Show that the function f : N → N defined by
Question: Show that the function $f: N \rightarrow N$ defined by $f(x)=x^{2}+x+1$ is one-one but not onto. Find the inverse of $f: N \rightarrow S$, where $S$ is range of $f$. Solution: Given: The function $f: N \rightarrow N$ defined by $f(x)=x^{2}+x+1$ To showf is one-one: Let $f\left(x_{1}\right)=f\left(x_{2}\right)$ $\Rightarrow x_{1}^{2}+x_{1}+1=x_{2}^{2}+x_{2}+1$ $\Rightarrow x_{1}^{2}+x_{1}=x_{2}^{2}+x_{2}$ $\Rightarrow x_{1}^{2}+x_{1}-x_{2}^{2}-x_{2}=0$ $\Rightarrow x_{1}^{2}-x_{2}^{2}+x...
Read More →Factorize:
Question: Factorize: $x^{2}-2 x+\frac{7}{16}$ Solution: We have : $x^{2}-2 x+\frac{7}{16}$ $=\frac{16 x^{2}-32 x+7}{16}$ $=\frac{1}{16}\left(16 x^{2}-32 x+7\right)$ Now, we have to split $(-32)$ into two numbers such that their sum is $(-32)$ and their product is 112 , i.e., $16 \times 7$. Clearly, $(-4)+(-28)=-32$ and $(-4) \times(-28)=112$. $\therefore x^{2}-2 x+\frac{7}{16}=\frac{1}{16}\left(16 x^{2}-32 x+7\right)$ $=\frac{1}{16}\left(16 x^{2}-4 x-28 x+7\right)$ $=\frac{1}{16}[4 x(4 x-1)-7(4 ...
Read More →Factorize:
Question: Factorize: $10 x^{2}-9 x-7$ Solution: We have: $10 x^{2}-9 x-7$ We have to split $(-9)$ into two numbers such that their sum is $(-9)$ and their product is $(-70)$, i.e., $10 \times(-7)$. Clearly, $(-14)+5=-9$ and $(-14) \times 5=-70$ $\therefore 10 x^{2}-9 x-7=10 x^{2}+5 x-14 x-7$ $=5 x(2 x+1)-7(2 x+1)$ $=(2 x+1)(5 x-7)$...
Read More →Let A = R – {2} and B = R – {1}.
Question: Let $A=R-\{2\}$ and $B=R-\{1\}$. If $f: A \rightarrow B$ is a function defined by $f(x)=\frac{x-1}{x-2}$, show that $f$ is one-one and onto. Find $f^{-}$. Solution: Given: $f(x)=\frac{x-1}{x-2}$ To showf is one-one: Let $f\left(x_{1}\right)=f\left(x_{2}\right)$ $\Rightarrow \frac{x_{1}-1}{x_{1}-2}=\frac{x_{2}-1}{x_{2}-2}$ $\Rightarrow\left(x_{1}-1\right)\left(x_{2}-2\right)=\left(x_{2}-1\right)\left(x_{1}-2\right)$ $\Rightarrow x_{1} x_{2}-2 x_{1}-x_{2}+2=x_{1} x_{2}-2 x_{2}-x_{1}+2$ $...
Read More →Factorise:
Question: Factorise: $9 x^{2}-3 x-20$ Solution: $9 x^{2}-3 x-20=9 x^{2}-15 x+12 x-20$ $=3 x(3 x-5)+4(3 x-5)$ $=(3 x-5)(3 x+4)$ Hence, factorisation of $9 x^{2}-3 x-20$ is $(3 x-5)(3 x+4)$....
Read More →Factorise:
Question: Factorise: $6 x^{2}-11 x-35$ Solution: $6 x^{2}-11 x-35=6 x^{2}-21 x+10 x-35$ $=3 x(2 x-7)+5(2 x-7)$ $=(2 x-7)(3 x+5)$ Hence, factorisation of $6 x^{2}-11 x-35$ is $(2 x-7)(3 x+5)$....
Read More →Let f : R
Question: Let $f: R-\left\{-\frac{4}{3}\right\} \rightarrow R$ be a function defined as $f(x)=\frac{4 x}{3 x+4}$. Show that $f: R-\left\{-\frac{4}{3}\right\} \rightarrow$ Rang $(f)$ is one-one and onto. Hence, find $f-1$. Solution: The function $f: \mathbf{R}-\left\{-\frac{4}{3}\right\} \rightarrow \mathbf{R}-\left\{\frac{4}{3}\right\}$ is given by $f(x)=\frac{4 x}{3 x+4}$. Injectivity: Let $x, y \in \mathbf{R}-\left\{-\frac{4}{3}\right\}$ be such that $f(x)=f(y)$ $\Rightarrow \frac{4 x}{3 x+4}=...
Read More →Factorize:
Question: Factorize: $5 x^{2}-16 x-21$ Solution: We have: $5 x^{2}-16 x-21$ We have to split $(-16)$ into two numbers such that their sum is $(-16)$ and their product is $(-105)$, i.e., $5 \times(-21)$. Clearly, $(-21)+5=-16$ and $(-21) \times 5=-105$ $\therefore 5 x^{2}-16 x-21=5 x^{2}+5 x-21 x-21$ $=5 x(x+1)-21(x+1)$ $=(x+1)(5 x-21)$...
Read More →Factorize:
Question: Factorize: $2 x^{2}-7 x-15$ Solution: We have: $2 x^{2}-7 x-15$ We have to split $(-7)$ into two numbers such that their sum is $(-7)$ and their product is $(-30)$, i.e., $2 \times(-15)$. Clearly, $(-10)+3=-7$ and $(-10) \times 3=-30$ $\therefore 2 x^{2}-7 x-15=2 x^{2}-10 x+3 x-15$ $=2 x(x-5)+3(x-5)$ $=(x-5)(2 x+3)$...
Read More →Factorize:
Question: Factorize: $6 x^{2}-5 x-21$ Solution: We have: $6 x^{2}-5 x-21$ We have to split $(-5)$ into two numbers such that their sum is $(-5)$ and their product is $(-126)$, i.e., $6 \times(-21)$. Clearly, $9+(-14)=-5$ and $9 \times(-14)=-126$. $\therefore 6 x^{2}-5 x-21=6 x^{2}+9 x-14 x-21$ $=3 x(2 x+3)-7(2 x+3)$ $=(2 x+3)(3 x-7)$...
Read More →Factorize:
Question: Factorize: $15 x^{2}-x-128$ Solution: We have: $15 x^{2}-x-28$ We have to split $(-1)$ into two numbers such that their sum is $(-1)$ and their product is $(-420)$, i.e., $15 \times(-28)$. Clearly, $(-21)+20=-1$ and $(-21) \times 20=-420$. $\therefore 15 x^{2}-x-28=15 x^{2}-21 x+20 x-28$ $=3 x(5 x-7)+4(5 x-7)$ $=(5 x-7)(3 x+4)$...
Read More →Factorize:
Question: Factorize: $2 x^{2}+3 \sqrt{3} x+3$ Solution: We have: $2 x^{2}+3 \sqrt{3} x+3$ We have to split $3 \sqrt{3}$ into two numbers such that their sum is $3 \sqrt{3}$ and their product is 6 , i.e., $2 \times 3$. Clearly, $2 \sqrt{3}+\sqrt{3}=3 \sqrt{3}$ and $2 \sqrt{3} \times \sqrt{3}=6$. $\therefore 2 x^{2}+3 \sqrt{3} x+3=2 x^{2}+2 \sqrt{3} x+\sqrt{3} x+3$ $=2 x(x+\sqrt{3})+\sqrt{3}(x+\sqrt{3})$ $=(x+\sqrt{3})(2 x+\sqrt{3})$...
Read More →Factorize:
Question: Factorize: $\sqrt{2} x^{2}+3 x+\sqrt{2}$ Solution: We have: $\sqrt{2} x^{2}+3 x+\sqrt{2}$ We have to split 3 into two numbers such that their sum is 3 and their product is 2 , i.e., $\sqrt{2} \times \sqrt{2}$. Clearly, $2+1=3$ and $2 \times 1=2$ $\therefore \sqrt{2} x^{2}+3 x+\sqrt{2}=\sqrt{2} x^{2}+2 x+x+\sqrt{2}$ $=\sqrt{2} x(x+\sqrt{2})+1(x+\sqrt{2})$ $=(x+\sqrt{2})(\sqrt{2} x+1)$...
Read More →Let f : N→N be a function defined as f (x)
Question: Let $f: N \rightarrow N$ be a function defined as $f(x)=9 x^{2}+6 x-5$. Show that $f: N \rightarrow S$, where $S$ is the range of $f$, is invertible. find the inverse of $f$ and hence find $f^{-1}(43)$ and $f^{-1}(163)$. Solution: We have, $f: N \rightarrow N$ is a function defined as $f(x)=9 x^{2}+6 x-5$ Let $y=f(x)=9 x^{2}+6 x-5$ $\Rightarrow y=9 x^{2}+6 x-5$ $\Rightarrow y=9 x^{2}+6 x+1-1-5$ $\Rightarrow y=\left(9 x^{2}+6 x+1\right)-6$ $\Rightarrow y=(3 x+1)^{2}-6$ $\Rightarrow y+6=...
Read More →The sum of a number and its positive square root is 6/25.
Question: The sum of a number and its positive square root is 6/25. Find the number. Solution: Let first numbers bex Then according to question $x+\sqrt{x}=\frac{6}{25}$ Let $x=y^{2}$ then $y^{2}+y=\frac{6}{25}$ $25 y^{2}+25 y=6$ $25 y^{2}+25 y-6=0$ $25 y^{2}+30 y-5 y-6=0$ $5 y(5 y+6)-1(5 y+6)=0$ $(5 y+6)(5 y-1)=0$ $(5 y+6)=0$ $y=\frac{-6}{5}$ Or $(5 y-1)=0$ $x=\frac{1}{5}$ Since, being a positive number, soycannot be negative. Therefore, $x=y^{2}$ $=\left(\frac{1}{5}\right)^{2}$ $=\left(\frac{1...
Read More →Factorise:
Question: Factorise: $\sqrt{3} x^{2}+10 x+8 \sqrt{3}$ Solution: $\sqrt{3} x^{2}+10 x+8 \sqrt{3}=\sqrt{3} x^{2}+6 x+4 x+8 \sqrt{3}$ $=\sqrt{3} x(x+2 \sqrt{3})+4(x+2 \sqrt{3})$ $=(x+2 \sqrt{3})(\sqrt{3} x+4)$ Hence, factorisation of $\sqrt{3} x^{2}+10 x+8 \sqrt{3}$ is $(x+2 \sqrt{3})(\sqrt{3} x+4)$....
Read More →Consider the function f : R+ →[−9,∞] given by f(x) = 5x2 + 6x − 9.
Question: Consider the function $f: \mathrm{R}^{+} \rightarrow[-9, \infty]$ given by $f(x)=5 x^{2}+6 x-9$. Prove that $f$ is invertible with $f^{-1}(y)=\frac{\sqrt{54+5 y-3}}{5}$ [CBSE 2015] Solution: We have, $f(x)=5 x^{2}+6 x-9$ Let $y=5 x^{2}+6 x-9$ $=5\left(x^{2}+\frac{6}{5} x-\frac{9}{5}\right)$ $=5\left(x^{2}+2 \times x \times \frac{3}{5}+\frac{9}{25}-\frac{9}{25}-\frac{9}{5}\right)$ $=5\left(\left(x+\frac{3}{5}\right)^{2}-\frac{9}{25}-\frac{9}{5}\right)$ $=5\left(x+\frac{3}{5}\right)^{2}-...
Read More →Factorize:
Question: Factorize: $5 \sqrt{5} x^{2}+20 x+3 \sqrt{5}$ Solution: We have: $5 \sqrt{5} x^{2}+20 x+3 \sqrt{5}$ We have to split 20 into two numbers such that their sum is 20 and their product is 75.Clearly, $15+5=20$ and $15 \times 5=75$ $\therefore 5 \sqrt{5} x^{2}+20 x+3 \sqrt{5}=5 \sqrt{5} x^{2}+15 x+5 x+3 \sqrt{5}$ $=5 x(\sqrt{5} x+3)+\sqrt{5}(\sqrt{5} x+3)$ $=(\sqrt{5} x+3)(5 x+\sqrt{5})$...
Read More →The sum of two numbers is 8 and 15 times the sum of their reciprocals is also 8.
Question: The sum of two numbers is 8 and 15 times the sum of their reciprocals is also 8. Find the numbers. Solution: Let first numbers be $x$ and other $(8-x)$ Then according to question $\left\{\frac{1}{x}+\frac{1}{(8-x)}\right\} \times 15=8$ $\left\{\frac{8-\not x+\not x}{x(8-x)}\right\}=\frac{8}{15}$ $x(8-x)=15$ $x^{2}-8 x+15=0$ $x^{2}-5 x-3 x+15=0$ $x(x-5)-3(x-5)=0$ $(x-5)(x-3)=0$ $x=5$ Or $(x-3)=0$ $x=3$ Thus, two consecutive number be 3,5...
Read More →The sum of two numbers is 8 and 15 times the sum of their reciprocals is also 8.
Question: The sum of two numbers is 8 and 15 times the sum of their reciprocals is also 8. Find the numbers. Solution: Let first numbers be $x$ and other $(8-x)$ Then according to question $\left\{\frac{1}{x}+\frac{1}{(8-x)}\right\} \times 15=8$ $\left\{\frac{8-\not x+\not x}{x(8-x)}\right\}=\frac{8}{15}$ $x(8-x)=15$ $x^{2}-8 x+15=0$ $x^{2}-5 x-3 x+15=0$ $x(x-5)-3(x-5)=0$ $(x-5)(x-3)=0$ $x=5$ Or $(x-3)=0$ $x=3$ Thus, two consecutive number be 3,5...
Read More →Solve the following
Question: Solve $\frac{1}{|x|-3} \leq \frac{1}{2}$ Solution: $A s, \frac{1}{|x|-3} \leq \frac{1}{2}$ $\Rightarrow \frac{1}{|x|-3}-\frac{1}{2} \leq 0$ $\Rightarrow \frac{2-(|x|-3)}{2(|x|-3)} \leq 0$ $\Rightarrow \frac{2-|x|+3}{2(|x|-3)} \leq 0$ $\Rightarrow \frac{5-|x|}{|x|-3} \leq 0$ Case I : When $x \geq 0,|x|=x$, $\frac{5-x}{x-3} \leq 0$ $\Rightarrow(5-x \leq 0$ and $x-30)$ or $(5-x \geq 0$ and $x-30)$ $\Rightarrow(x \geq 5$ and $x3)$ or $(x \leq 5$ and $x3)$ $\Rightarrow x \geq 5$ or $x3$ $\R...
Read More →Factorize:
Question: Factorize: $6 \sqrt{3} x^{2}-47 x+5 \sqrt{3}$ Solution: We have: $6 \sqrt{3} x^{2}-47 x+5 \sqrt{3}$ Now, we have to split $(-47)$ into two numbers such that their sum is $(-47)$ and their product is 90 . Clearly, $(-45)+(-2)=-47$ and $(-45) \times(-2)=90$ $\therefore 6 \sqrt{3} x^{2}-47 x+5 \sqrt{3}=6 \sqrt{3} x^{2}-2 x-45 x+5 \sqrt{3}$ $=2 x(3 \sqrt{3} x-1)-5 \sqrt{3}(3 \sqrt{3} x-1)$ $=(3 \sqrt{3} x-1)(2 x-5 \sqrt{3})$...
Read More →The sum of squares of two consecutive odd positive integers is 394.
Question: The sum of squares of two consecutive odd positive integers is 394. Find them. Solution: Let two consecutive odd positive integer be $(2 x-1)$ and other $(2 x+1)$ Then according to question $(2 x+1)^{2}+(2 x-1)^{2}=394$ $8 x^{2}+2=394$ $8 x^{2}=394-2$ $x^{2}=\frac{392}{8}$ $x^{2}=49$ $x=\sqrt{49}$ $=\pm 7$ Since,xbeing a positive number, soxcannot be negative. Therefore, When $x=7$ then odd positive $2 x-1=2 \times 7-1$ $=13$ And $2 x+1=2 \times 7+1$ $=15$ Thus, two consecutive odd pos...
Read More →Let A = R − {3} and B = R − {1}.
Question: Let $A=R-\{3\}$ and $B=R-\{1\}$. Consider the function $f: A \rightarrow B$ defined by $f(x)=\frac{x-2}{x-3}$. Show that $f$ is one-one and onto and hence find $f^{-1}$. [CBSE 2012, 2014] Solution: We have, $A=R-\{3\}$ and $B=R-\{1\}$ The function $f: A \rightarrow B$ defined by $f(x)=\frac{x-2}{x-3}$ Let $x, y \in A$ such that $f(x)=f(y)$. Then, $\frac{x-2}{x-3}=\frac{y-2}{y-3}$ $\Rightarrow x y-3 x-2 y+6=x y-2 x-3 y+6$ $\Rightarrow-x=-y$ $\Rightarrow x=y$ $\therefore f$ is one-one. L...
Read More →Solve the following
Question: Solve $\frac{|x-2|-1}{|x-2|-2} \leq 0$ [NCERT EXEMPLAR] Solution: As, $\frac{|x-2|-1}{|x-2|-2} \leq 0$ Case I: When $x \geq 2,|x-2|=x-2$, $\frac{x-2-1}{x-2-2} \leq 0$ $\Rightarrow \frac{x-3}{x-4} \leq 0$ $\Rightarrow(x-3 \leq 0$ and $x-40)$ or $(x-3 \geq 0$ and $x-40)$ $\Rightarrow(x \leq 3$ and $x4)$ or $(x \geq 3$ and $x4)$ $\Rightarrow \phi$ or $(3 \leq x4)$ $\Rightarrow 3 \leq x4$ So, $x \in[3,4)$ Case II : When $x \leq 2,|x-2|=2-x$, $\frac{2-x-1}{2-x-2} \leq 0$ $\Rightarrow \frac{...
Read More →