Solve the following
Question: $\frac{n^{11}}{11}+\frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{62}{165} n$ is a positive integer for all $n \in N$ Solution: Let P(n) be the given statement. Now, $P(n): \frac{n^{11}}{11}+\frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{62}{165} n$ is a positive integer for all $n \in N$. Step 1: $P(1)=\frac{1}{11}+\frac{1}{5}+\frac{1}{3}+\frac{62}{165}=\frac{15+33+55+62}{165}=\frac{165}{165}=1$ It is certainly a positive integer. Hence, $P(1)$ is true. Step 2 : Le $t P(m)$ be true. Then, $\frac{m^{11}...
Read More →Let
Question: LetA= {1, 2, 3}, and letR1= {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)},R2= {(2, 2), (3, 1), (1, 3)},R3= {(1, 3), (3, 3)}. Find whether or not each of the relationsR1,R2,R3on A is (i) reflexive (ii) symmetric (iii) transitive. Solution: (1)1R1Reflexivity:Here, $(1,1),(2,2),(3,3) \in R$ So, $R_{1}$ is reflexive. Symmetry: Here, $(2,1) \in R_{1}$, but $(1,2) \notin R_{1}$ So, $R_{1}$ is not symmetric. Transitivity: Here, $(2,1) \in R_{1}$ and $(1,3) \in R_{1}$, but $(2,3) \notin R_{...
Read More →2 (sin6 θ + cos6 θ) − 3 (sin4 θ + cos4 θ) is equal to
Question: $2\left(\sin ^{6} \theta+\cos ^{6} \theta\right)-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)$ is equal to (a) 0(b) 1(c) 1(d) None of these Solution: The given expression is $2\left(\sin ^{6} \theta+\cos ^{6} \theta\right)-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)$. Simplifying the given expression, we have $2\left(\sin ^{6} \theta+\cos ^{6} \theta\right)-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)$ $=2 \sin ^{6} \theta+2 \cos ^{6} \theta-3 \sin ^{4} \theta-3 \cos ^{4} \th...
Read More →Simplified value of
Question: Simplified value of $(16)^{-\frac{1}{4}} \times \sqrt[4]{16}$ is (a) 0 (b) 1 (c) 4 (d) 16 Solution: $(16)^{-\frac{1}{4}} \times \sqrt[4]{16}=\left(2^{4}\right)^{-\frac{1}{4}} \times\left(2^{4}\right)^{\frac{1}{4}}$ $=(2)^{-1} \times(2)^{1}$ $=2^{-1+1}$ $=2^{0}$ $=1$ $\therefore$ Simplified value of $(16)^{-\frac{1}{4}} \times \sqrt[4]{16}$ is 1 Hence, the correct option is (b)....
Read More →solve this
Question: $9^{3}+(-3)^{3}-6^{3}=?$ (a) 432(b) 270(c) 486(d) 540 Solution: $9^{3}+(-3)^{3}-6^{3}=(729)+(-27)-216$ $=729-27-216$ $=729-(27+216)$ $=729-243$ $=486$ $\therefore 9^{3}+(-3)^{3}-6^{3}=486$ Hence, the correct option is (c)....
Read More →Solve the following
Question: $\frac{n^{7}}{7}+\frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{n^{2}}{2}-\frac{37}{210} n$ is a positive integer for all $n \in N$. Solution: Let P(n) be the given statement. Now, $P(n): \frac{n^{7}}{7}+\frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{n^{2}}{2}-\frac{37}{210} n$ is a positive integer. Step 1: $P(1)=\frac{1}{7}+\frac{1}{5}+\frac{1}{3}+\frac{1}{2}-\frac{37}{210}=\frac{30+42+70+105-37}{210}=\frac{210}{210}=1$ It is a positive integer. Thus, $P(1)$ is true. Step 2 : Let $P(m)$ be true. Then,...
Read More →cot θcot θ−cot 3θ+tan θtan θ−tan 3θis equal to
Question: $\frac{\cot \theta}{\cot \theta-\cot 3 \theta}+\frac{\tan \theta}{\tan \theta-\tan 3 \theta}$ is equal to (a) 0(b) 1(c) 1(d) 2 Solution: The given expression is $\frac{\cot \theta}{\cot \theta-\cot 3 \theta}+\frac{\tan \theta}{\tan \theta-\tan 3 \theta}$. Simplifying the given expression, we have $\frac{\cot \theta}{\cot \theta-\cot 3 \theta}+\frac{\tan \theta}{\tan \theta-\tan 3 \theta}$ $=\frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta}{\sin \theta}-\frac{\cos 3 \theta}{\sin...
Read More →The value of
Question: The value of $(243)^{\frac{1}{5}}$ is (a) 3 (b) $-3$ (c) 5 (d) $\frac{1}{3}$ Solution: $(243)^{\frac{1}{5}}=\left(3^{5}\right)^{\frac{1}{5}}$ $=3^{1}$ $=3$ $\therefore$ The value of $(243)^{\frac{1}{5}}$ is 3 Hence, the correct option is (a)....
Read More →7 + 77 + 777 + ... + 777
Question: $7+77+777+\ldots+777 \ldots \ldots \ldots \ldots 7=\frac{7}{81}\left(10^{n+1}-9 n-10\right)$ Solution: Let P(n) be the given statement. Now, $P(n): 7+77+777+\ldots+777 \ldots n$ digits $\ldots 7=\frac{7}{81}\left(10^{n+1}-9 n-10\right)$ Step 1: $P(1)=7=\frac{7}{81}\left(10^{2}-9-10\right)=\frac{7}{81} \times 81$ Thus, $P(1)$ is true. Step 2 : Let $P(m)$ be true. Then, $7+77+777+\ldots+777 \ldots m$ digits $\ldots 7=\frac{7}{81}\left(10^{m+1}-9 m-10\right)$ We need to show that $P(m+1)$...
Read More →The value of
Question: The value of $\frac{2^{0}+7^{0}}{5^{0}}$ is (a) 0 (b) 2 (c) $\frac{9}{5}$ (d) $\frac{1}{5}$ Solution: $\frac{2^{0}+7^{0}}{5^{0}}=\frac{1+1}{1}$ $=\frac{2}{1}$ $=2$ $\therefore$ The value of $\frac{2^{0}+7^{0}}{5^{0}}$ is 2 . Hence, the correct option is (b)....
Read More →If x = a sec θ and y = b tan θ, then b2x2 − a2y2 =
Question: If $x=a \sec \theta$ and $y=b \tan \theta$, then $b^{2} x^{2}-a^{2} y^{2}=$ (a) $a b$ (b) $a^{2}-b^{2}$ (c) $a^{2}+b^{2}$ (d) $a^{2} b^{2}$ Solution: Given: $x=a \sec \theta, y=b \tan \theta$ So, $b^{2} x^{2}-a^{2} y^{2}$ $=b^{2}(a \sec \theta)^{2}-a^{2}(b \tan \theta)^{2}$ $=b^{2} a^{2} \sec ^{2} \theta-a^{2} b^{2} \tan ^{2} \theta$ $=b^{2} a^{2}\left(\sec ^{2} \theta-\tan ^{2} \theta\right)$ We know that, $\sec ^{2} \theta-\tan ^{2} \theta=1$ Therefore, $b^{2} x^{2}-a^{2} y^{2}=a^{2}...
Read More →If x = a cos θ and y = b sin θ, then b2x2 + a2y2 =
Question: If $x=a \cos \theta$ and $y=b \sin \theta$, then $b^{2} x^{2}+a^{2} y^{2}=$ (a) $a^{2} b^{2}$ (b) $a b$ (c) $a^{4} b^{4}$ (d) $a^{2}+b^{2}$ Solution: Given: $x=a \cos \theta, y=b \sin \theta$ So, $b^{2} x^{2}+a^{2} y^{2}$ $=b^{2}(a \cos \theta)^{2}+a^{2}(b \sin \theta)^{2}$ $=b^{2} a^{2} \cos ^{2} \theta+a^{2} b^{2} \sin ^{2} \theta$ $=b^{2} a^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)$ We know that, $\sin ^{2} \theta+\cos ^{2} \theta=1$ Therefore, $b^{2} x^{2}+a^{2} y^{2}=a^{2}...
Read More →The value of
Question: The value of $\sqrt[4]{(64)^{-2}}$ is (a) $\frac{1}{8}$ (b) $\frac{1}{2}$ (c) 8 (d) $\frac{1}{64}$ Solution: $\sqrt[4]{(64)^{-2}}=\left[(64)^{-2}\right]^{\frac{1}{4}}$ $=\left[\left(2^{6}\right)^{-2}\right]^{\frac{1}{4}}$ $=\left[2^{-12}\right]^{\frac{1}{4}}$ $=2^{-3}$ $=\frac{1}{2^{3}}$ $=\frac{1}{8}$ $\therefore$ The value of $\sqrt[4]{(64)^{-2}}$ is $\frac{1}{8}$. Hence, the correct option is (a)....
Read More →Prove that 1 + 2 + 2
Question: Prove that $1+2+2^{2}+\ldots+2^{n}=2^{n+1}-1$ for all $n \in \mathbf{N}$. Solution: Let $\mathrm{p}(n): 1+2+2^{2}+\ldots+2^{n}=2^{n+1}-1 \forall n \in \mathbf{N}$ Step I: For $n=1$, $\mathrm{LHS}=1+2^{1}=3$ $\mathrm{RHS}=2^{1+1}-1=2^{2}-1=4-1=3$ As, $\mathrm{LHS}=\mathrm{RHS}$ So, it is true for $n=1$. Step II : For $n=k$, Let $\mathrm{p}(k): 1+2+2^{2}+\ldots+2^{k}=2^{k+1}-1$ be true $\forall k \in \mathbf{N}$ $\mathrm{LHS}=1+2+2^{2}+\ldots+2^{k}+2^{k+1}$ $=2^{k+1}-1+2^{k+1} \quad($ Us...
Read More →After simplification,
Question: After simplification, $\frac{13^{\frac{1}{5}}}{13^{\frac{1}{3}}}$ is (a) $13^{\frac{2}{15}}$ (b) $13^{\frac{8}{15}}$ (c) $13^{\frac{1}{3}}$ (d) $13^{\frac{-2}{15}}$ Solution: $\frac{13 \frac{1}{5}}{13^{\frac{1}{3}}}$ $=13^{\frac{1}{5}-\frac{1}{3}} \quad\left(\frac{x^{a}}{x^{b}}=x^{a-b}\right)$ $=13^{\frac{3-5}{15}}$ $=13^{\frac{-2}{15}}$ Hence, the correct answer is option (d)....
Read More →Solve the following
Question: Prove that $n^{3}-7 n+3$ is divisible by 3 for all $n \in \mathbf{N}$. Solution: Let $\mathrm{p}(n)=n^{3}-7 n+3$ is divisible by $3 \forall n \in \mathbf{N}$. Step I : For $n=1$, $\mathrm{p}(1)=1^{3}-7 \times 1+3=1-7+3=-3$, which is clearly divisible by 3 So, it is true for $n=1$ Step II : For $n=k$, Let $\mathrm{p}(k)=k^{3}-7 k+3=3 m$, where $m$ is any integer, be true $\forall k \in \mathbf{N}$. Step III : For $n=k+1$ $\mathrm{p}(k+1)=(k+1)^{3}-7(k+1)+3$ $=k^{3}+3 k^{2}+3 k+1-7 k-7+3...
Read More →(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to
Question: $(\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)(\tan \theta+\cot \theta)$ is equal to (a) 0(b) 1(c) 1(d) None of these Solution: The given expression is $(\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)(\tan \theta+\cot \theta)$ Simplifying the given expression, we have $(\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)(\tan \theta+\cot \theta)$ $=\left(\frac{1}{\sin \theta}-\sin \theta\right)\left(\frac{1}{\cos \theta}-\cos \theta...
Read More →Test whether the following relations
Question: Test whether the following relationsR1,R2, andR3are (i) reflexive (ii) symmetric and (iii) transitive: (i) $R_{1}$ on $Q_{0}$ defined by $(a, b) \in R_{1} \Leftrightarrow a=1 / b$. (ii) $R_{2}$ on $Z$ defined by $(a, b) \in R_{2} \Leftrightarrow|a-b| \leq 5$ (iii) $R_{3}$ on $R$ defined by $(a, b) \in R_{3} \Leftrightarrow a^{2}-4 a b+3 b^{2}=0$. Solution: (i) Reflexivity:Letabe an arbitrary element ofR1. Then, $a \in R_{1}$ $\Rightarrow a \neq \frac{1}{a}$ for all $a \in \mathrm{Q}_{0...
Read More →The value of
Question: The value of $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}$ is (a) $(28)^{1 / 2}$ (b) $(56)^{1 / 2}$ (c) $(14)^{1 / 2}$ (d) $(42)^{1 / 2}$ Solution: $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}$ $=(7 \times 8)^{\frac{1}{2}} \quad\left[x^{a} \times y^{a}=(x \times y)^{a}\right]$ $=(56)^{\frac{1}{2}}$ Hence, the correct answer is option (b)....
Read More →Given a1=
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 $a0, A0 .$ 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: Let: $\mathrm{P}(\mathrm{n}): \frac{\mathrm{a}_{\mathrm{n}}-\sqrt{\mathrm{A}}}{\mathrm{a}_{\mathrm{n}}+\sqrt{\mathrm{A}}}=\left(\frac{\mathrm{a}_{1}-\sqrt{\mathrm{A...
Read More →tan θsec θ−1+tan θsec θ+1 is equal to
Question: $\frac{\tan \theta}{\sec \theta-1}+\frac{\tan \theta}{\sec \theta+1}$ is equal to (a) 2 tan (b) 2 sec (c) 2 cosec (d) 2 tan sec Solution: The given expression is $\frac{\tan \theta}{\sec \theta-1}+\frac{\tan \theta}{\sec \theta+1}$. Simplifying the given expression, we have $\frac{\tan \theta}{\sec \theta-1}+\frac{\tan \theta}{\sec \theta+1}$ $=\frac{\tan \theta(\sec \theta+1)+\tan \theta(\sec \theta-1)}{(\sec \theta-1)(\sec \theta+1)}$ $=\frac{\tan \theta \sec \theta+\tan \theta+\tan ...
Read More →Solve the following
Question: 11n+2+ 122n+1is divisible by 133 for allnN. Solution: Let P(n) be the given statement. Now, $P(n): 11^{n+2}+12^{2 n+1}$ is divisible by $133 .$ Step 1: $P(1)=11^{1+2}+12^{2+1}=1331+1728=3059$ It is divisible by 133 . Step 2 : Let $P(m)$ be divisible by 133 . Now, $11^{m+2}+12^{2 m+1}$ is divisible by 133 . Suppose : $11^{m+2}+12^{2 m+1}=133 \lambda \quad \ldots$ (1) We shall show that $P(m+1)$ is true whenever $P(m)$ is true. Now, $P(m+1)=11^{m+3}+12^{2 m+3}$ $=11^{m+2} \cdot 11+12^{2 ...
Read More →Solve the following
Question: 2.7n+ 3.5n 5 is divisible by 24 for allnN. Solution: LetP(n) be the given statement. Now, $P(n): 2.7^{n}+3.5^{n}-5$ is divisible by 24 . Step 1: $P(1): 2.7^{1}+3.5^{1}-5=24$ It is divisible by 24 . Thus, $P(1)$ is true. Step 2 : Let $P(m)$ be true. Then, 2. $7^{m}+3.5^{m}-5$ is divisible by 24 . Suppose : $2.7^{m}+3.5^{m}-5=24 \lambda \quad \ldots(1)$ We need to show that $P(m+1)$ is true whenever $P(m)$ is true. Now, $P(m+1)=2.7^{m+1}+3.5^{m+1}-5$ $=2.7^{m+1}+\left(24 \lambda+5-2.7^{m...
Read More →Solve the following
Question: 2.7n+ 3.5n 5 is divisible by 24 for allnN. Solution: LetP(n) be the given statement. Now, $P(n): 2.7^{n}+3.5^{n}-5$ is divisible by 24 . Step 1: $P(1): 2.7^{1}+3.5^{1}-5=24$ It is divisible by 24 . Thus, $P(1)$ is true. Step 2 : Let $P(m)$ be true. Then, 2. $7^{m}+3.5^{m}-5$ is divisible by 24 . Suppose : $2.7^{m}+3.5^{m}-5=24 \lambda \quad \ldots(1)$ We need to show that $P(m+1)$ is true whenever $P(m)$ is true. Now, $P(m+1)=2.7^{m+1}+3.5^{m+1}-5$ $=2.7^{m+1}+\left(24 \lambda+5-2.7^{m...
Read More →Three relations R1, R2 and R3 are defined on a set A = {a, b, c} as follows:
Question: Three relationsR1,R2andR3are defined on a setA= {a, b, c} as follows:R1= {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)}R2= {(a, a)}R3= {(b, c)}R4= {(a, b), (b, c), (c, a)}. Find whether or not each of the relationsR1,R2,R3,R4onAis (i) reflexive (ii) symmetric and (iii) transitive. Symmetric: Clearly $(a, a) \in R \Rightarrow(a, a) \in R .$ So, $\mathrm{R}_{2}$ is symmetric. Transitive: $R_{2}$ is clearly a transitive relation, since there is only one element in it. So...
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