f : R→R given by
Question: $f: R \rightarrow R$ given by $f(x)=x+\sqrt{x^{2}}$ is (a) injective (b) surjective (c) bijective (d) None of these Solution: $f(x)=x+\sqrt{x^{2}}=x \pm x=0$ or $2 x$ $\Rightarrow$ Each element of the domain has 2 images. $\Rightarrow f$ is not a function. So, the answer is (d)....
Read More →Find all pairs of consecutive even positive integers, both of which are larger than 5,
Question: Find all pairs of consecutive even positive integers, both of which are larger than 5, such that their sum is less than 23. Solution: Letxbe the smaller even integer. Then, the other even integer shall bex+ 2.Therefore, as per the given condition: $x5$ and $x+x+223$ $\Rightarrow x5$ and $2 x+223$ $\Rightarrow x5$ and $2 x21$ $\Rightarrow x5$ and $x\frac{21}{2}$ $\therefore x \in\{6,8,10\}$ Hence, the pairs are $(6,8),(8,10),(10,12)$....
Read More →There are three consecutive integers such that the square of the first
Question: There are three consecutive integers such that the square of the first increased by the product of the other two gives 154. What are the integers? Solution: Let three consecutive integer be $x,(x+1)$ and $(x+2)$ Then according to question $x^{2}+(x+1)(x+2)=154$ $x^{2}+x^{2}+3 x+2=154$ $2 x^{2}+3 x+2-154=0$ $2 x^{2}+3 x-152=0$ $2 x^{2}+3 x-152=0$ $2 x^{2}-16 x+19 x-152=0$ $2 x(x-8)+19(x-8)=0$ $(x-8)(2 x+19)=0$ $(x-8)=0$ $x=8$ Or $(2 x+19)=0$ $x=\frac{-19}{2}$ Since,xbeing a positive num...
Read More →Find all pairs of consecutive even positive integers, both of which are larger than 5,
Question: Find all pairs of consecutive even positive integers, both of which are larger than 5, such that their sum is less than 23. Solution: Letxbe the smaller even integer. Then, the other even integer shall bex+ 2.Therefore, as per the given condition: $x5$ and $x+x+223$ $\Rightarrow x5$ and $2 x+223$ $\Rightarrow x5$ and $2 x21$ $\Rightarrow x5$ and $x\frac{21}{2}$ $\therefore x \in\{6,8,10\}$ Hence, the pairs are $(6,8),(8,10),(10,12)$....
Read More →Let
Question: Let $A=\{x \in R:-1 \leq x \leq 1\}=B$ and $C=\{x \in R: x \geq 0\}$ and let $S=\left\{(x, y) \in A \times B: x^{2}+y^{2}=1\right\}$ and $S_{0}=\left\{(x, y) \in A \times C: x^{2}+y^{2}=1\right\} .$ Then, (a) $S$ defines a function from $A$ to $B$ (b) So defines a function from $A$ to $C$ (c) $S_{0}$ defines a function from $A$ to $B$ (d) $S$ defines a function from $A$ to $C$ Solution: (a)Sdefines a function fromAtoB Let $x \in A$ $\Rightarrow-1 \leq x \leq 1$ Now, $x^{2}+y^{2}=1$ $\R...
Read More →Find all pairs of consecutive odd natural number,
Question: Find all pairs of consecutive odd natural number, both of which are larger than 10, such that their sum is less than 40. Solution: Letxbe the smaller of the two odd natural numbers. Then, the other odd natural number will bex+ 2.Therefore, as per the given conditions: $x10$ and $x+x+240$ $\Rightarrow x10$ and $2 x+240$ $\Rightarrow x10$ and $x19$ $\Rightarrow 10x19$ $\therefore x \in\{11,13,15,17\}$ Hence, the pairs are $(11,13),(13,15),(15,17),(17,19)$....
Read More →Find all pairs of consecutive odd natural number,
Question: Find all pairs of consecutive odd natural number, both of which are larger than 10, such that their sum is less than 40. Solution: Letxbe the smaller of the two odd natural numbers. Then, the other odd natural number will bex+ 2.Therefore, as per the given conditions: $x10$ and $x+x+240$ $\Rightarrow x10$ and $2 x+240$ $\Rightarrow x10$ and $x19$ $\Rightarrow 10x19$ $\therefore x \in\{11,13,15,17\}$ Hence, the pairs are $(11,13),(13,15),(15,17),(17,19)$....
Read More →Find all pairs of consecutive odd positive integers,
Question: Find all pairs of consecutive odd positive integers, both of which are smaller than 10, such that their sum is more than 11. Solution: Let the smaller odd positive integer bex. Then, the other odd positive integer shall bex+ 2.Therefore, as per the given conditions: $x+210$ and $x+x+211$ $\Rightarrow x8$ and $2 x9$ $\Rightarrow x8$ and $x\frac{9}{2}$ Since $x$ is an odd integer, Therefore, $x=5,7$ Hence, pairs are $(5,7),(7,9)$....
Read More →Find all pairs of consecutive odd positive integers,
Question: Find all pairs of consecutive odd positive integers, both of which are smaller than 10, such that their sum is more than 11. Solution: Let the smaller odd positive integer bex. Then, the other odd positive integer shall bex+ 2.Therefore, as per the given conditions: $x+210$ and $x+x+211$ $\Rightarrow x8$ and $2 x9$ $\Rightarrow x8$ and $x\frac{9}{2}$ Since $x$ is an odd integer, Therefore, $x=5,7$ Hence, pairs are $(5,7),(7,9)$....
Read More →The sum of a number and its square is 63/4,
Question: The sum of a number and its square is 63/4, find the numbers. Solution: Let first numbers bex Then according to question $x+x^{2}=\frac{63}{4}$ Let $x=y^{2}$ then $4\left(x+x^{2}\right)=63$ $4 x^{2}+4 x-63=0$ $4 x^{2}+18 x-14 x-63=0$ $2 x(2 x+9)-7(2 x+9)=0$ $(2 x+9)(2 x-7)=0$ $(2 x+9)=0$ $x=-\frac{9}{2}$ Or $(2 x-7)=0$ $x=\frac{7}{2}$ Thus, the required number be $\frac{7}{2}, \frac{-9}{2}$...
Read More →Solve |3−4x|≥9
Question: Solve $|3-4 x| \geq 9$ Solution: As, $|3-4 x| \geq 9$ $\Rightarrow(3-4 x) \leq-9$ or $(3-4 x) \geq 9 \quad($ As, $|x| \geq a \Rightarrow x \leq-a$ or $x \geq a)$ $\Rightarrow-4 x \leq-9-3$ or $-4 x \geq 9-3$ $\Rightarrow-4 x \leq-12$ or $-4 x \geq 6$ $\Rightarrow x \geq \frac{-12}{-4}$ or $x \leq \frac{6}{-4}$ $\Rightarrow x \geq 3$ or $x \leq \frac{-3}{2}$ $\therefore x \in\left(-\infty, \frac{-3}{2}\right] \cup[3, \infty)$...
Read More →If f : A → A, g : A → A are two bijections, then prove that
Question: If $f: A \rightarrow A, g: A \rightarrow A$ are two bijections, then prove that (i) fog is an injection (ii) fog is a surjection Solution: Given: $A \rightarrow A, g: A \rightarrow A$ are two bijections. Then, fog: $A \rightarrow A$ (i) Injectivity offog:Letxandybe two elements of the domain (A), such that (ii) Surjectivity offog:Letzbe an element in the co-domain offog(A). Now, $z \in A$ (co-domain of $f$ ) and $f$ is a surjection. So, $z=f(y)$, where $y \in A$ (domain of $f$ ) ... (1...
Read More →Solve |3−4x|≥9
Question: Solve|34x|93-4x9 Solution: As, $|3-4 x| \geq 9$ $\Rightarrow(3-4 x) \leq-9$ or $(3-4 x) \geq 9 \quad($ As, $|x| \geq a \Rightarrow x \leq-a$ or $x \geq a)$ $\Rightarrow-4 x \leq-9-3$ or $-4 x \geq 9-3$ $\Rightarrow-4 x \leq-12$ or $-4 x \geq 6$ $\Rightarrow x \geq \frac{-12}{-4}$ or $x \leq \frac{6}{-4}$ $\Rightarrow x \geq 3$ or $x \leq \frac{-3}{2}$ $\therefore x \in\left(-\infty, \frac{-3}{2}\right] \cup[3, \infty)$...
Read More →Solve 1≤|x−2|≤3
Question: Solve $1 \leq|x-2| \leq 3$ Solution: As, $1 \leq|x-2| \leq 3$ $\Rightarrow|x-2|1$ and $|x-2|3$ $\Rightarrow((x-2) \leq-1$ or $(x-2) \geq 1)$ and $(-3 \leq(x-2) \leq 3)$ (As, $|x| \geq a \Rightarrow x \leq-a$ or $x \geq a ;$ and $|x| \leq a \Rightarrow-a \leq x \leq a$ ) $\Rightarrow(x \leq 1$ or $x \geq 3)$ and $(-3+2 \leq x \leq 3+2)$ $\Rightarrow(x \leq 1$ or $x \geq 3)$ and $(-1 \leq x \leq 5)$ $\Rightarrow x \in(-\infty, 1] \cup[3, \infty)$ and $x \in[-1,5]$ $\therefore x \in[-1,1]...
Read More →Solve |x+1|+|x|>3
Question: Solve $|x+1|+|x|3$ Solution: We have, $|x+1|+|x|3$ As, $|x+1|=\left\{\begin{array}{l}(x+1), x \geq-1 \\ -(x+1), x-1\end{array}\right.$ and $|x|=\left\{\begin{array}{l}x, x \geq 0 \\ -x, x0\end{array}\right.$ Case I: When $x-1$, $|x+1|+|x|3$ $\Rightarrow-(x+1)-x3$ $\Rightarrow-2 x-13$ $\Rightarrow-2 x4$ $\Rightarrow x\frac{4}{-2}$ $\Rightarrow x-2$ So, $x \in(-\infty,-2)$ Case II : When $-1 \leq x0$, $|x+1|+|x|3$ $\Rightarrow(x+1)-x3$ $\Rightarrow 13$, which is not possible So, $x \in \...
Read More →Let A and B be two sets, each with a finite number of elements.
Question: LetAandBbe two sets, each with a finite number of elements. Assume that there is an injective map fromAtoBand that there is an injective map fromBtoA. Prove that there is a bijection fromAtoB. Solution: $A$ and $B$ are two non empty sets. Let $f$ be a function from $A$ to $B$. It is given that there is injective map from $A$ to $B$. That means $f$ is one-one function. It is also given that there is injective map from $B$ to $A$. That means every element of set $B$ has its image in set ...
Read More →If A = {1, 2, 3, 4} and B = {a, b, c, d},
Question: If $A=\{1,2,3,4\}$ and $B=\{a, b, c, d\}$, define any four bijections from $A$ to $B$. Also give their inverse functions. Solution: $f_{1}=\{(1, a),(2, b),(3, c),(4, d)\} \Rightarrow f_{1}^{-1}=\{(a, 1),(b, 2),(c, 3),(d, 4)\}$ $f_{2}=\{(1, b),(2, a),(3, c),(4, d)\} \Rightarrow f_{2}^{-1}=\{(b, 1),(a, 2),(c, 3),(d, 4)\}$ $f_{3}=\{(1, a),(2, b),(4, c),(3, d)\} \Rightarrow f_{3}^{-1}=\{(a, 1),(b, 2),(c, 4),(d, 3)\}$ $f_{4}=\{(1, b),(2, a),(4, c),(3, d)\} \Rightarrow f_{4}^{-1}=\{(b, 1),(a...
Read More →Let f be a function from R to R, such that f(x) = cos (x + 2). Is f invertible? Justify your answer.
Question: Letfbe a function fromRtoR,such thatf(x) = cos (x+ 2). Isfinvertible? Justify your answer. Solution: Injectivity:Letxandybe two elements in the domain (R), such that $f(x)=f(y)$ $\Rightarrow \cos (x+2)=\cos (y+2)$ $\Rightarrow x+2=y+2$ or $x+2=2 \pi-(y+2)$ $\Rightarrow x=y$ or $x+2=2 \pi-y-2$ $\Rightarrow x=y$ or $x=2 \pi-y-4$ So, we cannot say that $x=y$ For example, $\cos \frac{\pi}{2}=\cos \frac{3 \pi}{2}=0$ So, $\frac{\pi}{2}$ and $\frac{3 \pi}{2}$ have the same image 0 . $\Rightar...
Read More →Factorise:
Question: Factorise: $\frac{2}{3} x^{2}-\frac{17}{3} x-28$ Solution: $\frac{2}{3} x^{2}-\frac{17}{3} x-28=\frac{2}{3} x^{2}-8 x+\frac{7}{3} x-28$ $=2 x\left(\frac{1}{3} x-4\right)+7\left(\frac{1}{3} x-4\right)$ $=\left(\frac{1}{3} x-4\right)(2 x+7)$ Hence, factorisation of $\frac{2}{3} x^{2}-\frac{17}{3} x-28$ is $\left(\frac{1}{3} x-4\right)(2 x+7)$....
Read More →Let A = {x &epsis; R | −1 ≤ x ≤ 1} and let f : A → A, g : A → A be two functions defined by
Question: Let $A=\{x$ \epsis; $R \mid-1 \leq x \leq 1\}$ and let $f: A \rightarrow A, g: A \rightarrow A$ be two functions defined by $f(x)=x^{2}$ and $g(x)=\sin (\pi x / 2)$. Show that $g^{-1}$ exists but $f^{-1}$ does not exist. Also, find $g-1$. Solution: fis not one-one because $f(-1)=(-1)^{2}=1$ and $f(1)=1^{2}=1$ $\Rightarrow-1$ and 1 have the same image under $f$. $\Rightarrow f$ is not a bijection. So, $f^{-1}$ does not exist. Injectivity of g: Let $x$ and $y$ be any two elements in the ...
Read More →Factorise:
Question: Factorise: $\frac{3}{2} x^{2}+16 x+10$ Solution: $\frac{3}{2} x^{2}+16 x+10=\frac{3}{2} x^{2}+15 x+x+10$ $=3 x\left(\frac{1}{2} x+5\right)+1(x+10)$ $=\frac{3}{2} x(x+10)+1(x+10)$ $=(x+10)\left(\frac{3}{2} x+1\right)$ Hence, factorisation of $\frac{3}{2} x^{2}+16 x+10$ is $(x+10)\left(\frac{3}{2} x+1\right)$...
Read More →Factorise:
Question: Factorise: $21 x^{2}-2 x+\frac{1}{21}$ Solution: $21 x^{2}-2 x+\frac{1}{21}=21 x^{2}-x-x+\frac{1}{21}$ $=21 x\left(x-\frac{1}{21}\right)-1\left(x-\frac{1}{21}\right)$ $=\left(x-\frac{1}{21}\right)(21 x-1)$ Hence, factorisation of $21 x^{2}-2 x+\frac{1}{21}$ is $\left(x-\frac{1}{21}\right)(21 x-1)$...
Read More →Factorise:
Question: Factorise: $x^{2}+\frac{12}{35} x+\frac{1}{35}$ Solution: $x^{2}+\frac{12}{35} x+\frac{1}{35}=\frac{35 x^{2}+12 x+1}{35}$ $=\frac{35 x^{2}+7 x+5 x+1}{35}$ $=\frac{7 x(5 x+1)+1(5 x+1)}{35}$ $=\frac{(5 x+1)(7 x+1)}{35}$ $=\frac{(5 x+1)(7 x+1)}{5 \times 7}$ $=\frac{(5 x+1)}{5} \times \frac{(7 x+1)}{7}$ $=\left(x+\frac{1}{5}\right)\left(x+\frac{1}{7}\right)$ Hence, factorisation of $x^{2}+\frac{12}{35} x+\frac{1}{35}$ is $\left(x+\frac{1}{5}\right)\left(x+\frac{1}{7}\right)$....
Read More →Let f : [−1, ∞) → [−1, ∞) be given by
Question: Let $f:[-1, \infty) \rightarrow[-1, \infty)$ be given by $f(x)=(x+1)^{2}-1, x \geq-1$. Show that $f$ is invertible. Also, find the set $S=\left\{x: f(x)=f^{-1}(x)\right\}$. Solution: Injectivity : Let $\mathrm{x}$ and $\mathrm{y} \in[-1, \infty)$, such that $f(x)=f(y)$ $\Rightarrow(x+1)^{2}-1=(y+1)^{2}-1$ $\Rightarrow(x+1)^{2}=(y+1)^{2}$ $\Rightarrow(x+1)=(y+1)$ $\Rightarrow x=y$ So, $f$ is a injection. Surjectivity : Let $\mathrm{y} \in[-1, \infty)$. Then, $\mathrm{f}(\mathrm{x})=\mat...
Read More →Factorise
Question: Factorise $\frac{1}{3} x^{2}-2 x-9$ Solution: $\frac{1}{3} x^{2}-2 x-9=\frac{x^{2}-6 x-27}{3}$ $=\frac{x^{2}-9 x+3 x-27}{3}$ $=\frac{x(x-9)+3(x-9)}{3}$ $=\frac{(x-9)(x+3)}{3}$ $=\frac{(x-9)}{3} \times \frac{(x+3)}{1}$ $=\left(\frac{1}{3} x-3\right)(x+3)$ Hence, factorisation of $\frac{1}{3} x^{2}-2 x-9$ is $\left(\frac{1}{3} x-3\right)(x+3)$....
Read More →