1)^{2}=\frac{1}{...

If A and B are two events such that P (A) ≠ 0 and P(B|A) = 1, then.

Question: If A and B are two events such that P (A) 0 and P(B|A) = 1, then. (A) $A \subset B$ (B) $B \subset A$ (C) $B=\Phi$ (D) $A=\Phi$ Solution: $P(A) \neq 0$ and $P(B \mid A)=1$ $P(B \mid A)=\frac{P(B \cap A)}{P(A)}$ $\mathrm{l}=\frac{\mathrm{P}(\mathrm{B} \cap \mathrm{A})}{\mathrm{P}(\mathrm{A})}$ $P(A)=P(B \cap A)$ $\Rightarrow A \subset B$ Thus, the correct answer is A....

An irrational number between

Question: An irrational number between $\sqrt{2}$ and $\sqrt{3}$ is (a) $(\sqrt{2}+\sqrt{3})$ (b) $\sqrt{2} \times \sqrt{3}$ (c) $5^{1 / 4}$ (d) $6^{1 / 4}$ Solution: (d) $6^{1 / 4}$ An irrational number between $\sqrt{2}$ and $\sqrt{3}$ : $\sqrt{\sqrt{2} \times \sqrt{3}}=6^{\frac{1}{4}}$...

Question: $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}$ Solution: LetP(n) be the given statement. Now, $P(n): \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}$ Step 1; $P(1)=\frac{1}{2}=1-\frac{1}{2^{1}}$ Thus, $P(1)$ is true. Step 2 : Suppose $P(m)$ is true. Then, $\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{m}}=1-\frac{1}{2^{m}}$ To show: $P(m+1)$ is true whenever $P(m)$ is true. That is, $\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{m+...

Question: $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}$ Solution: LetP(n) be the given statement. Now, $P(n): \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}$ Step 1; $P(1)=\frac{1}{2}=1-\frac{1}{2^{1}}$ Thus, $P(1)$ is true. Step 2 : Suppose $P(m)$ is true. Then, $\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{m}}=1-\frac{1}{2^{m}}$ To show: $P(m+1)$ is true whenever $P(m)$ is true. That is, $\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{m+...

Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls.

Question: Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black. Solution: Let $E_{1}$ and $E_{2}$ respectively denote the events that a red ball is transferred from bag I to II and a black ball is transferred from bag I to II. $P\left(E_{1}\right)=\frac{3}{7}$ and $P\left(E_{2}\rig...

An irrational number between 5 and 6 is

Question: An irrational number between 5 and 6 is (a) $\frac{1}{2}(5+6)$ (b) $\sqrt{5+6}$ (c) $\sqrt{5 \times 6}$ (d) none of these Solution: (c) $\sqrt{5 \times 6}$ An irrational number between $a$ and $b$ is given as $\sqrt{a b}$,...

If cos A=725, find the value of tan A + cot A.

Question: If $\cos A=\frac{7}{25}$, find the value of $\tan \mathrm{A}+\cot \mathrm{A} .$ Solution: Given: $\cos A=\frac{7}{25}$ We know that, $\sin ^{2} A+\cos ^{2} A=1$ $\Rightarrow \sin ^{2} A+\left(\frac{7}{25}\right)^{2}=1$ $\Rightarrow \sin ^{2} A+\frac{49}{625}=1$ $\Rightarrow \sin ^{2} A=1-\frac{49}{625}$ $\Rightarrow \sin ^{2} A=\frac{576}{625}$ $\Rightarrow \sin A=\frac{24}{25}$ Therefore, $\tan A+\cot A=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}$ $=\frac{\left(\frac{24}{25}\right)}{\...

The simplest form of

Question: The simplest form of $0.12 \overline{3}$ is (a) $\frac{41}{330}$ (b) $\frac{37}{330}$ (c) $\frac{41}{333}$ (d) none of these Solution: (d) none of theseLetx= 0.12333333333... ...(i)Multiplying by 100 on both sides, we get:100x= 12.33333333... ...(ii)Multiplying by 10 on both sides, we get:1000x= 123.33333333... ...(iii)Subtracting (ii) from (iii), we get:900x= 111 $\Rightarrow x=\frac{111}{900}$...

1.2 + 2.3 + 3.4 + ... + n (n + 1)

Question: $1.2+2.3+3.4+\ldots+n(n+1)=\frac{n(n+1)(n+2)}{3}$ Solution: LetP(n) be the given statement. Now, $P(n)=1.2+2.3+3.4+\ldots+n(n+1)=\frac{n(n+1)(n+2)}{3}$ Step 1: $P(1)=1.2=2=\frac{1(1+1)(1+2)}{3}$ Hence, $P(1)$ is true. Step 2: Let $P(m)$ be true. Then, $1.2+2.3+\ldots+m(m+1)=\frac{m(m+1)(m+2)}{3}$ To prove: $P(m+1)$ is true. That is, $1.2+2.3+\ldots+(m+1)(m+2)=\frac{(m+1)(m+2)(m+3)}{3}$ Now, $P(m)$ is $1.2+2.3+\ldots+m(m+1)=\frac{m(m+1)(m+2)}{3}$ $\Rightarrow 1.2+2.3+\ldots+m(m+1)+(m+1)...

An electronic assembly consists of two subsystems,

Question: An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known: P(A fails) = 0.2 P(B fails alone) = 0.15 P(A and B fail) = 0.15 Evaluate the following probabilities (i) P(A fails| B has failed) (ii) P(A fails alone) Solution: Let the event in which $A$ fails and $B$ fails be denoted by $E_{A}$ and $E_{B}$. $P\left(E_{A}\right)=0.2$ $P\left(E_{A} \cap E_{B}\right)=0.15$ $P$ (B fails alone) $=P\left(E...

1.2 + 2.3 + 3.4 + ... + n (n + 1)

Question: $1.2+2.3+3.4+\ldots+n(n+1)=\frac{n(n+1)(n+2)}{3}$ Solution: LetP(n) be the given statement. Now, $P(n)=1.2+2.3+3.4+\ldots+n(n+1)=\frac{n(n+1)(n+2)}{3}$ Step 1: $P(1)=1.2=2=\frac{1(1+1)(1+2)}{3}$ Hence, $P(1)$ is true. Step 2: Let $P(m)$ be true. Then, $1.2+2.3+\ldots+m(m+1)=\frac{m(m+1)(m+2)}{3}$ To prove: $P(m+1)$ is true. That is, $1.2+2.3+\ldots+(m+1)(m+2)=\frac{(m+1)(m+2)(m+3)}{3}$ Now, $P(m)$ is $1.2+2.3+\ldots+m(m+1)=\frac{m(m+1)(m+2)}{3}$ $\Rightarrow 1.2+2.3+\ldots+m(m+1)+(m+1)...

What is the value of (1 + tan2 θ) (1 − sin θ) (1 + sin θ)?

Question: What is the value of $\left(1+\tan ^{2} \theta\right)(1-\sin \theta)(1+\sin \theta) ?$ Solution: We have, $\left(1+\tan ^{2} \theta\right)(1-\sin \theta)(1+\sin \theta)=\left(1+\tan ^{2} \theta\right)\{(1-\sin \theta)(1+\sin \theta)\}$ $=\left(1+\tan ^{2} \theta\right)\left(1-\sin ^{2} \theta\right)$ We know that, $\sec ^{2} \theta-\tan ^{2} \theta=1$ $\Rightarrow \sec ^{2} \theta=1+\tan ^{2} \theta$ $\sin ^{2} \theta+\cos ^{2} \theta=1$ $\Rightarrow \cos ^{2} \theta=1-\sin ^{2} \theta...

The simplest form of

Question: The simplest form of $0.3 \overline{2}$ is (a) $\frac{16}{45}$ (b) $\frac{32}{99}$ (c) $\frac{29}{90}$ (d) none of these Solution: (c) $\frac{29}{90}$ Letx= 0.3222222222... ...(i)Multiplying by 10 on both sides, we get:10x= 3.222222222... ...(ii)Again, multiplying by 10 on both sides, we get:100x= 32.222222222... ...(iii)On subtracting (ii) from (iii), we get:90x= 29 $\therefore x=\frac{29}{90}$...

The simplest form of

Question: The simplest form of $0 . \overline{54}$ is (a) $\frac{27}{50}$ (b) $\frac{6}{11}$ (c) $\frac{4}{7}$ (d) none of these Solution: (b) $\frac{6}{11}$ Letx= 0.545454... ...(i)Multiplying both sides by 100, we get:100x= 54.5454545... ...(ii)Subtracting (i) from (ii), we get:99x= 540 $\Rightarrow x=\frac{54}{99}=\frac{6}{11}$...

1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1)

Question: $1.3+3.5+5.7+\ldots+(2 n-1)(2 n+1)=\frac{n\left(4 n^{2}+6 n-1\right)}{3}$ Solution: LetP(n) be the given statement. Now, $P(n)=1.3+3.5+5.7+\ldots+(2 n-1)(2 n+1)=\frac{n\left(4 n^{2}+6 n-1\right)}{3}$Step 1: Step 1: $P(1)=1.3=3=\frac{1\left(4 \times(1)^{2}+6 \times 1-1\right)}{3}$ Hence, $P(1)$ is true. Step 2: Let $P(m)$ be true. Then; $1.3+3.5+\ldots+(2 m-1)(2 m+1)=\frac{m\left(4 m^{2}+6 m-1\right)}{3}$ To prove : $P(m+1)$ is true. That is, $1.3+3.5+\ldots+(2 m+1)(2 m+3)=\frac{(m+1)\l...

What is the value of tan2 θ−sec2 θcot2 θ−cosec2 θ.

Question: What is the value of $\frac{\tan ^{2} \theta-\sec ^{2} \theta}{\cot ^{2} \theta-\operatorname{cosec}^{2} \theta}$. Solution: We have, $\frac{\tan ^{2} \theta-\sec ^{2} \theta}{\cot ^{2} \theta-\operatorname{cosec}^{2} \theta}=\frac{-1\left(\sec ^{2} \theta-\tan ^{2} \theta\right)}{-1\left(\operatorname{cosec}^{2} \theta-\cot ^{2} \theta\right)}$ $=\frac{\sec ^{2} \theta-\tan ^{2} \theta}{\operatorname{cosec}^{2} \theta-\cot ^{2} \theta}$ We know that, $\sec ^{2} \theta-\tan ^{2} \theta...

If each element of a second order determinant is either zero or one,

Question: If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability $\frac{1}{2}$ ). Solution: The total number of determinants of second order with each element being 0 or 1 is $(2)^{4}=16$ The value of determinant is positive in the following cases. $\left|\begin{array}{ll}1 0 \\ 0 1\end{arra...

The simplest for of

Question: The simplest for of $1 . \overline{6}$ is (a) $\frac{833}{500}$ (b) $\frac{8}{5}$ (C) $\frac{5}{3}$ (d) none of these Solution: (C) $\frac{5}{3}$ Letx= 1.6666666... ...(i)Multiplying by 10 on both sides, we get:10x= 16.6666666... ...(ii)Subtracting (i) from (ii), we get:9x= 15 $\Rightarrow x=\frac{15}{9}=\frac{5}{3}$...

What is the value of 6 tan2θ−6cos2θ.

Question: What is the value of $6 \tan ^{2} \theta-\frac{6}{\cos ^{2} \theta}$. Solution: We have, $6 \tan ^{2} \theta-\frac{6}{\cos ^{2} \theta}=6 \tan ^{2} \theta-6 \sec ^{2} \theta$ $=-6\left(\sec ^{2} \theta-\tan ^{2} \theta\right)$ We know that, $\sec ^{2} \theta-\tan ^{2} \theta=1$ Therefore, $6 \tan ^{2} \theta-\frac{6}{\cos ^{2} \theta}=-6$...

Assume that the chances of the patient having a heart attack are 40%.

Question: Assume that the chances of the patient having a heart attack are 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditatio...

1.3 + 2.4 + 3.5 + ... + n.

Question: $1.3+2.4+3.5+\ldots+n .(n+2)=\frac{1}{6} n(n+1)(2 n+7)$ Solution: LetP(n) be the given statement. Now, $P(n)=1.3+2.4+3.5+\ldots+n .(n+2)=\frac{1}{6} n(n+1)(2 n+7)$ Step 1: $P(1)=1.3=3=\frac{1}{6} \times 1(1+1)(2 \times 1+7)$ Hence, $P(1)$ is true. Step 2 : Let $P(m)$ be true. Then, 1. $3+2 \cdot 4+\ldots+m \cdot(m+2)=\frac{1}{6} m(m+1)(2 m+7)$ To prove: $P(m+1)$ is true. That is, $1.3+2.4+\ldots+(m+1)(m+3)=\frac{1}{6}(m+1)(m+2)(2 m+9)$ $P(m)$ is equal to $1.3+2.4+\ldots+m(m+2)=\frac{1}...

What is the value of 9cot2 θ − 9cosec2 θ?

Question: What is the value of $9 \cot ^{2} \theta-9 \operatorname{cosec}^{2} \theta ?$ Solution: We have, $9 \cot ^{2} \theta-9 \operatorname{cosec}^{2} \theta=9\left(\cot ^{2} \theta-\operatorname{cosec}^{2} \theta\right)$ $=-9\left(\operatorname{cosec}^{2} \theta-\cot ^{2} \theta\right)$ We know that, $\operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1$ Therefore, $9 \cot ^{2} \theta-9 \operatorname{cosec}^{2} \theta=-9$...

1.3 + 2.4 + 3.5 + ... + n.

Question: $1.3+2.4+3.5+\ldots+n .(n+2)=\frac{1}{6} n(n+1)(2 n+7)$ Solution: LetP(n) be the given statement. Now, $P(n)=1.3+2.4+3.5+\ldots+n .(n+2)=\frac{1}{6} n(n+1)(2 n+7)$ Step 1: $P(1)=1.3=3=\frac{1}{6} \times 1(1+1)(2 \times 1+7)$ Hence, $P(1)$ is true. Step 2 : Let $P(m)$ be true. Then, 1. $3+2 \cdot 4+\ldots+m \cdot(m+2)=\frac{1}{6} m(m+1)(2 m+7)$ To prove: $P(m+1)$ is true. That is, $1.3+2.4+\ldots+(m+1)(m+3)=\frac{1}{6}(m+1)(m+2)(2 m+9)$ $P(m)$ is equal to $1.3+2.4+\ldots+m(m+2)=\frac{1}...

The value of

Question: The value of $0 . \overline{2}$ in the form $\frac{p}{g}$, where $p$ and $q$ are integers and $q \neq 0$, is (a) $\frac{1}{5}$ (b) $\frac{2}{9}$ (c) $\frac{2}{5}$ (d) $\frac{1}{8}$ Solution: Let $x=0 . \overline{2}=0.222 \ldots \quad \ldots$ (1) Multiplying both sides by 10, we get $10 x=2 . \overline{2} \quad \ldots \ldots(2)$ Subtracting (1) from (2), we get $10 x-x=2 . \overline{2}-0 . \overline{2}$ $\Rightarrow 9 x=2$ $\Rightarrow x=\frac{2}{9}$ $\therefore 0 . \overline{2}=\frac{2...