The number of telephone calls received at an exchange per interval for 250 successive one-minute intervals are given in the following frequency table:
Question: The number of telephone calls received at an exchange per interval for 250 successive one-minute intervals are given in the following frequency table: Compute the mean number of calls per intervals. Solution: Let the assume mean be $A=3$. We know that mean, $\bar{X}=A+\frac{1}{N} \sum_{i=1}^{n} f_{i} d_{i}$ Here, we have $N=\sum f_{i}=250, \sum f_{i} d_{i}=135$ and $A=3$. Putting the values in the formula, we get $\bar{X}=A+\frac{1}{N} \sum_{i=1}^{n} f_{i} d_{i}$ $=3+\frac{1}{250} \tim...
Read More →The rationalisation factor of
Question: The rationalisation factor of $\frac{1}{2 \sqrt{3}-\sqrt{5}}$ is (a) $\sqrt{5}-2 \sqrt{3}$ (b) $\sqrt{3}+2 \sqrt{5}$ (c) $(\sqrt{3}+\sqrt{5})$ (d) $\sqrt{12}+\sqrt{5}$ Solution: Rationalisation factor of $\frac{1}{2 \sqrt{3}-\sqrt{5}}$ will be $2 \sqrt{3}+\sqrt{5}=\sqrt{4 \times 3}+\sqrt{5}=\sqrt{12}+\sqrt{5}$. Hence, the correct answer is option (d)....
Read More →Is it true that every relation which is symmetric and transitive is also reflexive? Give reasons.
Question: Is it true that every relation which is symmetric and transitive is also reflexive? Give reasons. Solution: No, it is not true. Consider a set A = {1, 2, 3} and relationRonAsuch thatR= {(1, 2), (2, 1), (2, 3), (1, 3)}The relationRonAis symmetric and transitive. However, it is not reflexive. $(1,1),(2,2)$ and $(3,3) \notin R$ Hence,Ris not reflexive....
Read More →The simplest rationalisation ractor of
Question: The simplest rationalisation ractor of $(2 \sqrt{2}-\sqrt{3})$ is (a) $2 \sqrt{2}+3$ (b) $2 \sqrt{2}+\sqrt{3}$ (c) $\sqrt{2}+\sqrt{3}$ (d) $\sqrt{2}-\sqrt{3}$ Solution: Simplest rationalisation ractor of $(2 \sqrt{2}-\sqrt{3})$ is $2 \sqrt{2}+\sqrt{3}$. Hence, the correct answer is option (b)....
Read More →The simplest rationalisation factor of
Question: The simplest rationalisation factor of $\sqrt[3]{500}$ is (a) $\sqrt{5}$ (b) $\sqrt{3}$ (c) $\sqrt[3]{5}$ (d) $\sqrt[3]{2}$ Solution: $\sqrt[3]{500}=\sqrt[3]{5 \times 5 \times 2 \times 5 \times 2}=\sqrt[3]{5^{3} \times 2^{2}}=5 \sqrt[3]{4}$ So, the simplest rationalisation factor of $\sqrt[3]{500}$ is $\sqrt[3]{2}$. Hence, the correct answer is option (d)....
Read More →Let R be a relation defined on the set of natural numbers N as
Question: LetRbe a relation defined on the set of natural numbersNasR= {(x,y) :x,yN, 2x+y= 41}Find the domain and range ofR. Also, verify whetherRis (i) reflexive, (ii) symmetric (iii) transitive. Solution: Domain ofRis the values ofxand range ofRis the values ofythat together should satisfy 2x+y= 41.So,Domain ofR= {1, 2, 3, 4, ... , 20}Range ofR= {1, 3, 5, ... , 37, 39} Reflexivity: Letxbe an arbitrary element ofR. Then, $x \in R$ $\Rightarrow 2 x+x=41$ cannot be true. $\Rightarrow(x, x) \notin...
Read More →Let P(n) be the statement:
Question: Let $\mathrm{P}(n)$ be the statement $: 2^{n} \geq 3 n$. If $\mathrm{P}(r)$ is true, then show that $\mathrm{P}(r+1)$ is true. Do you conclude that $\mathrm{P}(n)$ is true for all $n \in \mathbf{N}$ ? Solution: Since, for $n=1$ i.e. P $(1)$ : LHS $=2^{1}=2$ RHS $=3 \times 1=3$ As, LHS $$ RHS So, it is not true for $n=1$. Hence, we conclude that $\mathrm{P}(n)$ is not true for all $n \in \mathbf{N}$....
Read More →1+tan2 A1+cot2 Ais equal to
Question: $\frac{1+\tan ^{2} A}{1+\cot ^{2} A}$ is equal to (a) $\sec ^{2} A$ (b) $-1$ (c) $\cot ^{2} A$ (d) $\tan ^{2} A$ Solution: Given: $\frac{1+\tan ^{2} A}{1+\cot ^{2} A}$ $=\frac{1+\frac{\sin ^{2} A}{\cos ^{2} A}}{1+\frac{\cos ^{2} A}{\sin ^{2} A}}$ $=\frac{\frac{\cos ^{2} A+\sin ^{2} A}{\cos ^{2} A}}{\frac{\sin ^{2} A+\cos ^{2} A}{\sin ^{2} A}}$ $=\frac{\frac{1}{\cos ^{2} A}}{\frac{1}{\sin ^{2} A}}$ $=\frac{\sin ^{2} A}{\cos ^{2} A}$ $=\tan ^{2} A$ Therefore, the correct option is (d)....
Read More →On simplification, the expression
Question: On simplification, the expression $\frac{5^{n+2}-6 \times 5^{n+1}}{13 \times 5^{n}-2 \times 5^{n+1}}$ equals (a) $\frac{5}{3}$ (b) $-\frac{5}{3}$ (C) $\frac{3}{5}$ (d) $-\frac{3}{5}$ Solution: $\frac{5^{n+2}-6 \times 5^{n+1}}{13 \times 5^{n}-2 \times 5^{n+1}}$ $=\frac{5^{n} \times 5^{2}-6 \times 5^{n} \times 5}{13 \times 5^{n}-2 \times 5^{n} \times 5}$ $=\frac{5^{n} \times 5[5-6]}{5^{n}[13-2 \times 5]}$ $=\frac{-5}{3}$ Hence, the correct answer is option (b)....
Read More →Solve the following
Question: Given $a_{1}=\frac{1}{2}\left(a_{0}+\frac{A}{a_{0}}\right), a_{2}=\frac{1}{2}\left(a_{1}+\frac{A}{a_{1}}\right)$ and $a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{A}{a_{n}}\right)$ for $n \geq 2$, where $a0, A$ $0$. Prove that $\frac{a_{n}-\sqrt{A}}{a_{n}+\sqrt{A}}=\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right) 2^{n-1}$. Solution: Given: $a_{1}=\frac{1}{2}\left(a_{0}+\frac{A}{a_{0}}\right), a_{2}=\frac{1}{2}\left(a_{1}+\frac{A}{a_{1}}\right)$ and $a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{A}{a_...
Read More →Solve the following
Question: Given $a_{1}=\frac{1}{2}\left(a_{0}+\frac{A}{a_{0}}\right), a_{2}=\frac{1}{2}\left(a_{1}+\frac{A}{a_{1}}\right)$ and $a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{A}{a_{n}}\right)$ for $n \geq 2$, where $a0, A$ $0$. Prove that $\frac{a_{n}-\sqrt{A}}{a_{n}+\sqrt{A}}=\left(\frac{a_{1}-\sqrt{A}}{a_{1}+\sqrt{A}}\right) 2^{n-1}$. Solution: Given: $a_{1}=\frac{1}{2}\left(a_{0}+\frac{A}{a_{0}}\right), a_{2}=\frac{1}{2}\left(a_{1}+\frac{A}{a_{1}}\right)$ and $a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{A}{a_...
Read More →(sec A + tan A) (1 − sin A) =
Question: $(\sec A+\tan A)(1-\sin A)=$ (a) sec A(b) sin A(c) cosec A(d) cos A Solution: The given expression is $(\sec A+\tan A)(1-\sin A)$. Simplifying the given expression, we have $(\sec A+\tan A)(1-\sin A)$ $=\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)(1-\sin A)$ $=\left(\frac{1+\sin A}{\cos A}\right) \times(1-\sin A)$ $=\frac{1-\sin ^{2} A}{\cos A}$ $=\frac{\cos ^{2} A}{\cos A}$ $=\cos A$ Therefore, the correct option is (d)....
Read More →Solve this
Question: If $\left(3^{3}\right)^{2}=9^{x}$ then $5^{x}=?$ (a) 1 (b) 5 (c) 25 (d) 125 Solution: $\left(3^{3}\right)^{2}=9^{x}$ $\Rightarrow\left(3^{2}\right)^{3}=\left(3^{2}\right)^{x}$ $\Rightarrow x=3$ Now $5^{x}=5^{3}=125$ Hence, the correct answer is option (d)....
Read More →(1 + tan θ + sec θ) (1 + cot θ − cosec θ) =
Question: $(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)=$ (a) 0(b) 1(c) 1(d) 1 Solution: The given expression is $(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)$. Simplifying the given expression, we have $(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)$ $=\left(1+\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right)\left(1+\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta}\right)$ $=\frac{\cos \theta+\sin \theta+...
Read More →If A = {1, 2, 3, 4} define relations on A which have properties of being
Question: IfA= {1, 2, 3, 4} define relations onAwhich have properties of being(i) reflexive, transitive but not symmetric(ii) symmetric but neither reflexive nor transitive(iii) reflexive, symmetric and transitive. Solution: (i) The relation onAhaving properties of being reflexive, transitive, but not symmetric isR= {(1, 1), (2, 2), (3, 3), (4, 4), (2, 1)} Relation $R$ satisfies reflexivity and transitivity. $\Rightarrow(1,1),(2,2),(3,3) \in R$ and $(1,1),(2,1) \in R \Rightarrow(1,1) \in R$ Howe...
Read More →Prove that every identity relation on a set is reflexive, but the converse is not necessarily true.
Question: Prove that every identity relation on a set is reflexive, but the converse is not necessarily true. Solution: LetAbe a set. Then, Identity relation IA $=I_{A}$ is reflexive, since $(a, a) \in A \forall a$ The converse of it need not be necessarily true.Consider the setA= {1, 2, 3} Here,RelationR= {(1, 1), (2, 2) , (3, 3), (2, 1), (1, 3)} is reflexive onA.However,Ris not an identity relation....
Read More →solve this
Question: If $\left(\frac{2}{3}\right)^{x}\left(\frac{3}{2}\right)^{2 x}=\frac{81}{16}$ then $x=?$ (a) 1 (b) 2 (c) 3 (d) 4 Solution: $\left(\frac{2}{3}\right)^{x}\left(\frac{3}{2}\right)^{2 x}=\frac{81}{16}$ $\Rightarrow\left(\frac{2}{3}\right)^{x}\left(\frac{2}{3}\right)^{-2 x}=\frac{81}{16}$ $\Rightarrow\left(\frac{2}{3}\right)^{x-2 x}=\frac{81}{16}$ $\Rightarrow\left(\frac{2}{3}\right)^{-x}=\frac{3^{4}}{2^{4}}$ $\Rightarrow\left(\frac{3}{2}\right)^{x}=\frac{3^{4}}{2^{4}}$ $\Rightarrow x=4$ He...
Read More →Solve the following
Question: Prove that $\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2 n}\frac{13}{24}$, for all natural numbers $n1$.[NCERT EXEMPLAR] Solution: Let $\mathrm{p}(n): \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2 n}\frac{13}{24}$, for all natural numbers $n1$. Step I: For $n=2$, $\mathrm{LHS}=\frac{1}{2+1}+\frac{1}{2 \times 2}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}=\frac{14}{24}\frac{13}{24}=\mathrm{RHS}$ As, LHS $$ RHS So, it is true for $n=2$. Step II : For $n=k$, Let $\mathrm{p}(k): \frac{1}{k+1}+\f...
Read More →9 sec2 A − 9 tan2 A is equal to
Question: $9 \sec ^{2} A-9 \tan ^{2} A$ is equal to (a) 1(b) 9(c) 8(d) 0 Solution: Given: $9 \sec ^{2} A-9 \tan ^{2} A$ $=9\left(\sec ^{2} A-\tan ^{2} A\right)$ We know that, $\sec ^{2} A-\tan ^{2} A=1$ Therefore, $9 \sec ^{2} A-9 \tan ^{2} A=9$ Hence, the correct option is (b)....
Read More →Check whether the relation $R$ on
Question: Check whether the relation $R$ on $\mathbf{R}$ defined by $R=\left\{(a, b): a \leq b^{3}\right\}$ is reflexive, symmetric or transitive. Solution: Reflexivity: Since $\frac{1}{2}\left(\frac{1}{2}\right)^{3}$ $\left(\frac{1}{2}, \frac{1}{2}\right) \notin R$ So, $R$ is not reflexive. Symmetry: Since $\left(\frac{1}{2}, 2\right) \in R$, $\frac{1}{2}2^{3}$ But $2\left(\frac{1}{2}\right)^{3}$ $\Rightarrow\left(2, \frac{1}{2}\right) \in R$ So, $R$ is not symmetric. Transitivity: Since $(7,3)...
Read More →If a cos θ − b sin θ = c, then a sin θ + b cos θ =
Question: If $a \cos \theta-b \sin \theta=c$, then $a \sin \theta+b \cos \theta=$ (a) $\pm \sqrt{a^{2}+b^{2}+c^{2}}$ (b) $\pm \sqrt{a^{2}+b^{2}-c^{2}}$ (c) $\pm \sqrt{c^{2}-a^{2}-b^{2}}$ (d) None of these Solution: Given: $a \cos \theta-b \sin \theta=c$ Squaring on both sides, we have $(a \cos \theta-b \sin \theta)^{2}=c^{2}$ $\Rightarrow a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta-2 \cdot a \cos \theta \cdot b \sin \theta=c^{2}$ $\Rightarrow a^{2}\left(1-\sin ^{2} \theta\right)+b^{2}\left(1-\...
Read More →solve this
Question: $\sqrt[3]{2} \times \sqrt[4]{2} \times \sqrt[12]{32}=?$ (a) 2 (b) $\sqrt{2}$ (c) $2 \sqrt{2}$ (d) $4 \sqrt{2}$ Solution: $\sqrt[3]{2} \times \sqrt[4]{2} \times \sqrt[12]{32}=(2)^{\frac{1}{3}} \times(2)^{\frac{1}{4}} \times(32)^{\frac{1}{12}}$ $=(2)^{\frac{1}{3}} \times(2)^{\frac{1}{4}} \times\left(2^{5}\right)^{\frac{1}{12}}$ $=(2)^{\frac{1}{3}} \times(2)^{\frac{1}{4}} \times(2)^{\frac{5}{12}}$ $=(2)^{\frac{1}{3}+\frac{1}{4}+\frac{5}{12}}$ $=(2)^{\frac{4+3+5}{12}}$ $=(2)^{\frac{12}{12}...
Read More →Prove that cosα+cos(α+β)+cos(α+2β)+...
Question: Prove that $\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos [\alpha+(n-1) \beta]=\frac{\cos \left\{\alpha+\left(\frac{n-1}{2}\right) \beta\right\} \sin \left(\frac{n \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$ for all $n \in \mathbf{N}$. [NCERT EXEMPLAR] Solution: Let $\mathrm{p}(n): \cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos [\alpha+(n-1) \beta]=\frac{\cos \left\{\alpha+\left(\frac{n-1}{2}\right) \beta\right\} \sin \left(\frac{n \beta...
Read More →If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, then x2a2+y2b2=
Question: If $x=a \sec \theta \cos \phi, y=b \sec \theta \sin \phi$ and $z=c \tan \theta$, then $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=$ (a) $\frac{z^{2}}{c^{2}}$ (b) $1-\frac{z^{2}}{c^{2}}$ (C) $\frac{z^{2}}{c^{2}}-1$ (d) $1+\frac{z^{2}}{c^{2}}$ Solution: Given: $x=a \sec \theta \cos \phi$ $\Rightarrow \frac{x}{a}=\sec \theta \cos \phi$ $y=b \sec \theta \sin \phi$ $\Rightarrow \frac{y}{h}=\sec \theta \sin \phi$ $z=c \tan \theta$ $\Rightarrow \frac{z}{c}=\tan \theta$ Now, $\left(\frac{x}{a}\ri...
Read More →The value of
Question: The value of $\sqrt{p^{-1} q} \cdot \sqrt{q^{-1} r} \cdot \sqrt{r^{-1} p}$ is (a) $-1$ (b) 0 (C) 1 (d) 2 Solution: $\sqrt{p^{-1} q} \cdot \sqrt{q^{-1} r} \cdot \sqrt{r^{-1} p}=\left(p^{-1} q\right)^{\frac{1}{2}} \cdot\left(q^{-1} r\right)^{\frac{1}{2}} \cdot\left(r^{-1} p\right)^{\frac{1}{2}}$ $=\left(p^{-\frac{1}{2}} q^{\frac{1}{2}}\right) \cdot\left(q^{-\frac{1}{2}} r^{\frac{1}{2}}\right) \cdot\left(r^{-\frac{1}{2}} p^{\frac{1}{2}}\right)$ $=p^{-\frac{1}{2}+\frac{1}{2}} q^{\frac{1}{2...
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