Check whether the relation R defined on the set
Question: Check whether the relationRdefined on the setA= {1, 2, 3, 4, 5, 6} asR= {(a, b) :b = a+ 1} is reflexive, symmetric or transitive. Solution: Reflexivity: Let $a$ be an arbitrary element of R. Then, $a=a+1$ cannot be true for all $a \in A$ $\Rightarrow(a, a) \notin R$ So, $R$ is not reflexive on $A$. Symmetry: Let $(a, b) \in R$ $\Rightarrow b=a+1$ $\Rightarrow-a=-b+1$ $\Rightarrow a=b-1$ Thus, $(b, a) \notin R$ So, $R$ is not symmetric on $A$. Transitivity: Let $(1,2)$ and $(2,3) \in R$...
Read More →sin x+ sin 3x+...+sin
Question: $\sin x+\sin 3 x+\ldots+\sin (2 n-1) x=\frac{\sin ^{2} n x}{\sin x}$ Solution: Let P(n) be the given statement. $P(n): \sin x+\sin 3 x+\ldots+\sin (2 n-1) x=\frac{\sin ^{2} n x}{\sin x}$ Step 1: $P(1): \sin x=\frac{\sin ^{2} x}{\sin x}$ Thus, $P(1)$ is true. Step $2:$ Let $P(m)$ be true. $\therefore \sin x+\sin 3 x+\ldots+\sin (2 m-1) x=\frac{\sin ^{2} m x}{\sin x}$ We shall show that $P(m+1)$ is true. We know that $P(m)$ is true. $\therefore \sin x+\sin 3 x+\ldots+\sin (2 m-1)=\frac{\...
Read More →sin x+ sin 3x+...+sin
Question: $\sin x+\sin 3 x+\ldots+\sin (2 n-1) x=\frac{\sin ^{2} n x}{\sin x}$ Solution: Let P(n) be the given statement. $P(n): \sin x+\sin 3 x+\ldots+\sin (2 n-1) x=\frac{\sin ^{2} n x}{\sin x}$ Step 1: $P(1): \sin x=\frac{\sin ^{2} x}{\sin x}$ Thus, $P(1)$ is true. Step $2:$ Let $P(m)$ be true. $\therefore \sin x+\sin 3 x+\ldots+\sin (2 m-1) x=\frac{\sin ^{2} m x}{\sin x}$ We shall show that $P(m+1)$ is true. We know that $P(m)$ is true. $\therefore \sin x+\sin 3 x+\ldots+\sin (2 m-1)=\frac{\...
Read More →The value of
Question: The value of $x^{p-q} \cdot x^{q-r} \cdot x^{r-p}$ is equal to (a) 0 (b) 1 (C) $X$ (d) $x^{09 r}$ Solution: $x^{p-q} \cdot x^{q-r} \cdot x^{r-p}=x^{p-q+q-r+r-p}$ $=x^{0}$ $=1$ $\therefore$ The value of $x^{p-q} \cdot x^{q-r} \cdot x^{r-p}$ is equal to 1 . Hence, the correct option is (b)....
Read More →The value of
Question: The value of $\left[\frac{256 x^{16}}{81 y^{4}}\right]^{-\frac{1}{4}}$ is (a) $\frac{3 y}{8 x^{4}}$ (b) $\frac{3 y}{4 x^{4}}$ (C) $\frac{4 y}{5 x^{4}}$ (d) $\frac{4 x^{4}}{3 y}$ Solution: $\left[\frac{256 x^{16}}{81 y^{4}}\right]^{-\frac{1}{4}}=\left[\frac{2^{8} x^{16}}{3^{4} y^{4}}\right]^{-\frac{1}{4}}$ $=\left[\frac{3^{4} y^{4}}{2^{8} x^{16}}\right]^{\frac{1}{4}}$ $=\left[\frac{\left(3^{4}\right)^{\frac{1}{4}}\left(y^{4}\right)^{\frac{1}{4}}}{\left(2^{8}\right)^{\frac{1}{4}}\left(x^...
Read More →Solve the following
Question: x2n1+y2n1is divisible byx+yfor allnN. Solution: LetP(n) be the given statement. Now, $P(n): x^{2 n-1}+y^{2 n-1}$ is divisible by $x+y$ Step 1: $P(1): x^{2-1}+y^{2-1}=x+y$ is divisible by $x+y$ Step 2 : Let $P(m)$ be true. Also, $x^{2 m-1}+y^{2 m-1}$ is divisible by $x+y .$ Suppose : $x^{2 m-1}+y^{2 m-1}=\lambda(x+y)$ where $\lambda \in N \quad \ldots(1)$ We shall show that $P(m+1)$ is true whenever $P(m)$ is true. Now, $P(m+1)=x^{2 m+1}+y^{2 m+1}$ $=x^{2 m+1}+y^{2 m+1}-x^{2 m-1} \cdot ...
Read More →The following relations are defined on the set of real numbers.
Question: The following relations are defined on the set of real numbers. (i) $a R b$ if $a-b0$ (ii) $a R b$ if $1+a b0$ (iii) $a R b$ if $|a| \leq b$ Find whether these relations are reflexive, symmetric or transitive. Solution: (i)Reflexivity: Letabe an arbitrary element ofR. Then, $a \in R$ But $a-a=0 \ngtr 0$ So, this relation is not reflexive. Symmetry: Let $(a, b) \in R$ $\Rightarrow a-b0$ $\Rightarrow-(b-a)0$ $\Rightarrow b-a0$ So, the given relation is not symmetric. Transitivity: Let $(...
Read More →Solve the following
Question: x2n1+y2n1is divisible byx+yfor allnN. Solution: LetP(n) be the given statement. Now, $P(n): x^{2 n-1}+y^{2 n-1}$ is divisible by $x+y$ Step 1: $P(1): x^{2-1}+y^{2-1}=x+y$ is divisible by $x+y$ Step 2 : Let $P(m)$ be true. Also, $x^{2 m-1}+y^{2 m-1}$ is divisible by $x+y .$ Suppose : $x^{2 m-1}+y^{2 m-1}=\lambda(x+y)$ where $\lambda \in N \quad \ldots(1)$ We shall show that $P(m+1)$ is true whenever $P(m)$ is true. Now, $P(m+1)=x^{2 m+1}+y^{2 m+1}$ $=x^{2 m+1}+y^{2 m+1}-x^{2 m-1} \cdot ...
Read More →then the value of p is
Question: If $x=\frac{\sqrt{7}}{5}$ and $\frac{5}{x}=p \sqrt{7}$ then the value of $p$ is (a) $\frac{7}{25}$ (b) $\frac{25}{7}$ (c) $\frac{7}{15}$ (d) $\frac{15}{7}$ Solution: $\frac{5}{x}=p \sqrt{7}$ $\Rightarrow \frac{5}{\frac{\sqrt{7}}{5}}=p \sqrt{7} \quad\left(\because x=\frac{\sqrt{7}}{5}\right)$ $\Rightarrow \frac{5 \times 5}{\sqrt{7}}=p \sqrt{7}$ $\Rightarrow \frac{25}{\sqrt{7}}=p \sqrt{7}$ $\Rightarrow \frac{25}{\sqrt{7} \times \sqrt{7}}=p$ $\Rightarrow \frac{25}{7}=p$ $\therefore$ The v...
Read More →Solve the following
Question: $1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\ldots+\frac{1}{n^{2}}2-\frac{1}{n}$ for all $n \geq 2, n \in N$ Solution: LetP(n) be the given statement. Thus, we have: $P(n): 1+\frac{1}{4}+\frac{1}{9}+\ldots+\frac{1}{n^{2}}2-\frac{1}{n}$ Step $1: P(2): \frac{1}{2^{2}}=\frac{1}{4}2-\frac{1}{2}$ Thus, $P(2)$ is true. [We have not taken $n=1$ because it is not possible. We will start this function from $n=2$ onwards.] Step 2 : Let $P(m)$ be true. Now, $1+\frac{1}{4}+\frac{1}{9}+\ldots+\frac{1}{...
Read More →If cos A + cos2 A = 1, then sin2 A + sin4 A =
Question: If $\cos A+\cos ^{2} A=1$, then $\sin ^{2} A+\sin ^{4} A=$ (a) 1(b) 0(c) 1(d) None of these Solution: Given: $\cos A+\cos ^{2} A=1$ $\Rightarrow 1-\cos ^{2} A=\cos A$ So, $\sin ^{2} A+\sin ^{4} A$ $=\sin ^{2} A+\sin ^{2} A \sin ^{2} A$ $=\sin ^{2} A+\left(1-\cos ^{2} A\right)\left(1-\cos ^{2} A\right)$ $=\sin ^{2} A+\cos A \cos A$ $=\sin ^{2} A+\cos ^{2} A$ $=1$ Hence, the correct option is (c)....
Read More →Solve the following
Question: $\frac{(2 n) !}{2^{2 n}(n !)^{2}} \leq \frac{1}{\sqrt{3 n+1}}$ for all $n \in N$ Solution: LetP(n) be the given statement. Thus, we have: $P(n): \frac{(2 n) !}{2^{2 n}(n !)^{2}} \leq \frac{1}{\sqrt{3 n+1}}$ Step 1: $P(1): \frac{2 !}{2^{2} .1}=\frac{1}{2} \leq \frac{1}{\sqrt{3+1}}$ Thus, $P(1)$ is true. Step 2 : Let $P(m)$ be true. Thus, we have : $\frac{(2 m) !}{2^{2 m}(m !)^{2}} \leq \frac{1}{\sqrt{3 m+1}}$ We need to prove that $P(m+1)$ is true. Now, $P(m+1):$ $\frac{(2 m+2) !}{2^{2 ...
Read More →If a cos θ + b sin θ = m and a sin θ − b cos θ = n, then a2 + b2 =
Question: If $a \cos \theta+b \sin \theta=m$ and $a \sin \theta-b \cos \theta=n$, then $a^{2}+b^{2}=$ (a) $m^{2}-n^{2}$ (b) $m^{2} n^{2}$ (C) $n^{2}-m^{2}$ (d) $m^{2}+n^{2}$ Solution: Given: $a \cos \theta+b \sin \theta=m$ $a \sin \theta-b \cos \theta=n$ Squaring and adding these equations, we have $(a \cos \theta+b \sin \theta)^{2}+(a \sin \theta-b \cos \theta)^{2}=(m)^{2}+(n)^{2}$ $\Rightarrow\left(a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+2 a \cos \theta \cdot b \sin \theta\right)+\left(a...
Read More →There is a number x such that
Question: There is a number $x$ such that $x^{2}$ is irrational but $x^{4}$ is rational. Then, $x$ can be (a) $\sqrt{5}$ (b) $\sqrt{2}$ (c) $\sqrt[3]{2}$ (d) $\sqrt[4]{2}$ Solution: (a) Let $x=\sqrt{5}$. $x^{2}=(\sqrt{5})^{2}=5$, which is a rational number. (b) Let $x=\sqrt{2}$. $x^{2}=(\sqrt{2})^{2}=2$, which is a rational number. (c) Let $x=\sqrt[3]{2}$ $x^{2}=(\sqrt[3]{2})^{2}=(2)^{\frac{2}{3}}$, which is an irrational number. $x^{4}=(\sqrt[3]{2})^{4}=(2)^{\frac{4}{3}}$, which is also an irra...
Read More →Solve the following
Question: $\frac{(2 n) !}{2^{2 n}(n !)^{2}} \leq \frac{1}{\sqrt{3 n+1}}$ for all $n \in N$ Solution: LetP(n) be the given statement. Thus, we have: $P(n): \frac{(2 n) !}{2^{2 n}(n !)^{2}} \leq \frac{1}{\sqrt{3 n+1}}$ Step 1: $P(1): \frac{2 !}{2^{2} .1}=\frac{1}{2} \leq \frac{1}{\sqrt{3+1}}$ Thus, $P(1)$ is true. Step 2 : Let $P(m)$ be true. Thus, we have : $\frac{(2 m) !}{2^{2 m}(m !)^{2}} \leq \frac{1}{\sqrt{3 m+1}}$ We need to prove that $P(m+1)$ is true. Now, $P(m+1):$ $\frac{(2 m+2) !}{2^{2 ...
Read More →If sin θ + sin2 θ = 1, then cos2 θ + cos4 θ =
Question: If $\sin \theta+\sin ^{2} \theta=1$, then $\cos ^{2} \theta+\cos ^{4} \theta=$ (a) 1(b) 1(c) 0(d) None of these Solution: Given: $\sin \theta+\sin ^{2} \theta=1$ $\Rightarrow 1-\sin ^{2} \theta=\sin \theta$ Now, $\cos ^{2} \theta+\cos ^{4} \theta$ $=\cos ^{2} \theta+\cos ^{2} \theta \cos ^{2} \theta$ $=\cos ^{2} \theta+\left(1-\sin ^{2} \theta\right)\left(1-\sin ^{2} \theta\right)$ $=\cos ^{2} \theta+\sin \theta \sin \theta$ $=\cos ^{2} \theta+\sin ^{2} \theta$ $=1$ Hence, the correct ...
Read More →Let P(n) be the statement :
Question: LetP(n) be the statement : 2n 3n. IfP(r) is true, show thatP(r+ 1) is true. Do you conclude thatP(n) is true for allnN? Solution: $P(n): 2^{n} \geq 3 n$ We know that $P(r)$ is true. Th $u s$, we have : $2^{r} \geq 3 r$ To show: $P(r+1)$ is true. We know : $P(r)$ is true. $\therefore 2^{r} \geq 3 r$ $\Rightarrow 2^{r} .2 \geq 3 r .2 \quad$ [Multiplying both sides by 2] $\Rightarrow 2^{r+1} \geq 6 r$ $\Rightarrow 2^{r+1} \geq 3 r+3 r$ $=2^{r+1} \geq 3 r+3 \quad[$ Since $3 r \geq 3$ for a...
Read More →If x = r sin θ cos ϕ, y = r sin θ sin ϕ and z = r cos θ, then
Question: If $x=r \sin \theta \cos \phi, y=r \sin \theta \sin \phi$ and $z=r \cos \theta$, then (a) $x^{2}+y^{2}+z^{2}=r^{2}$ (b) $x^{2}+y^{2}-z^{2}=r^{2}$ (c) $x^{2}-y^{2}+z^{2}=r^{2}$ (d) $z^{2}+y^{2}-x^{2}=r^{2}$ Solution: Given: $x=r \sin \theta \cos \phi$, $y=r \sin \theta \sin \phi$, $z=r \cos \theta$ Squaring and adding these equations, we get $x^{2}+y^{2}+z^{2}=(r \sin \theta \cos \phi)^{2}+(r \sin \theta \sin \phi)^{2}+(r \cos \theta)^{2}$ $\Rightarrow x^{2}+y^{2}+z^{2}=r^{2} \sin ^{2} ...
Read More →The value of
Question: The value of $\left[(81)^{\frac{1}{2}}\right]^{\frac{1}{2}}$ is (a) 3 (b) $-3$ (C) 9 (d) $\frac{1}{3}$ Solution: $\left[(81)^{\frac{1}{2}}\right]^{\frac{1}{2}}=\left[\left(3^{4}\right)^{\frac{1}{2}}\right]^{\frac{1}{2}}$ $=\left[3^{2}\right]^{\frac{1}{2}}$ $=3$ $\therefore$ The value of $\left[(81)^{\frac{1}{2}}\right]^{\frac{1}{2}}$ is 3 Hence, the correct option is (a)....
Read More →The value of sin2 29° + sin2 61° is
Question: The value of $\sin ^{2} 29^{\circ}+\sin ^{2} 61^{\circ}$ is (a) 1 (b) 0 (c) $2 \sin ^{2} 29^{\circ}$ (d) $2 \cos ^{2} 61^{\circ}$ Solution: The given expression is $\sin ^{2} 29^{\circ}+\sin ^{2} 61^{\circ}$. $\sin ^{2} 29^{\circ}+\sin ^{2} 61^{\circ}$ $=\sin ^{2} 29^{\circ}+\left(\sin 61^{\circ}\right)^{2}$ $=\sin ^{2} 29^{\circ}+\left\{\sin \left(90^{\circ}-29^{\circ}\right)\right\}^{2}$ $=\sin ^{2} 29^{\circ}+\left(\cos 29^{\circ}\right)^{2}$ $=\sin ^{2} 29^{\circ}+\cos ^{2} 29^{\ci...
Read More →Solve the following
Question: $\frac{1}{2} \tan \left(\frac{x}{2}\right)+\frac{1}{4} \tan \left(\frac{x}{4}\right)+\ldots+\frac{1}{2^{n}} \tan \left(\frac{x}{2^{n}}\right)=\frac{1}{2^{n}} \cot \left(\frac{x}{2^{n}}\right)-\cot x$ for all $n \in N$ and $0x\frac{\pi}{2}$ Solution: We need to prove $\frac{1}{2} \tan \left(\frac{x}{2}\right)+\frac{1}{4} \tan \left(\frac{x}{4}\right)+\ldots+\frac{1}{2^{n}} \tan \left(\frac{x}{2^{n}}\right)=\frac{1}{2^{n}} \cot \left(\frac{x}{2^{n}}\right)-\cot x$ for all $n \in N$ and $...
Read More →If a cot θ + b cosec θ = p and b cot θ − a cosec θ = q, then p2 − q2 =
Question: If $a \cot \theta+b \operatorname{cosec} \theta=p$ and $b \cot \theta-a \operatorname{cosec} \theta=q$, then $p^{2}-q^{2}=$ (a) $a^{2}-b^{2}$ (b) $b^{2}-a^{2}$ (C) $a^{2}+b^{2}$ (d) $b-a$ Solution: Given: $a \cot \theta+b \operatorname{cosec} \theta=p$ $b \cot \theta+a \operatorname{cosec} \theta=q$ Squaring both the equations and then subtracting the second from the first, we have $(p)^{2}-(q)^{2}=(a \cot \theta+b \operatorname{cosec} \theta)^{2}-(b \cot \theta+a \operatorname{cosec} ...
Read More →Simplified value of
Question: Simplified value oi $(25)^{\frac{1}{3}} \times 5^{\frac{1}{3}}$ is (a) 25 (b) 3 (c) 1 (d) 5 Solution: $(25)^{\frac{1}{2}} \times 5^{\frac{1}{3}}=\left(5^{2}\right)^{\frac{1}{3}} \times 5^{\frac{1}{3}}$ $=5^{\frac{2}{3}} \times 5^{\frac{1}{3}}$ $=5^{\frac{2}{3}+\frac{1}{3}}$ $=5^{\frac{2+1}{3}}$ $=5^{\frac{3}{3}}$ $=5$ $\therefore$ Simplified value of $(25)^{\frac{1}{3}} \times 5^{\frac{1}{3}}$ is $5 .$ Hence, the correct option is (d)....
Read More →The value of
Question: The value of $\sqrt[4]{\sqrt[3]{2^{2}}}$ is (a) $2^{\frac{-1}{6}}$ (b) $2^{-6}$ (c) $2^{\frac{1}{6}}$ (d) $2^{6}$ Solution: $\sqrt[4]{\sqrt[3]{2^{2}}}=\left[\left(2^{2}\right)^{\frac{1}{3}}\right]^{\frac{1}{4}}$ $=\left[2^{\frac{2}{3}}\right]^{\frac{1}{4}}$ $=2^{\frac{2}{12}}$ $=2^{\frac{1}{6}}$ $\therefore$ The value of $\sqrt[4]{\sqrt[3]{2^{2}}}$ is $2^{\frac{1}{6}}$. Hence, the correct option is (c)....
Read More →If a cos θ + b sin θ = 4 and a sin θ − b sin θ = 3, then a2 + b2 =
Question: If $a \cos \theta+b \sin \theta=4$ and $a \sin \theta-b \sin \theta=3$, then $a^{2}+b^{2}=$ (a) 7(b) 12(c) 25(d) None of these Solution: Given: $a \cos \theta+b \sin \theta=4$, $a \sin \theta-b \cos \theta=3$ Squaring and then adding the above two equations, we have $(a \cos \theta+b \sin \theta)^{2}+(a \sin \theta-b \cos \theta)^{2}=(4)^{2}+(3)^{2}$ $\Rightarrow\left(a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+2 \cdot a \cos \theta \cdot b \sin \theta\right)+\left(a^{2} \sin ^{2} \t...
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