Factorize:
Question: Factorize: $7 x^{2}+2 \sqrt{14} x+2$ Solution: We have: $7 x^{2}+2 \sqrt{14} x+2$ We have to split $2 \sqrt{14}$ into two numbers such that their sum is $2 \sqrt{14}$ and product is 14 . Clearly, $\sqrt{14}+\sqrt{14}=2 \sqrt{14}$ and $\sqrt{14} \times \sqrt{14}=14$ $\therefore 7 x^{2}+2 \sqrt{14} x+2=7 x^{2}+\sqrt{14} x+\sqrt{14} x+2$ $=\sqrt{7} x(\sqrt{7} x+\sqrt{2})+\sqrt{2}(\sqrt{7} x+\sqrt{2})$ $=(\sqrt{7} x+\sqrt{2})(\sqrt{7} x+\sqrt{2})$ $=(\sqrt{7} x+\sqrt{2})^{2}$...
Read More →Find two consecutive natural numbers whose product is 20.
Question: Find two consecutive natural numbers whose product is 20. Solution: Let two consecutive numbers be $x$ and $(x+1)$ Then according to question $x(x+1)=20$ $x^{2}+x-20=0$ $x^{2}+5 x-4 x-20=0$ $x(x+5)-4(x+5)=0$ $(x+5)(x-4)=0$ $(x+5)=0$ $x=-5$ Or $(x-4)=0$ $x=4$ Since,xbeing a natural number, Therefore negative value is not possible So when $x=4$ then $x+1=4+1$ $=5$ Thus, two consecutive numbers are 4,5...
Read More →Solve |x−1|+|x−2|+|x−3|≥6
Question: Solve $|x-1|+|x-2|+|x-3| \geq 6$ Solution: We have, $|x-1|+|x-2|+|x-3| \geq 6 \quad \ldots$ (i) As, $|x-1|=\left\{\begin{array}{l}x-1, x \geq 1 \\ 1-x, x1\end{array}\right.$ $|x-2|=\left\{\begin{array}{l}x-2, x \geq 2 \\ 2-x, x2\end{array}\right.$ and $|x-3|=\left\{\begin{array}{l}x-3, x \geq 3 \\ 3-x, x3\end{array}\right.$ Now, Case I : When $x1$, $1-x+2-x+3-x \geq 6$ $\Rightarrow 6-3 x \geq 6$ $\Rightarrow 3 x \leq 0$ $\Rightarrow x \leq 0$ So, $x \in(-\infty, 0]$ Case II : When $1 \...
Read More →Factorize:
Question: Factorize: $2 \sqrt{3} x^{2}+x-5 \sqrt{3}$ Solution: We have: $2 \sqrt{3} x^{2}+x-5 \sqrt{3}$ We have to split 1 into two numbers such that their sum is 1 and product is 30 , i.e., $2 \sqrt{3} \times(-5 \sqrt{3})$. Clearly, $6+(-5)=1$ and $6 \times(-5)=-30$ $\therefore 2 \sqrt{3} x^{2}+x-5 \sqrt{3}=2 \sqrt{3} x^{2}+6 x-5 x-5 \sqrt{3}$ $=2 \sqrt{3} x(x+\sqrt{3})-5(x+\sqrt{3})$ $=(x+\sqrt{3})(2 \sqrt{3} x-5)$...
Read More →Factorize:
Question: Factorize: $\sqrt{5} x^{2}+2 x-3 \sqrt{5}$ Solution: We have: $\sqrt{5} x^{2}+2 x-3 \sqrt{5}$ We have to split 2 into two numbers such that their sum is 2 and product is $(-15)$, i.e., $\sqrt{5} \times(-3 \sqrt{5})$. Clearly, $5+(-3)=2$ and $5 \times(-3)=-15$. $\therefore \sqrt{5} x^{2}+2 x-3 \sqrt{5}=\sqrt{5} x^{2}+5 x-3 x-3 \sqrt{5}$ $=\sqrt{5} x(x+\sqrt{5})-3(x+\sqrt{5})$ $=(x+\sqrt{5})(\sqrt{5} x-3)$...
Read More →Solve the following
Question: Solve $\left|\frac{2 x-1}{x-1}\right|2$ Solution: As, $\left|\frac{2 x-1}{x-1}\right|2$ $\Rightarrow \frac{2 x-1}{x-1}-2$ or $\frac{2 x-1}{x-1}2 \quad($ As, $|x|2 \Rightarrow x-2$ or $x2)$ $\Rightarrow \frac{2 x-1}{x-1}+20$ or $\frac{2 x-1}{x-1}-20$ $\Rightarrow \frac{2 x-1+2 x-2}{x-1}0$ or $\frac{2 x-1-2 x+2}{x-1}0$ $\Rightarrow \frac{4 x-3}{x-1}0$ or $\frac{1}{x-1}0$ $\Rightarrow \frac{4 x-3}{x-1}0$ or $x-10$ $\Rightarrow[(4 x-30$ and $x-10)$ or $(4 x-20$ and $x-10)]$ or $[x-10]$ $\R...
Read More →If f : Q → Q, g : Q → Q are two functions defined by f(x) = 2 x and g(x) = x + 2,
Question: If $f: Q \rightarrow Q, g: Q \rightarrow Q$ are two functions defined by $f(x)=2 x$ and $g(x)=x+2$, show that $f$ and $g$ are bijective maps. Verify that $(g \circ f)^{-1}=f^{-1}$ og $^{-1}$. Solution: Injectivity of $f$. Let $x$ and $y$ be two elements of domain $(Q)$, such that $f(x)=f(y)$ $\Rightarrow 2 x=2 y$ $\Rightarrow x=y$ So, $f$ is one-one. Surjectivity off:Letybe in the co-domain (Q), such thatf(x) = y. $\Rightarrow 2 x=y$ $\Rightarrow x=\frac{y}{2} \in Q \quad$ (domain) $\R...
Read More →Solve the following
Question: Solve $\frac{|x+2|-x}{x}2$ Solution: As, $\frac{|x+2|-x}{x}2$ $\Rightarrow \frac{|x+2|-x}{x}-20$ $\Rightarrow \frac{|x+2|-x-2 x}{x}0$ $\Rightarrow \frac{|x+2|-3 x}{x}0$ Case I : When $x \geq-2,|x+2|=(x+2)$, $\frac{(x+2)-3 x}{x}0$ $\Rightarrow \frac{2-2 x}{x}0$ $\Rightarrow \frac{-2(x-1)}{x}0$ $\Rightarrow \frac{x-1}{x}0$ $\Rightarrow(x-10$ and $x0)$ or $(x-10$ and $x0)$ $\Rightarrow(x1$ and $x0)$ or $(x1$ and $x0)$ $\Rightarrow x1$ or $x0$ $\Rightarrow x \in[-2,0) \cup(1, \infty)$ Case...
Read More →Solve the following
Question: Solve $\frac{1}{|x|-3}\frac{1}{2}$ Solution: As, $\frac{1}{|x|-3}\frac{1}{2}$ $\Rightarrow \frac{1}{|x|-3}-\frac{1}{2}0$ $\Rightarrow \frac{2-(|x|-3)}{2(|x|-3)}0$ $\Rightarrow \frac{2-|x|+3}{|x|-3}0$ $\Rightarrow \frac{5-|x|}{|x|-3}0$ Case I: When $x \geq 0,|x|=x$ $\frac{5-x}{x-3}0$ $\Rightarrow(5-x0$ and $x-30)$ or $(5-x0$ and $x-30)$ $\Rightarrow(x5$ and $x3)$ or $(x5$ and $x3)$ $\Rightarrow x5$ and $x3$ $\Rightarrow x \in[0,3) \cup(5, \infty)$ Case II: When $x \leq 0,|x|=-x$, $\frac...
Read More →Find the whole number which when decreased by 20 is equal to 69
Question: Find the whole number which when decreased by 20 is equal to 69 times the reciprocal of the number. Solution: Let the whole numbers bex. Then according to question $(x-20)=69 \times \frac{1}{x}$ $x(x-20)=69$ $x^{2}-20 x-69=0$ $x^{2}-23 x+3 x-69=0$ $x(x-23)+3(x-23)=0$ $(x-23)(x+3)=0$ $(x-23)=0$ $x=23$ Or $(x+3)=0$ $x=-3$ Since, whole numbers being a positive, soxcannot be negative. Thus, whole numbers be...
Read More →A function f : R → R is defined as f(x)
Question: A function $f: R \rightarrow R$ is defined as $f(x)=x^{3}+4$. Is it a bijection or not? In case it is a bijection, find $f^{-1}(3)$. Solution: Injectivity of $f$. Let $x$ and $y$ be two elements of domain $(R)$, such that $f(x)=f(y)$ $\Rightarrow x^{3}+4=y^{3}+4$ $\Rightarrow x^{3}=y^{3}$ $\Rightarrow x=y$ So,fis one-one. Surjectivity off:Letybe in the co-domain (R), such thatf(x) = y. $\Rightarrow x^{3}+4=y$ $\Rightarrow x^{3}=y-4$ $\Rightarrow x=\sqrt[3]{y-4} \in R$ (domain) $\Righta...
Read More →Factorize:
Question: Factorize: $2 x^{2}+3 x-90$ Solution: We have: $2 x^{2}+3 x-90$ We have to split 3 into two numbers such that their sum is 3 and their product is $(-180)$, i.e., $2 \times(-90)$. Clearly, $-12+15=3$ and $-12 \times 15=-180$ $\therefore 2 x^{2}+3 x-90=2 x^{2}-12 x+15 x-90$ $=2 x(x-6)+15(x-6)$ $=(x-6)(2 x+15)$...
Read More →Solve the following
Question: Solve $\frac{|x-2|}{x-2}0$ Solution: We have, $\frac{|x-2|}{x-2}0$ As, $|x-2|=\left\{\begin{array}{l}x-2, x \geq 2 \\ 2-x, x2\end{array}\right.$ And $\frac{|x-2|}{x-2}0$ for $x2$ So, $x2$ $\therefore x \in(2, \infty)$...
Read More →Factorise:
Question: Factorise: $3 x^{2}-14 x+8$ Solution: $3 x^{2}-14 x+8=3 x^{2}-12 x-2 x+8$ $=3 x(x-4)-2(x-4)$ $=(x-4)(3 x-2)$ Hence, factorisation of $3 x^{2}-14 x+8$ is $(x-4)(3 x-2)$....
Read More →Solve the following
Question: Solve $\left|\frac{3 x-4}{2}\right| \leq \frac{5}{12}$ Solution: As, $\left|\frac{3 x-4}{2}\right| \leq \frac{5}{12}$ $\Rightarrow-\frac{5}{12} \leq \frac{3 x-4}{2} \leq \frac{5}{12} \quad($ As,$|x| \leq a \Rightarrow-a \leq x \leq a)$ $\Rightarrow-\frac{5}{6} \leq 3 x-4 \leq \frac{5}{6}$ $\Rightarrow-\frac{5}{6}+4 \leq 3 x \leq \frac{5}{6}+4$ $\Rightarrow \frac{-5+24}{6} \leq 3 x \leq \frac{5+24}{6}$ $\Rightarrow \frac{19}{6} \leq 3 x \leq \frac{29}{6}$ $\Rightarrow \frac{19}{18} \leq...
Read More →Factorize:
Question: Factorize: $24 x^{2}-41 x+12$ Solution: We have: $24 x^{2}-41 x+12$ We have to split $(-41)$ into two numbers such that their sum is $(-41)$ and their product is 288 , i.e., $24 \times 12$. Clearly, $(-32)+(-9)=-41$ and $(-32) \times(-9)=288$ $\therefore 24 x^{2}-41 x+12=24 x^{2}-32 x-9 x+12$ $=8 x(3 x-4)-3(3 x-4)$ $=(3 x-4)(8 x-3)$...
Read More →If f : R → R be defined by
Question: If $f: R \rightarrow R$ be defined by $f(x)=x^{3}-3$, then prove that $f^{-1}$ exists and find a formula for $f^{-1}$. Hence, find $f^{-1}(24)$ and $f^{-1}(5)$. Solution: Injectivity of $f$ : Let $x$ and $y$ be two elements in domain $(R)$, such that, $x^{3}-3=y^{3}-3$ $\Rightarrow x^{3}=y^{3}$ $\Rightarrow x=y$ So, $f$ is one-one. Surjectivity of $f$ : Let $y$ be in the co-domain (R) such that $f(x)=y$ $\Rightarrow x^{3}-3=y$ $\Rightarrow x^{3}=y+3$ $\Rightarrow x=\sqrt[3]{y+3} \in R$...
Read More →If an integer is added to its square, the sum is 90.
Question: If an integer is added to its square, the sum is 90. Find the integer with the help of quadratic equation. Solution: Let an integer bex. Then according to question $x+x^{2}=90$ $x^{2}+x-90=0$ $x^{2}+10 x-9 x-90=0$ $x(x+10)-9(x+10)=0$ $(x+10)(x-9)=0$ $(x+10)=0$ $x=-10$ Or $(x-9)=0$ $x=9$ Thus, an integer be $-10,9$...
Read More →Solve the following
Question: Solve $|4-x|+13$ Solution: As, $|4-x|+13$ $\Rightarrow|4-x|3-1$ $\Rightarrow|4-x|2$ $\Rightarrow-24-x2 \quad($ As,$|x|a \Rightarrow-axa)$ $\Rightarrow-2-4-x2-4$ $\Rightarrow-6-x-2$ $\Rightarrow 2x6$ $\therefore x \in(2,6)$...
Read More →Factorise:
Question: Factorise: $21 x^{2}+5 x-6$ Solution: $21 x^{2}+5 x-6$ $=21 x^{2}+14 x-9 x-6$ $=7 x(3 x+2)-3(3 x+2)$ $=(3 x+2)(7 x-3)$...
Read More →Solve the following
Question: Solve $\left|x+\frac{1}{3}\right|\frac{8}{3}$ Solution: As, $\left|x+\frac{1}{3}\right|\frac{8}{3}$ $\Rightarrow\left(x+\frac{1}{3}\right)\frac{-8}{3}$ or $\left(x+\frac{1}{3}\right)\frac{8}{3} \quad($ As, $|x|a \Rightarrow x-a$ or $xa)$ $\Rightarrow x-\frac{1}{3}-\frac{8}{3}$ or $x\frac{8}{3}-\frac{1}{3}$ $\Rightarrow x-\frac{9}{3}$ or $x\frac{7}{3}$ $\Rightarrow x-3$ or $x\frac{7}{3}$ $\therefore x \in(-\infty,-3) \cup\left(\frac{7}{3}, \infty\right)$...
Read More →The sum of two numbers is 48 and their product is 432. Find the numbers.
Question: The sum of two numbers is 48 and their product is 432. Find the numbers. Solution: Let first numbers be $x$ and other $(48-x)$ Then according to question $x(48-x)=432$ $48 x-x^{2}=432$ $x^{2}-48 x+432=0$ $x^{2}-36 x-12 x+432=0$ $x(x-36)-12(x-36)=0$ $(x-36)(x-12)=0$ $(x-36)=0$ $x=36$ Or $(x-12)=0$ $x=12$ Thus, two number be 36,12...
Read More →Solve each of the following system of equations in R.
Question: Solve each of the following system of equations in R. $\frac{4}{x+1} \leq 3 \leq \frac{6}{x+1}, x0$ Solution: $\frac{4}{x+1} \leq 3 \leq \frac{6}{x+1}, x0$ $\Rightarrow \frac{4}{x+1} \leq 3$ and $3 \leq \frac{6}{x+1}$ Now, $\frac{4}{x+1} \leq 3$ $\Rightarrow \frac{4}{x+1}-3 \leq 0$ $\Rightarrow \frac{4-3 x-3}{x+1} \leq 0$ $\Rightarrow \frac{1-3 x}{x+1} \leq 0$ $\Rightarrow \frac{3 x-1}{x+1} \geq 0$ $\Rightarrow x \in(-\infty,-1) \cup\left[\frac{1}{3}, \infty\right)$ Thus, the solution ...
Read More →Factorize:
Question: Factorize: $15 x^{2}+2 x-8$ Solution: We have: $15 x^{2}+2 x-8$ We have to split 2 into two numbers such that their sum is 2 and their product is $(-120)$, i.e., $15 \times(-8)$. Clearly, $12+(-10)=2$ and $12 \times(-10)=-120$ $\therefore 15 x^{2}+2 x-8=15 x^{2}+12 x-10 x-8$ $=3 x(5 x+4)-2(5 x+4)$ $=(5 x+4)(3 x-2)$...
Read More →Two squares have sides x cm and (x + 4) cm.
Question: Two squares have sides $x \mathrm{~cm}$ and $(x+4) \mathrm{cm}$. The sum of their areas is $656 \mathrm{~cm}^{2}$. Find the sides of the squares. Solution: Given that the sides of two square be $x \mathrm{~cm}$ and $(x+4) \mathrm{cm}$ Then according to question $x^{2}+(x+4)^{2}=656$ $x^{2}+x^{2}+8 x+16=656$ $2 x^{2}+8 x+16-656=0$ $2 x^{2}+8 x-640=0$ $x^{2}+4 x-320=0$ $x^{2}+20 x-16 x-320=0$ $x(x+20)-16(x+20)=0$ $(x+20)(x-16)=0$ $(x+20)=0$ $x=-20$ Or $(x-16)=0$ $x=16$ Since, sides of th...
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