Let f be a real function given by
Question: Let $f$ be a real function given by $f(x)=\sqrt{x-2}$. Find each of the following:(i)fof(ii)fofof(iii) (fofof) (38)(iv)f2 Also, show that $f \circ f \neq f^{2}$. Solution: $f(x)=\sqrt{x-2}$ For domain, $x-2 \geq 0$ $\Rightarrow x \geq 2$ Domain of $f=[2, \infty)$ Since $f$ is a square-root function, range of $f=(0, \infty)$ So, $f:[2, \infty) \rightarrow(0, \infty)$ (i) fof Range of $f$ is not a subset of the domain of $f$. $\Rightarrow$ Domain $(f o f)=\{x: x \in$ domain of $f$ and $f...
Read More →Factorize:
Question: Factorize: $x^{2}+18 x+32$ Solution: We have: $x^{2}+18 x+32$ We have to split 18 into two numbers such that their sum is 18 and their product is 32. Clearly, $16+2=18$ and $16 \times 2=32$ $\therefore x^{2}+18 x+32=x^{2}+16 x+2 x+32$ $=x(x+16)+2(x+16)$ $=(x+16)(x+2)$...
Read More →If the roots of the equation (b−c)x2+(c−a)x+(a−b)=0 are equal,
Question: If the roots of the equation $(b-c) x^{2}+(c-a) x+(a-b)=0$ are equal, then prove that $2 b=a+c$. Solution: The given quadric equation is $(b-c) x^{2}+(c-a) x+(a-b)=0$, and roots are real Then prove that $2 b=a+c$. Here, $a=(b-c), b=(c-a)$ and, $c=(a-b)$ As we know that $D=b^{2}-4 a c$ Putting the value of $a=(b-c), b=(c-a)$ and, $c=(a-b)$ $D=b^{2}-4 a c$ $=(c-a)^{2}-4 \times(b-c) \times(a-b)$ $=c^{2}-2 c a+a^{2}-4\left(a b-b^{2}-c a+b c\right)$ $=c^{2}-2 c a+a^{2}-4 a b+4 b^{2}+4 c a-4...
Read More →Solve the following
Question: $\frac{x-1}{x+3}2$ Solution: We have, $\frac{x-1}{x+3}2$ $\Rightarrow \frac{x-1}{x+3}-20$ $\Rightarrow \frac{x-1-2(x+3)}{x+3}0$ $\Rightarrow \frac{x-1-2 x-6}{x+3}0$ $\Rightarrow \frac{-x-7}{x+3}0$ $\Rightarrow \frac{x+7}{x+3}0$ $\therefore x \in(-7,-3)$...
Read More →Solve the following
Question: $\frac{5 x+8}{4-x}2$ Solution: We have, $\frac{5 x+8}{4-x}2$ $\Rightarrow \frac{5 x+8}{4-x}-20$ $\Rightarrow \frac{5 x+8-2(4-x)}{4-x}0$ $\Rightarrow \frac{5 x+8-8+2 x}{4-x}0$ $\Rightarrow \frac{7 x}{4-x}0$ $\Rightarrow \frac{7 x}{x-4}0$ $\therefore x \in(-\infty, 0) \cup(4, \infty)$...
Read More →Factorize:
Question: Factorize: $x^{2}+11 x+30$ Solution: We have: $x^{2}+11 x+30$ We have to split 11 into two numbers such that their sum of is 11 and their product is 30. Clearly, $5+6=11$ and $5 \times 6=30$ $\therefore x^{2}+11 x+30=x^{2}+5 x+6 x+30$ $=x(x+5)+6(x+5)$ $=(x+5)(x+6)$...
Read More →Find the values of k for which the given quadratic equation has real and distinct roots:
Question: Find the values ofkfor which the given quadratic equation has real and distinct roots: (a) $k x^{2}+2 x+1=0$ (b) $k x^{2}+6 x+1=0$ (c) $x^{2}-k x+9=0$ Solution: (i) The given quadric equation is $k x^{2}+2 x+1=0$, and roots are real and distinct Then find the value of $k$. Here, $a=k, b=2$ and,$c=1$ As we know that $D=b^{2}-4 a c$ Putting the value of $a=k, b=2$ and, $c=1$ $D=(2)^{2}-4 \times k \times 1$ $=4-4 k$ The given equation will have real and distinct roots, if $D0$ $4-4 k0$ No...
Read More →Solve the following
Question: $\frac{5 x-6}{x+6}1$ Solution: $\frac{5 x-6}{x+6}1$ $\Rightarrow \frac{5 x-6}{x+6}-10$ $\Rightarrow \frac{5 x-6-x-6}{x+6}0$ $\Rightarrow \frac{4 x-12}{x+6}0$ $\Rightarrow \frac{x-3}{x+6}0$ $\therefore x \in(-6,3)$...
Read More →Solve the following
Question: $\frac{4 x+3}{2 x-5}6$ Solution: $\frac{4 x+3}{2 x-5}6$ $\Rightarrow \frac{4 x+3}{2 x-5}-60$ $\Rightarrow \frac{4 x+3-6(2 x-5)}{2 x-5}0$ $\Rightarrow \frac{4 x+3-12 x+30}{2 x-5}0$ $\Rightarrow \frac{-8 x+33}{2 x-5}0$ $\Rightarrow \frac{8 x-33}{2 x-5}0$ $x \in\left(-\infty, \frac{5}{2}\right) \cup\left(\frac{33}{8}, \infty\right)$...
Read More →Solve the following
Question: $\frac{1}{x-1} \leq 2$ Solution: $\frac{1}{x-1} \leq 2$ $\Rightarrow \frac{1}{x-1}-2 \leq 0$ $\Rightarrow \frac{1-2 x+2}{x-1} \leq 0$ $\Rightarrow \frac{-2 x+3}{x-1} \leq 0$ $\Rightarrow \frac{2 x-3}{x-1} \geq 0$ $\therefore x \in(-\infty, 1) \cup\left[\frac{3}{2}, \infty\right)$...
Read More →Solve the following
Question: $\frac{3}{x-2}1$ Solution: $\frac{3}{x-2}1$ $\Rightarrow \frac{3}{x-2}-10$ $\Rightarrow \frac{3-x+2}{x-2}0$ $\Rightarrow \frac{-x+5}{x-2}0$ $\Rightarrow \frac{x-5}{x-2}0$ $\therefore x \in(-\infty, 2) \cup(5, \infty)$...
Read More →Solve the following
Question: $\frac{2 x-3}{3 x-7}0$ Solution: $\frac{2 x-3}{3 x-7}0$ Equating $2 x-3$ and $3 x-7$ to zero, we obtain $x=\frac{3}{2}, \frac{7}{3}$ as the critical points. $\therefore x \in\left(-\infty, \frac{3}{2}\right) \cup\left(\frac{7}{3}, \infty\right)$...
Read More →Solve the following
Question: $\frac{6 x-5}{4 x+1}0$ Solution: $\frac{6 x-5}{4 x+1}0$ Equating $6 x-5$ and $4 x+1$ to zero, we obtain $x=\frac{5}{6}$ and $-\frac{1}{4}$ as the critical points. $\therefore x \in\left(\frac{-1}{4}, \frac{5}{6}\right)$...
Read More →Solve the following
Question: $x-2 \leq \frac{5 x+8}{3}$ Solution: $x-2 \leq \frac{5 x+8}{3}$ $\Rightarrow 3(x-2) \leq 5 x+8 \quad$ [Multiplying both the sides by 3$]$ $\Rightarrow 3 x-6 \leq 5 x+8$ $\Rightarrow 5 x+8 \geq 3 x-6$ $\Rightarrow 5 x-3 x \geq-6-8 \quad$ [Transposing $3 x$ to the LHS and 8 to the RHS] $\Rightarrow 2 x \geq-14$ $\Rightarrow x \geq-7 \quad$ [Dividing both the sides by 2] Hence, the solution of the given inequation is $[-7, \infty)$....
Read More →Find the least positive value of k for which the equation
Question: Find the least positive value of $k$ for which the equation $x^{2}+k x+4=0$ has real roots. Solution: The given quadric equation is $x^{2}+k x+4=0$, and roots are real. Then find the value of $k$. Here, $a=1, b=k$ and,$c=4$ As we know that $D=b^{2}-4 a c$ Putting the value of $a=1, b=k$ and, $c=4$ $=(k)^{2}-4 \times 1 \times 4$ $=k^{2}-16$ The given equation will have real and equal roots, if $D=0$ $k^{2}-16=0$ Now factorizing of the above equation $k^{2}-16=0$ $k^{2}=16$ $k=\sqrt{16}$...
Read More →Solve the following
Question: $\frac{2 x+3}{5}-2\frac{3(x-2)}{5}$ Solution: $\frac{2 x+3}{5}-2\frac{3(x-2)}{5}$ $\Rightarrow \frac{2 x+3}{5}-\frac{3 x-6}{5}2 \quad\left[\right.$ Transposing $\frac{3(x-2)}{5}$ to the LHS and $-2$ to the RHS $]$ $\Rightarrow \frac{2 \mathrm{x}+3-3 \mathrm{x}+6}{5}2$ $\Rightarrow 2 x+3-3 x+610 \quad$ [Multiplying both the sides by 5 ] $\Rightarrow-\mathrm{x}+910$ $\Rightarrow-\mathrm{x}1$ $\Rightarrow x-1$ [Multiplying both the sides by $-1]$ Hence, the solution set of the given inequ...
Read More →For what value of k,
Question: For what value of $k,(4-k) x^{2}+(2 k+4) x+(8 k+1)=0$, is a perfect square. Solution: The given quadric equation is $(4-k) x^{2}+(2 k+4) x+(8 k+1)=0$, and roots are real and equal Then find the value of $k$. Here, $a=(4-k), b=(2 k+4)$ and,$c=(8 k+1)$ As we know that $D=b^{2}-4 a c$ Putting the value of $a=(4-k), b=(2 k+4)$ and, $c=(8 k+1)$ $=(2 k+4)^{2}-4 \times(4-k) \times(8 k+1)$ $=4 k^{2}+16 k+16-4\left(4+31 k-8 k^{2}\right)$ $=4 k^{2}+16 k+16-16-124 k+32 k^{2}$ $=36 k^{2}-108 k+0$ ...
Read More →Solve the following
Question: $\frac{4+2 x}{3} \geq \frac{x}{2}-3$ Solution: $\frac{4+2 x}{3} \geq \frac{x}{2}-3$ $\Rightarrow \frac{4+2 x}{3}-\frac{x}{2} \geq-3$ $\Rightarrow \frac{8+4 x-3 x}{6} \geq-3$ $\Rightarrow 8+x \geq-18 \quad$ [Multiplying both the sides by 6$]$ $\Rightarrow x \geq-26 \quad$ [Transposing 8 to the RHS] Thus, the solution set of the given inequation is $[-26, \infty)$....
Read More →Solve the following
Question: $\frac{5-2 x}{3}\frac{x}{6}-5$ Solution: $\frac{5-2 x}{3}\frac{x}{6}-5$ $\Rightarrow \frac{5-2 x}{3}-\frac{x}{6}-5 \quad\left[\right.$ Transposing $\frac{x}{6}$ to the LHS $]$ $\Rightarrow \frac{2(5-2 \mathrm{x})-\mathrm{x}}{6}-5$ $\Rightarrow \frac{10-4 \mathrm{x}-\mathrm{x}}{6}-5$ $\Rightarrow \frac{10-5 \mathrm{x}}{6}-5$ $\Rightarrow 10-5 x-30$ $\Rightarrow 10+305 x$ $\Rightarrow 405 x$ $\Rightarrow 5 x40$ $\Rightarrow x\frac{40}{5}$ $\Rightarrow x8$ Hence, the solution set of the g...
Read More →Solve the following
Question: $\frac{2 x+3}{4}-3\frac{x-4}{3}-2$ Solution: $\frac{2 x+3}{4}-3\frac{x-4}{3}-2$ $\Rightarrow \frac{2 x+3}{4}-\frac{x-4}{3}-2+3 \quad$ (Transposing $\frac{x-4}{3}$ to the LHS and $-3$ to the RHS) $\Rightarrow \frac{3(2 \mathrm{x}+3)-4(\mathrm{x}-4)}{12}1$ $\Rightarrow 3(2 x+3)-4(x-4)12 \quad$ (Multiplying both the sides by 12$)$ $\Rightarrow 6 x+9-4 x+1612$ $\Rightarrow 2 x+2512$ $\Rightarrow 2 x12-25$ $\Rightarrow 2 x-13$ $\Rightarrow x-\frac{13}{2} \quad$ (Dividing both the sides by 2...
Read More →Solve the following
Question: $\frac{x-1}{3}+4\frac{x-5}{5}-2$ Solution: $\frac{x-1}{3}+4\frac{x-5}{5}-2$ $\Rightarrow \frac{x-1}{3}-\frac{x-5}{5}-2-4 \quad\left[\right.$ Transposing 4 to the RHS and $\frac{\mathrm{x}-5}{5}$ to the LHS $]$ $\Rightarrow \frac{5(x-1)-3(x-5)}{15}-6$ $\Rightarrow \frac{5 x-5-3 x+15}{15}-6$ $\Rightarrow \frac{2 x+10}{15}-6$ $\Rightarrow 2 x+10-90$ $\Rightarrow 2 x-90-10$[Transposing 10 to the RHS] $\Rightarrow 2 x-100$ $\Rightarrow x-\frac{100}{2}$ [Dividing both the sides by 2 ] $\Righ...
Read More →Solve the following
Question: $\frac{5 x}{2}+\frac{3 x}{4} \geq \frac{39}{4}$ Solution: $\frac{5 x}{2}+\frac{3 x}{4} \geq \frac{39}{4}$ $\Rightarrow \frac{10 x+3 x}{4} \geq \frac{39}{4}$ $\Rightarrow 10 x+3 x \geq 39$ $\Rightarrow 13 x \geq 39$ $\Rightarrow x \geq \frac{39}{13} \quad$ (Dividing both the sides by 13$)$ $\Rightarrow x \geq 3$ Hence, the solution set of the given inequation is $[3, \infty)$....
Read More →In the following, determine the set values of k for
Question: In the following, determine the set values ofkfor which the given quadratic equation has real roots: (i) $2 x^{2}+3 x+k=0$ (ii) $2 x^{2}+k x+3=0$ (iii) $2 x^{2}-5 x-k=0$ (iv) $k x^{2}+6 x+1=0$ (v) $x^{2}-k x+9=0$ (vi) $2 x^{2}+k x+2=0$ (vii) $3 x^{2}+2 x+k=0$ (viii) $4 x^{2}-3 k x+1=0$ (ix) $2 x^{2}+k x-4=0$ Solution: (i) The given quadric equation is $2 x^{2}+3 x+k=0$, and roots are real. Then find the value ofk. Here, $a=2, b=3$ and, $c=k$ As we know that $D=b^{2}-4 a c$ Putting the ...
Read More →Solve the following
Question: $\frac{2(x-1)}{5} \leq \frac{3(2+x)}{7}$ Solution: $\frac{2(x-1)}{5} \leq \frac{3(2+x)}{7}$ $\Rightarrow 14(x-1) \leq 15(2+x)$ $\Rightarrow 14 x-14 \leq 30+15 x$ $\Rightarrow 30+15 x \geq 14 x-14$ $\Rightarrow 15 x-14 x \geq-14-30 \quad$ (Transposing $14 x$ to the LHS and 30 to the RHS) $\Rightarrow x \geq-44$ Hence, the solution to the given inequation is $[-44, \infty)$....
Read More →