If f : A → B and g : B → C are onto functions, show that gof is a onto function.
Question: Iff:ABandg:BCare onto functions, show thatgofis a onto function. Solution: Given,f:ABandg:B C are onto. Then,gof :AC Let us take an elementzin the co-domain (C). Now, $z$ is in $C$ and $g: B \rightarrow C$ is onto. So, there exists some element $y$ in $B$, such that $g(y)=z$ Now, $y$ is in $B$ and $f: A \rightarrow B$ is onto. So, there exists some $x$ in $A$, such that $f(x)=y \ldots$ (2) From (1) and (2), $z=g(y)=g(f(x))=(g \circ f)(x)$ So, $z=(g \circ f)(x)$, where $x$ is in $A$. He...
Read More →If α, β are the roots of the equation ax
Question: If $\alpha, \beta$ are the roots of the equation $a x^{2}+b x+c=0$, then $\frac{1}{a \alpha+b}+\frac{1}{a \beta+b}=$ (a)c/ab (b)a/bc (c)b/ac (d) none of these. Solution: (c)b/ac Given equation: $a x^{2}+b x+c=0$ Also, $\alpha$ and $\beta$ are the roots of the given equation. Then, sum of the roots $=\alpha+\beta=-\frac{b}{a}$ Product of the roots $=\alpha \beta=\frac{c}{a}$ $\therefore \frac{1}{a \alpha+b}+\frac{1}{a \beta+b}=\frac{a \beta+b+a \alpha+b}{(a \alpha+b)(a \beta+b)}$ $=\fra...
Read More →If α, β are the roots of the equation ax
Question: If $\alpha, \beta$ are the roots of the equation $a x^{2}+b x+c=0$, then $\frac{1}{a \alpha+b}+\frac{1}{a \beta+b}=$ (a)c/ab (b)a/bc (c)b/ac (d) none of these. Solution: (c)b/ac Given equation: $a x^{2}+b x+c=0$ Also, $\alpha$ and $\beta$ are the roots of the given equation. Then, sum of the roots $=\alpha+\beta=-\frac{b}{a}$ Product of the roots $=\alpha \beta=\frac{c}{a}$ $\therefore \frac{1}{a \alpha+b}+\frac{1}{a \beta+b}=\frac{a \beta+b+a \alpha+b}{(a \alpha+b)(a \beta+b)}$ $=\fra...
Read More →The number of real roots of the equation
Question: The number of real roots of the equation $\left(x^{2}+2 x\right)^{2}-(x+1)^{2}-55=0$ is (a) 2 (b) 1 (c) 4 (d) none of these Solution: (a) 2 $\left(x^{2}+2 x\right)^{2}-(x+1)^{2}-55=0$ $\Rightarrow\left(x^{2}+2 x+1-1\right)^{2}-(x+1)^{2}-55=0$ $\Rightarrow\left\{(x+1)^{2}-1\right\}^{2}-(x+1)^{2}-55=0$ $\Rightarrow\left\{(x+1)^{2}\right\}^{2}+1-3(x+1)^{2}-55=0$ $\Rightarrow\left\{(x+1)^{2}\right\}^{2}-3(x+1)^{2}-54=0$ Let $p=(x+1)^{2}$ $\Rightarrow p^{2}-3 p-54=0$ $\Rightarrow p^{2}-9 p+...
Read More →The number of real roots of the equation
Question: The number of real roots of the equation $\left(x^{2}+2 x\right)^{2}-(x+1)^{2}-55=0$ is (a) 2 (b) 1 (c) 4 (d) none of these Solution: (a) 2 $\left(x^{2}+2 x\right)^{2}-(x+1)^{2}-55=0$ $\Rightarrow\left(x^{2}+2 x+1-1\right)^{2}-(x+1)^{2}-55=0$ $\Rightarrow\left\{(x+1)^{2}-1\right\}^{2}-(x+1)^{2}-55=0$ $\Rightarrow\left\{(x+1)^{2}\right\}^{2}+1-3(x+1)^{2}-55=0$ $\Rightarrow\left\{(x+1)^{2}\right\}^{2}-3(x+1)^{2}-54=0$ Let $p=(x+1)^{2}$ $\Rightarrow p^{2}-3 p-54=0$ $\Rightarrow p^{2}-9 p+...
Read More →Let $A=\{-1,0,1\}$ and $f=\left\{\left(x, x^{2}\right): x \in A\right\}$. Show that $f: A \rightarrow A$ is neither one-one nor onto.
[question] Question. Let $A=\{-1,0,1\}$ and $f=\left\{\left(x, x^{2}\right): x \in A\right\}$. Show that $f: A \rightarrow A$ is neither one-one nor onto. [/question] [solution] Solution: $A=\{-1,0,1\}$ and $f=\left\{\left(x, x^{2}\right): x \in A\right\}$ Given, $f(x)=x^{2}$ Given, $f(x)=x^{2}$ Injectivity: $f(1)=1^{2}=1$ and $f(-1)=(-1)^{2}=1$ $\Rightarrow 1$ and $-1$ have the same images. So, $f$ is not one-one. Surjectivity: Co-domain of $f=\{-1,0,1\}$ $f(1)=1^{2}=1$ $f(-1)=(-1)^{2}=1$ and $...
Read More →If f : A → B and g : B → C are one-one functions, show that gof is a one-one function.
Question: Iff:ABandg:BCare one-one functions, show thatgofis a one-one function. Solution: Given,f:ABandg:B C are one - one.Then,gof :ABLet us take two elementsxandyfromA,such that $(go f)(x)=(g o f)(y)$ $\Rightarrow g(f(x))=g(f(y))$ $\Rightarrow f(x)=f(y)$ (As, $g$ is one-one) $\Rightarrow x=y$ (As, $f$ is one-one) Hence,gofis one-one....
Read More →The values of x satisfying log
Question: The values of $x$ satisfying $\log _{3}\left(x^{2}+4 x+12\right)=2$ are (a) 2, 4 (b) 1, 3 (c) 1, 3 (d) 1, 3 Solution: (d) 1, 3 The given equation is $\log _{3}\left(x^{2}+4 x+12\right)=2$. $\Rightarrow x^{2}+4 x+12=3^{2}=9$ $\Rightarrow x^{2}+4 x+3=0$ $\Rightarrow(x+1)(x+3)=0$ $\Rightarrow x=-1,-3$...
Read More →The values of x satisfying log
Question: The values of $x$ satisfying $\log _{3}\left(x^{2}+4 x+12\right)=2$ are (a) 2, 4 (b) 1, 3 (c) 1, 3 (d) 1, 3 Solution: (d) 1, 3 The given equation is $\log _{3}\left(x^{2}+4 x+12\right)=2$. $\Rightarrow x^{2}+4 x+12=3^{2}=9$ $\Rightarrow x^{2}+4 x+3=0$ $\Rightarrow(x+1)(x+3)=0$ $\Rightarrow x=-1,-3$...
Read More →If α, β are roots of the equation
Question: If $\alpha, \beta$ are roots of the equation $4 x^{2}+3 x+7=0$, then $1 / \alpha+1 / \beta$ is equal to (a) 7/3 (b) 7/3 (c) 3/7 (d) 3/7 Solution: (d) 3/7 Given equation: $4 x^{2}+3 x+7=0$ Also, $\alpha$ and $\beta$ are the roots of the equation. Sum of the roots $=\alpha+\beta=\frac{-C \text { oefficient of } x}{C \text { oefficient of } x^{2}}=-\frac{3}{4}$ Product of the roots $=\alpha \beta=\frac{C \text { onstant term }}{C \text { oefficient of } x^{2}}=\frac{7}{4}$ $\therefore \qu...
Read More →If α, β are roots of the equation
Question: If $\alpha, \beta$ are roots of the equation $4 x^{2}+3 x+7=0$, then $1 / \alpha+1 / \beta$ is equal to (a) 7/3 (b) 7/3 (c) 3/7 (d) 3/7 Solution: (d) 3/7 Given equation: $4 x^{2}+3 x+7=0$ Also, $\alpha$ and $\beta$ are the roots of the equation. Sum of the roots $=\alpha+\beta=\frac{-C \text { oefficient of } x}{C \text { oefficient of } x^{2}}=-\frac{3}{4}$ Product of the roots $=\alpha \beta=\frac{C \text { onstant term }}{C \text { oefficient of } x^{2}}=\frac{7}{4}$ $\therefore \qu...
Read More →Factorize:
Question: Factorize: $3 x^{3}-48 x$ Solution: $3 x^{3}-48 x=3 x\left(x^{2}-16\right)$ $=3 x\left(x^{2}-4^{2}\right)$ $=3 x(x-4)(x+4)$...
Read More →Give examples of two functions
Question: Give examples of two functions $f: N \rightarrow Z$ and $g: Z \rightarrow Z$, such that $g$ of is injective but $g$ is not injective. Solution: Let $f: N \rightarrow Z$ be given by $f(x)=x$, which is injective. (If we takef(x) = f(y), then it givesx = y) Let $g: Z \rightarrow Z$ be given by $g(x)=|x|$, which is not injective. If we take $f(x)=f(y)$, we get: $|x|=|y|$ $\Rightarrow x=\pm y$ Now, gof: $N \rightarrow Z$. $(g \circ f)(x)=g(f(x))=g(x)=|x|$ Let us take two elementsxandyin the...
Read More →If a, b are the roots of the equation
Question: If $a, b$ are the roots of the equation $x^{2}+x+1=0$, then $a^{2}+b^{2}=$ (a) 1 (b) 2 (c) 1 (d) 3 Solution: (c) 1 Given equation: $x^{2}+x+1=0$ Also, $a$ and $b$ are the roots of the given equation. Sum of the roots $=a+b=\frac{-C \text { oefficient of } x}{C \text { oefficient of } x^{2}}=-\frac{1}{1}=-1$ Product of the roots $=a b=\frac{C \text { onstant term }}{C \text { oefficient of } x^{2}}=\frac{1}{1}=1$ $\therefore(a+b)^{2}=a^{2}+b^{2}+2 a b$ $\Rightarrow(-1)^{2}=a^{2}+b^{2}+2...
Read More →If a, b are the roots of the equation
Question: If $a, b$ are the roots of the equation $x^{2}+x+1=0$, then $a^{2}+b^{2}=$ (a) 1 (b) 2 (c) 1 (d) 3 Solution: (c) 1 Given equation: $x^{2}+x+1=0$ Also, $a$ and $b$ are the roots of the given equation. Sum of the roots $=a+b=\frac{-C \text { oefficient of } x}{C \text { oefficient of } x^{2}}=-\frac{1}{1}=-1$ Product of the roots $=a b=\frac{C \text { onstant term }}{C \text { oefficient of } x^{2}}=\frac{1}{1}=1$ $\therefore(a+b)^{2}=a^{2}+b^{2}+2 a b$ $\Rightarrow(-1)^{2}=a^{2}+b^{2}+2...
Read More →Factorize:
Question: Factorize: $3 a^{3} b-243 a b^{3}$ Solution: $3 a^{3} b-243 a b^{3}=3 a b\left(a^{2}-81 b^{2}\right)$ $=3 a b\left[a^{2}-(9 b)^{2}\right]$ $=3 a b(a-9 b)(a+9 b)$...
Read More →Factorise:
Question: Factorise: $2 x^{4}-32$ Solution: $2 x^{4}-32$ $=2\left(x^{4}-16\right)$ $=2\left[\left(x^{2}\right)^{2}-4^{2}\right]$ $=2\left(x^{2}+4\right)\left(x^{2}-4\right) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$ $=2\left(x^{2}+4\right)\left(x^{2}-2^{2}\right)$ $=2\left(x^{2}+4\right)(x+2)(x-2) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$...
Read More →For the equation
Question: For the equation $|x|^{2}+|x|-6=0$, the sum of the real roots is (a) 1 (b) 0 (c) 2 (d) none of these Solution: (b) 0 Let $p=|x|$ $\Rightarrow p^{2}+p-6=0$ $\Rightarrow p^{2}+3 p-2 p-6=0$ $\Rightarrow(p+3)(p-2)=0$ $\Rightarrow p=-3,2$ Also, $|x|=p$ $\Rightarrow|x|=2$, or $|x|=-3$ Modulus can not be negative,$\therefore|x|=2$ $\Rightarrow x=\pm 2$ $\Rightarrow x=2$ or $-2$ Sum of the roots of $x$ is 0...
Read More →For the equation
Question: For the equation $|x|^{2}+|x|-6=0$, the sum of the real roots is (a) 1 (b) 0 (c) 2 (d) none of these Solution: (b) 0 Let $p=|x|$ $\Rightarrow p^{2}+p-6=0$ $\Rightarrow p^{2}+3 p-2 p-6=0$ $\Rightarrow(p+3)(p-2)=0$ $\Rightarrow p=-3,2$ Also, $|x|=p$ $\Rightarrow|x|=2$, or $|x|=-3$ Modulus can not be negative,$\therefore|x|=2$ $\Rightarrow x=\pm 2$ $\Rightarrow x=2$ or $-2$ Sum of the roots of $x$ is 0...
Read More →Give examples of two functions
Question: Give examples of two functions $f: N \rightarrow N$ and $g: N \rightarrow N$, such that $g$ of is onto but $f$ is not onto. Solution: Let us consider a function $f: N \rightarrow N$ given by $f(x)=x+1$, which is not onto. [This not onto because if we take 0 inN(co-domain), then, 0=x+1 $\Rightarrow x=-1 \notin N]$ Let us consider $g: N \rightarrow N$ given by $g(x)=\left\{\begin{array}{l}x-1, \text { if } x1 \\ 1, \text { if } x=1\end{array}\right.$ Now, let us find $(g o f)(x)$ Case 1:...
Read More →Factorise:
Question: Factorise: $5-20 x^{2}$ Solution: $5-20 x^{2}$ $=5\left(1-4 x^{2}\right)$ $=5\left[1^{2}-(2 x)^{2}\right]$ $=5(1+2 x)(1-2 x) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$...
Read More →The complete set of values of k, for which the quadratic equation
Question: The complete set of values of $k$, for which the quadratic equation $x^{2}-k x+k+2=0$ has equal roots, consists of (a) $2+\sqrt{12}$ (b) $2 \pm \sqrt{12}$ (c) $2-\sqrt{12}$ (d) $-2-\sqrt{12}$ Solution: (b) $2 \pm \sqrt{12}$ Since the equation has real roots. $\Rightarrow \mathrm{D}=0$ $\Rightarrow \mathrm{b}^{2}-4 \mathrm{ac}=0$ $\Rightarrow \mathrm{k}^{2}-4(1)(\mathrm{k}+2)=0$ $\Rightarrow \mathrm{k}^{2}-4 \mathrm{k}-8=0$ $\Rightarrow \mathrm{k}=\frac{4 \pm \sqrt{16-4(1)(-8)}}{2(1)}$ ...
Read More →Write the discriminant of the following quadratic equations:
Question: Write the discriminant of the following quadratic equations: (i) $2 x^{2}-5 x+3=0$ (ii) $x^{2}+2 x+4=0$ (iii) $(x-1)(2 x-1)=0$ (iv) $x^{2}-2 x+k=0, k \in \mathrm{R}$ (v) $\sqrt{3} x^{2}+2 \sqrt{2} x-2 \sqrt{3}=0$ (vi) $x^{2}-x+1=0$ Solution: We have to find the discriminant of the following quadratic equations (i) We have been given, $2 x^{2}-5 x+3=0$ Now we also know that for an equation $a x^{2}+b x+c=0$, the discriminant is given by the following equation: $D=b^{2}-4 a c$ Now, accor...
Read More →Factorise:
Question: Factorise: $81-16 x^{2}$ Solution: $81-16 x^{2}$ $=9^{2}-(4 x)^{2}$ $=(9+4 x)(9-4 x) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$...
Read More →If $\left(x^{3}+a x^{2}+b x+6\right)$ has $(x-2)$ as a factor and leaves a remainder 3 when divided by
[question] Question. If $\left(x^{3}+a x^{2}+b x+6\right)$ has $(x-2)$ as a factor and leaves a remainder 3 when divided by $(x-3)$, find the values of $a$ and $b$. [/question] [solution] Solution: Let: $f(x)=x^{3}+a x^{2}+b x+6$ $(x-2)$ is a factor of $f(x)=x^{3}+a x^{2}+b x+6$ $\Rightarrow f(2)=0$ $\Rightarrow 2^{3}+a \times 2^{2}+b \times 2+6=0$ $\Rightarrow 14+4 a+2 b=0$ $\Rightarrow 4 a+2 b=-14$ $\Rightarrow 2 a+b=-7 \quad \ldots(1)$ Now, $x-3=0 \Rightarrow x=3$ By the factor theorem, we ca...
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