Solve the following quadratic equations:
Question: Solve the following quadratic equations: (i) $x^{2}-(3 \sqrt{2}+2 i) x+6 \sqrt{2 i}=0$ (ii) $x^{2}-(5-i) x+(18+i)=0$ (iii) $(2+i) x^{2}-(5-i) x+2(1-i)=0$ (iv) $x^{2}-(2+i) x-(1-7 i)=0$ (v) $i x^{2}-4 x-4 i=0$ (vi) $x^{2}+4 i x-4=0$ (vii) $2 x^{2}+\sqrt{15} i x-i=0$ (viii) $x^{2}-x+(1+i)=0$ (ix) $i x^{2}-x+12 i=0$ (x) $x^{2}-(3 \sqrt{2}-2 i) x-\sqrt{2} i=0$ (xi) $x^{2}-(\sqrt{2}+i) x+\sqrt{2} i=0$ (xii) $2 x^{2}-(3+7 i) x+(9 i-3)=0$ Solution: (i) $x^{2}-(3 \sqrt{2}+2 i) x+6 \sqrt{2} i=0...
Read More →Solve the following quadratic equations:
Question: Solve the following quadratic equations: (i) $x^{2}-(3 \sqrt{2}+2 i) x+6 \sqrt{2 i}=0$ (ii) $x^{2}-(5-i) x+(18+i)=0$ (iii) $(2+i) x^{2}-(5-i) x+2(1-i)=0$ (iv) $x^{2}-(2+i) x-(1-7 i)=0$ (v) $i x^{2}-4 x-4 i=0$ (vi) $x^{2}+4 i x-4=0$ (vii) $2 x^{2}+\sqrt{15} i x-i=0$ (viii) $x^{2}-x+(1+i)=0$ (ix) $i x^{2}-x+12 i=0$ (x) $x^{2}-(3 \sqrt{2}-2 i) x-\sqrt{2} i=0$ (xi) $x^{2}-(\sqrt{2}+i) x+\sqrt{2} i=0$ (xii) $2 x^{2}-(3+7 i) x+(9 i-3)=0$ Solution: (i) $x^{2}-(3 \sqrt{2}+2 i) x+6 \sqrt{2} i=0...
Read More →Factorise:
Question: Factorise: $\left(\frac{25}{4} x^{2}-\frac{1}{9} y^{2}\right)$ Solution: $\left(\frac{25}{4} x^{2}-\frac{1}{9} y^{2}\right)$ $=\left(\frac{5}{2} x\right)^{2}-\left(\frac{1}{3} y\right)^{2}$ $=\left(\frac{5}{2} x+\frac{1}{3} y\right)\left(\frac{5}{2} x-\frac{1}{3} y\right) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$...
Read More →Prove that the function $f: N \rightarrow N$, defined by $f(x)=x^{2}+x+1$, is one-one but not onto.
[question] Question. Prove that the function $f: N \rightarrow N$, defined by $f(x)=x^{2}+x+1$, is one-one but not onto. [solution] Solution: $f: N \rightarrow N$, defined by $f(x)=x^{2}+x+1$ Injectivity: Let x and y be any two elements in the domain (N), such that f(x) = f(y). $\Rightarrow x^{2}+x+1=y^{2}+y+1$ $\Rightarrow\left(x^{2}-y^{2}\right)+(x-y)=0$ $\Rightarrow(x+y)(x-y)+(x-y)=0$ $\Rightarrow(x-y)(x+y+1)=0$ $\Rightarrow x-y=0 \quad[(\mathrm{x}+\mathrm{y}+1)$ cannot be zero because $x$ an...
Read More →Consider f : N → N, g : N → N and h : N → R defined as f(x) = 2x, g(y) = 3y + 4
Question: Consider $f: N \rightarrow N, g: N \rightarrow N$ and $h: N \rightarrow R$ defined as $f(x)=2 x, g(y)=3 y+4$ and $h(z)=\sin z$ for all $x, y, z \in N$. Show that ho $(g \circ f)=($ hog $)$ of. Solution: Given, $f: N \rightarrow N, g: N \rightarrow N$ and $h: N \rightarrow R$ $\Rightarrow$ gof $: N \rightarrow N$ and $h o g: N \rightarrow R$ $\Rightarrow$ ho $(g \circ f): N \rightarrow R$ and $(h o g)$ of $: N \rightarrow R$ So, both have the same domains. $(go f)(x)=g(f(x))=g(2 x)=3(2 ...
Read More →Factorise:
Question: Factorise: $9 x^{2}-16 y^{2}$ Solution: $9 x^{2}-16 y^{2}$ $=(3 x)^{2}-(4 y)^{2}$ $=(3 x+4 y)(3 x-3 y) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$...
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Question: Factorize: $x^{2}+\frac{1}{x^{2}}-2-3 x+\frac{3}{x}$ Solution: We have : $x^{2}+\frac{1}{x^{2}}-2-3 x+\frac{3}{x}$ $=x^{2}-2+\frac{1}{x^{2}}-3 x+\frac{3}{x}$ $=(x)^{2}-2 \times x \times \frac{1}{x}+\left(\frac{1}{x}\right)^{2}-3\left(x-\frac{1}{x}\right)$ $=\left(x-\frac{1}{x}\right)^{2}-3\left(x-\frac{1}{x}\right)$ $=\left(x-\frac{1}{x}\right)\left(x-\frac{1}{x}-3\right)$...
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Question: Factorize: $x^{2}-(a+b) x+a b$ Solution: We have: $x^{2}-(a+b) x+a b=x^{2}-a x-b x+a b$ $=\left(x^{2}-a x\right)-(b x-a b)$ $=x(x-a)-b(x-a)$ $=(x-a)(x-b)$...
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Question: Factorize: $a b\left(x^{2}+1\right)+x\left(a^{2}+b^{2}\right)$ Solution: We have: $a b\left(x^{2}+1\right)+x\left(a^{2}+b^{2}\right)=a b x^{2}+a b+a^{2} x+b^{2} x$ $=\left(a b x^{2}+a^{2} x\right)+\left(b^{2} x+a b\right)$ $=a x(b x+a)+b(b x+a)$ $=(b x+a)(a x+b)$...
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Question: Factorize: $a^{2} x^{2}+\left(a x^{2}+1\right) x+a$ Solution: We have: $a^{2} x^{2}+\left(a x^{2}+1\right) x+a=\left(a x^{2}+1\right) x+\left(a^{2} x^{2}+a\right)$ $=x\left(a x^{2}+1\right)+a\left(a x^{2}+1\right)$ $=\left(a x^{2}+1\right)(x+a)$...
Read More →Verify associativity for the following three mappings
Question: Verify associativity for the following three mappings: $f: N \rightarrow Z_{0}$ (the set of non-zero integers), $g: Z_{0} \rightarrow Q$ and $h: Q \rightarrow R$ given by $f(x)=2 x, g(x)=1 / x$ and $h(x)=e^{x}$. Solution: Given that $f: N \rightarrow Z_{0}, g: Z_{0} \rightarrow Q$ and $h: Q \rightarrow R$. $g \circ f: N \rightarrow Q$ and $h o g: Z_{0} \rightarrow R$ $\Rightarrow h \circ(gof): N \rightarrow R$ and (hog) of $N \rightarrow R$ So, both have the same domains. $(g o f)(x)=g...
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Question: Factorize: $a(a-2 b-c)+2 b c$ Solution: We have: $a(a-2 b-c)+2 b c=a^{2}-2 a b-a c+2 b c$ $=\left(a^{2}-2 a b\right)-(a c-2 b c)$ $=a(a-2 b)-c(a-2 b)$ $=(a-2 b)(a-c)$...
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Question: Factorize: $a(a+b-c)-b c$ Solution: We have: $a(a+b-c)-b c=a^{2}+a b-a c-b c$ $=\left(a^{2}-a c\right)+(a b-b c)$ $=a(a-c)+b(a-c)$ $=(a-c)(a+b)$...
Read More →Factorize:
Question: Factorize: $(a x+b y)^{2}+(b x-a y)^{2}$ Solution: We have: $(a x+b y)^{2}+(b x-a y)^{2}=\left[(a x)^{2}+2 \times a x \times b y+(b y)^{2}\right]+\left[(b x)^{2}-2 \times b x \times a y+(a y)^{2}\right]$ $=a^{2} x^{2}+2 a b x y+b^{2} y^{2}+b^{2} x^{2}-2 a b x y+a^{2} y^{2}$ $=a^{2} x^{2}+b^{2} y^{2}+b^{2} x^{2}+a^{2} y^{2}$ $=\left(a^{2} x^{2}+b^{2} x^{2}\right)+\left(a^{2} y^{2}+b^{2} y^{2}\right)$ $=x^{2}\left(a^{2}+b^{2}\right)+y^{2}\left(a^{2}+b^{2}\right)$ $=\left(a^{2}+b^{2}\righ...
Read More →Find the roots of the following quadratic equations
Question: Find the roots of the following quadratic equations (if they exist) by the method of completing the square. $x^{2}-4 a x+4 a^{2}-b^{2}=0$ Solution: We have to find the roots of given quadratic equation by the method of completing the square. We have, $x^{2}-4 a x+4 a^{2}-b^{2}=0$ Now shift the constant to the right hand side, $x^{2}-4 a x=b^{2}-4 a^{2}$ Now add square of half of coefficient ofon both the sides, $x^{2}-2(2 a) x+(2 a)^{2}=b^{2}-4 a^{2}+(2 a)^{2}$ We can now write it in t...
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Question: Factorize: $2 a^{2}+b c-2 a b-a c 2$ Solution: We have: $2 a^{2}+b c-2 a b-a c=\left(2 a^{2}-2 a b\right)-(a c-b c)$ $=2 a(a-b)-c(a-b)$ $=(a-b)(2 a-c)$...
Read More →Let f : R → R and g : R → R be defined by
Question: Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x)=x+1$ and $g(x)=x-1 .$ Show that fog $=g \circ f=I_{R}$. Solution: Given, $f: R \rightarrow R$ and $g: R \rightarrow R$ $\Rightarrow$ fog $: R \rightarrow R$ and gof $: R \rightarrow R$ (Also, we know that $I_{R}: R \rightarrow R$ ) So, the domains of allfog, gofandIRare the same. $(fog)(x)=f(g(x))=f(x-1)=x-1+1=x=I_{R}(x) \quad \ldots(1)$ $(gof)(x)=g(f(x))=g(x+1)=x+1-1=x=I_{R}(x) \quad \ldots(2)$ From $(1)$ and $(2)$,...
Read More →Find the roots of the following quadratic equations
Question: Find the roots of the following quadratic equations (if they exist) by the method of completing the square. $x^{2}-(\sqrt{2}+1) x+\sqrt{2}=0$ Solution: We have been given that, $x^{2}-(\sqrt{2}+1) x+\sqrt{2}=0$ Now take the constant term to the RHS and we get $x^{2}-(\sqrt{2}+1) x=-\sqrt{2}$ Now add square of half of co-efficient of x on both the sides. We have, $x^{2}-(\sqrt{2}+1) x+\left(\frac{\sqrt{2}+1}{2}\right)^{2}=\left(\frac{\sqrt{2}+1}{2}\right)^{2}-\sqrt{2}$ $x^{2}+\left(\fra...
Read More →Factorize:
Question: Factorize: $a^{3}+a b(1-2 a)-2 b^{2}$ Solution: We have: $a^{3}+a b(1-2 a)-2 b^{2}=a^{3}+a b-2 a^{2} b-2 b^{2}$ $=\left(a^{3}-2 a^{2} b\right)+\left(a b-2 b^{2}\right)$ $=a^{2}(a-2 b)+b(a-2 b)$ $=(a-2 b)\left(a^{2}+b\right)$...
Read More →Find the roots of the following quadratic equations
Question: Find the roots of the following quadratic equations (if they exist) by the method of completing the square. $\sqrt{3} x^{2}+10 x+7 \sqrt{3}=0$ Solution: We have been given that, $\sqrt{3} x^{2}+10 x+7 \sqrt{3}=0$ Now divide throughout by $\sqrt{3}$. We get, $x^{2}+\frac{10}{\sqrt{3}} x+7=0$ Now take the constant term to the RHS and we get $x^{2}+\frac{10}{\sqrt{3}} x=-7$ Now add square of half of co-efficient of x on both the sides. We have, $x^{2}+\frac{10}{\sqrt{3}} x+\left(\frac{10}...
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Question: Factorize: $a^{2}+a b(b+1)+b^{3}$ Solution: We have: $a^{2}+a b(b+1)+b^{3}=a^{2}+a b^{2}+a b+b^{3}$ $=\left(a^{2}+a b^{2}\right)+\left(a b+b^{3}\right)$ $=a\left(a+b^{2}\right)+b\left(a+b^{2}\right)$ $=\left(a+b^{2}\right)(a+b)$...
Read More →Let f : R → R and g : R → R be defined by
Question: Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x)=x^{2}$ and $g(x)=x+1$. Show that fog $\neq g$ of. Solution: Given, $f: R \rightarrow R$ and $g: R \rightarrow R$. So, the domains of $f$ and $g$ are the same. $(f o g)(x)=f(g(x))=f(x+1)=(x+1)^{2}=x^{2}+1+2 x$ $(g o f)(x)=g(f(x))=g\left(x^{2}\right)=x^{2}+1$ So,fog gof...
Read More →Factorize:
Question: Factorize: $a b\left(x^{2}+y^{2}\right)-x y\left(a^{2}+b^{2}\right)$ Solution: We have: $a b\left(x^{2}+y^{2}\right)-x y\left(a^{2}+b^{2}\right)=a b x^{2}+a b y^{2}-a^{2} x y-b^{2} x y$ $=\left(a b x^{2}-a^{2} x y\right)-\left(b^{2} x y-a b y^{2}\right)$ $=a x(b x-a y)-b y(b x-a y)$ $=(b x-a y)(a x-b y)$...
Read More →Find the roots of the following quadratic equations
Question: Find the roots of the following quadratic equations (if they exist) by the method of completing the square. $\sqrt{2} x^{2}-3 x-2 \sqrt{2}=0$ Solution: We have been given that, $\sqrt{2} x^{2}-3 x-2 \sqrt{2}=0$ Now divide throughout by $\sqrt{2}$. We get, $x^{2}-\frac{3}{\sqrt{2}} x-2=0$ Now take the constant term to the RHS and we get $x^{2}-\frac{3}{\sqrt{2}} x=2$ Now add square of half of co-efficient of x on both the sides. We have, $x^{2}-\frac{3}{\sqrt{2}} x+\left(\frac{3}{2 \sqr...
Read More →Solving the following quadratic equations by factorization method:
Question: Solving the following quadratic equations by factorization method: (i) $x^{2}+10 i x-21=0$ (ii) $x^{2}+(1-2 i) x-2 i=0$ (iii) $x^{2}-(2 \sqrt{3}+3 i) x+6 \sqrt{3} i=0$ (iv) $6 x^{2}-17 i x-12=0$ Solution: (i) $x^{2}+10 i x-21=0$ $\Rightarrow x^{2}+7 i x+3 i x-21=0$ $\Rightarrow x(x+7 i)+3 i(x+7 i)=0$ $\Rightarrow(x+7 i)(x+3 i)=0$ $\Rightarrow(x+7 i)=0$ or $(x+3 i)=0$ $\Rightarrow x=-7 i,-3 i$ So, the roots of the given quadratic equation are $-3 i$ and $-7 i$. (ii) $x^{2}+(1-2 i) x-2 i...
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