Write the number of quadratic equations, with real roots,
Question: Write the number of quadratic equations, with real roots, which do not change by squaring their roots. Solution: Let $\alpha$ and $\beta$ be the real roots of the quadratic equation $a x^{2}+b x+c=0$. On squaring these roots, we get: $\alpha=\alpha^{2} \quad$ and $\quad \beta=\beta^{2}$ $\Rightarrow \alpha(1-\alpha)=0$ and $\beta(1-\beta)=0$ $\Rightarrow \alpha=0, \alpha=1$ and $\beta=0,1$ Three cases arise: (i) $\alpha=0, \beta=0$ (ii) $\alpha=1, \beta=0$ (iii) $\alpha=1, \beta=1$ $(i...
Read More →If a and b are roots of the equation x
Question: If $a$ and $b$ are roots of the equation $x^{2}-x+1=0$, then write the value of $a^{2}+b^{2}$. Solution: Given: $x^{2}-x+1=0$ Also, $a$ and $b$ are the roots of the equation. Then, sum of the roots $=a+b=-\left(\frac{-1}{1}\right)=1$ Product of the roots $=a b=\frac{1}{1}=1$ $\therefore(a+b)^{2}=a^{2}+b^{2}+2 a b$ $\Rightarrow 1^{2}=a^{2}+b^{2}+2 \times 1$ $\Rightarrow a^{2}+b^{2}=1-2=-1$ $\Rightarrow a^{2}+b^{2}=-1$...
Read More →Write roots of the equation
Question: Write roots of the equation $(a-b) x^{2}+(b-c) x+(c-a)=0$. Solution: Given: $(a-b) x^{2}+(b-c) x+(c-a)=0$ $\Rightarrow x^{2}+\frac{b-c}{a-b} x+\frac{c-a}{a-b}=0$ $\Rightarrow x^{2}-\frac{c-a}{a-b} x-x+\frac{c-a}{a-b}=0 \quad\left[\because \frac{b-c}{a-b}=\frac{-c+a-a+b}{a-b}=-\frac{c-a}{a-b}-1\right]$ $\Rightarrow x\left(x-\frac{c-a}{a-b}\right)-1\left(x+\frac{c-a}{a-b}\right)=0$ $\Rightarrow\left(x-\frac{c-a}{a-b}\right)(x-1)=0$ $\Rightarrow x-\frac{c-a}{a-b}=0$ or $x-1=0$ $\Rightarro...
Read More →If the difference between the roots of the equation
Question: If the difference between the roots of the equation $x^{2}+a x+8=0$ is 2 , write the values of $a$. Solution: Given: $x^{2}+a x+8=0$ Let $\alpha$ and $\beta$ are the roots of the equation. Sum of the roots $=\alpha+\beta=\frac{-a}{1}=-a$. Product of the roots $=\alpha \beta=\frac{8}{1}=8$ Given: $\alpha-\beta=2$ Then, $(\alpha+\beta)^{2}-(\alpha-\beta)^{2}=4 \alpha \beta$ $\Rightarrow(\alpha+\beta)^{2}-2^{2}=4 \times 8$ $\Rightarrow(\alpha+\beta)^{2}-4=32$ $\Rightarrow(\alpha+\beta)^{2...
Read More →Solve the following
Question: If $2+\sqrt{3}$ is root of the equation $x^{2}+p x+q=0$, than write the values of $p$ and $q$. Solution: Irrational roots always occur in conjugate pairs. If $2+\sqrt{3}$ is a root and $2-\sqrt{3}$ is its conjugate root. $\Rightarrow(2+\sqrt{3}+2-\sqrt{3})=-p$ $\Rightarrow 4=-9$ $\Rightarrow p=-4$ Also, $(2+\sqrt{3})(2-\sqrt{3})=q$ $\Rightarrow 4-3=q$ $\Rightarrow q=1$...
Read More →Solve the following
Question: If $2+\sqrt{3}$ is root of the equation $x^{2}+p x+q=0$, than write the values of $p$ and $q$. Solution: Irrational roots always occur in conjugate pairs. If $2+\sqrt{3}$ is a root and $2-\sqrt{3}$ is its conjugate root. $\Rightarrow(2+\sqrt{3}+2-\sqrt{3})=-p$ $\Rightarrow 4=-9$ $\Rightarrow p=-4$ Also, $(2+\sqrt{3})(2-\sqrt{3})=q$ $\Rightarrow 4-3=q$ $\Rightarrow q=1$...
Read More →If roots α, β of the equation x
Question: If roots $\alpha, \beta$ of the equation $x^{2}-p x+16=0$ satisfy the relation $\alpha^{2}+\beta^{2}=9$, then write the value $p$. Solution: Given equation: $x^{2}-p x+16=0$ Also, $\alpha$ and $\beta$ are the roots of the equation satisfying $\alpha^{2}+\beta^{2}=9$. From the equation. we have: Sum of the roots $=\alpha+\beta=-\left(\frac{-p}{1}\right)=p$ Product of the roots $=\alpha \beta=\frac{16}{1}=16$ Now, $(\alpha+\beta)^{2}=\alpha^{2}+\beta^{2}+2 \alpha \beta$ $\Rightarrow p^{2...
Read More →Find the value
Question: $81 x^{4}-y^{4}$ Solution: $81 x^{4}-y^{4}$ $=\left(9 x^{2}\right)^{2}-\left(y^{2}\right)^{2}$ $=\left(9 x^{2}+y^{2}\right)\left(9 x^{2}-y^{2}\right) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$ $=\left(9 x^{2}+y^{2}\right)\left[(3 x)^{2}-y^{2}\right]$ $=\left(9 x^{2}+y^{2}\right)(3 x+y)(3 x-y) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$...
Read More →If a and b are roots of the equation x
Question: If $a$ and $b$ are roots of the equation $x^{2}-p x+q=0$, than write the value of $\frac{1}{a}+\frac{1}{b}$. Solution: Given: $x^{2}-p x+q=0$ Also, $a$ and $b$ are the roots of the given equation. Sum of the roots $=a+b=p$ ...(1) Product of the roots $=a b=q$ .....(2) Now, $\frac{1}{a}+\frac{1}{b}=\frac{b+a}{a b}=\frac{p}{q}$ [Using equation (1) and (2)] Hence, the value of $\frac{1}{a}+\frac{1}{b}$ is $\frac{p}{q}$....
Read More →Factorise:
Question: Factorise: $16 x^{4}-1$ Solution: $16 x^{4}-1$ $=\left(4 x^{2}\right)^{2}-1^{2}$ $=\left(4 x^{2}+1\right)\left(4 x^{2}-1\right) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$ $=\left(4 x^{2}+1\right)\left[(2 x)^{2}-1^{2}\right]$ $=\left(4 x^{2}+1\right)(2 x+1)(2 x-1) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$...
Read More →Write the number of real roots of the equation
Question: Write the number of real roots of the equation $(x-1)^{2}+(x-2)^{2}+(x-3)^{2}=0$. Solution: $(x-1)^{2}+(x-2)^{2}+(x-3)^{2}=0$ $\Rightarrow x^{2}+1-2 x+x^{2}+4-4 x+x^{2}+9-6 x=0$ $\Rightarrow 3 x^{2}-12 x+14=0$ Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=3, b=-12$ and $c=14$. $D=b^{2}-4 a c=(-12)^{2}-4 \times 3 \times 14=144-168=-24$ Since the value of $D$ is less than 0 , the given equation has no real roots....
Read More →If the quadratic equation 2x
Question: If the quadratic equation 2x2 (a3+ 8a 1)x+a2 4a= 0 possesses roots of opposite signs, then a lies in the interval ____________. Solution: For $2 x^{2}-\left(a^{3}+8 a-1\right) x+a^{2}-4 a=0$ has roots of opposite sign. Let $\alpha$ and $\beta$ be 2 roots of above equation. Then, according to give condition, $\alpha \beta0$ also, since sum of roots $=\frac{a^{3}+8 a-1}{2}$ $\Rightarrow \alpha+\beta=\frac{a^{3}+8 a-1}{2}$ and product of roots is $a^{2}-4 a$ i. e. $\alpha \beta=a^{2}-4 a$...
Read More →Factorise:
Question: Factorise: $x^{8}-1$ Solution: $x^{8}-1$ $=\left(x^{4}\right)^{2}-1^{2}$ $=\left(x^{4}+1\right)\left(x^{4}-1\right) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$ $=\left(x^{4}+1\right)\left[\left(x^{2}\right)^{2}-1^{2}\right]$ $=\left(x^{4}+1\right)\left(x^{2}+1\right)\left(x^{2}-1\right) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$ $=\left(x^{4}+1\right)\left(x^{2}+1\right)\left(x^{2}-1\right)^{2}$ $=\left(x^{4}+1\right)\left(x^{2}+1\right)(x+1)(x-1) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$...
Read More →Find the values of k for which the roots are real and equal in each of the following equations:
Question: Find the values ofkfor which the roots are real and equal in each of the following equations: (i) $k x^{2}+4 x+1=0$ (ii) $k x^{2}-2 \sqrt{5} x+4=0$ (iii) $3 x^{2}-5 x+2 k=0$ (iv) $4 x^{2}+k x+9=0$ (v) $2 k x^{2}-40 x+25=0$ (vi) $9 x^{2}-24 x+k=0$ (vii) $4 x^{2}-3 k x+1=0$ (viii) $x^{2}-2(5+2 k) x+3(7+10 k)=0$ (ix) $(3 k+1) x^{2}+2(k+1) x+k=0$ (x) $k x^{2}+k x+1=-4 x^{2}-x$ (xi) $(k+1) x^{2}+2(k+3) x+(k+8)=0$ (xii) $x^{2}-2 k x+7 k-12=0$ (xiii) $(k+1) x^{2}-2(3 k+1) x+8 k+1=0$ (xiv) $(2...
Read More →Solve the following
Question: If the equations $x^{2}+x+a=0$ and $x^{2}+a x+1=0, a \neq 1$, have a common root, then $a=$ ____________________ Solution: Given: $x^{2}+x+a=0$ and $x^{2}+a x+1=0$ have a common root where $a \neq 1$. Solving above 2 equations, we get $x(1-a)+a-1=0$ i. e. $x(1-a)=+1-a$ i. e. $x=1$ $(\because a \neq 1)$ given $\therefore a$ can be obtained form $x^{2}+x+a=0$ Since $x=1$ satisfy $x^{2}+x+a=0$ i. e. $(1)^{2}+1+a=0$ i. e. $a=-2$...
Read More →Factorise:
Question: Factorise: $x^{4}+\frac{4}{x^{4}}$ Solution: $x^{4}+\frac{4}{x^{4}}$ $=x^{4}+\frac{4}{x^{4}}+4-4$ $=\left[\left(x^{2}\right)^{2}+\left(\frac{2}{x^{2}}\right)^{2}+2 \times\left(x^{2}\right) \times\left(\frac{2}{x^{2}}\right)\right]-2^{2}$ $=\left(x^{2}+\frac{2}{x^{2}}\right)^{2}-2^{2} \quad\left[a^{2}+2 a b+b^{2}=(a+b)^{2}\right]$ $=\left(x^{2}+\frac{2}{x^{2}}+2\right)\left(x^{2}+\frac{2}{x^{2}}-2\right) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$...
Read More →If the roots of the equation
Question: If the roots of the equation $x^{2}-8 x+a^{2}-6 a=0$ are real, then ' $a$ ' lies in the interval _____________ Solution: For $x^{2}-8 x+a^{2}-6 a=0$ roots are real Here $a=1, b=-8, c=a^{2}-6 a$ i.e. $D \geq 0$ (Discriminant) i. e. $b^{2}-4 a c \geq 0$ i. e. $64-4(1)\left(a^{2}-6 a\right) \geq 0$ i.e. $64-4 a^{2}+24 a \geq 0$ i.e. $-4 a^{2}+24 a+64 \geq 0$ Now, multiplying by $-\frac{1}{4}$ i. e. $\frac{4 a^{2}}{4}-\frac{24 a}{4}-\frac{64}{4} \leq 0$ i. e. $a^{2}-6 a-16 \leq 0$ i. e. $a...
Read More →Factorise:
Question: Factorise: $x^{2}-2+\frac{1}{x^{2}} y^{2}$ Solution: $x^{2}-2+\frac{1}{x^{2}}-y^{2}$ $=\left[x^{2}-2 \times x \times \frac{1}{x}+\left(\frac{1}{x}\right)^{2}\right]-y^{2}$ $=\left(x-\frac{1}{x}\right)^{2}-y^{2} \quad\left[a^{2}-2 a b+b^{2}=(a-b)^{2}\right]$ $=\left(x-\frac{1}{x}+y\right)\left(x-\frac{1}{x}-y\right) \quad\left[a^{2}-b^{2}=(a-b)(a+b)\right]$ Disclaimer: The expression of the question should be $x^{2}-2+\frac{1}{x^{2}}-y^{2} .$ The same has been done before solving the qu...
Read More →Factorise:
Question: Factorise: $x^{2}+\frac{1}{x^{2}}-3$ Solution: $x^{2}+\frac{1}{x^{2}}-3$ $=x^{2}+\frac{1}{x^{2}}-2-1$ $=\left[x^{2}+\left(\frac{1}{x}\right)^{2}-2 \times x \times \frac{1}{x}\right]-1$ $=\left(x-\frac{1}{x}\right)^{2}-1^{2} \quad\left[a^{2}-2 a b+b^{2}=(a-b)^{2}\right]$ $=\left(x-\frac{1}{x}+1\right)\left(x-\frac{1}{x}-1\right) \quad\left[a^{2}-b^{2}=(a-b)(a+b)\right]$...
Read More →Factorize:
Question: Factorize: $9 a^{2}+3 a-8 b-64 b^{2}$ Solution: $9 a^{2}+3 a-8 b-64 b^{2}=9 a^{2}-64 b^{2}+3 a-8 b$ $=(3 a)^{2}-(8 b)^{2}+(3 a-8 b)$ $=(3 a-8 b)(3 a+8 b)+1(3 a-8 b)$ $=(3 a-8 b)(3 a+8 b+1)$...
Read More →Solve the following
Question: If the equations $p x^{2}+2 q x+r=0$ and $q x^{2}-2 \sqrt{p r} x+q=0$ have real roots, then $q^{2}=$ ______________________ Solution: Given $p x^{2}+2 q x+r=0$ and $q x^{2}-2 \sqrt{p r} x+q=0$ have real roots for $p x^{2}+2 q x+r=0$ Since roots are real $\Rightarrow$ Discriminant $D \geq 0$ i.e. $\mathrm{b}^{2}-4 a c \geq 0$ i. e. $(2 q)^{2}-4(p)(r) \geq 0$ i. e. $4 q^{2}-4 p r \geq 0$ i. e. $q^{2} \geq p r$ ....(i) also for $q x^{2}-2 \sqrt{p r} x+q=0$ $D \geq 0$ i. e. $b^{2}-4 a c \g...
Read More →Factorize:
Question: Factorize: $4 x^{2}-9 y^{2}-2 x-3 y$ Solution: $4 x^{2}-9 y^{2}-2 x-3 y=\left(4 x^{2}-9 y^{2}\right)-(2 x+3 y)$ $=\left[(2 x)^{2}-(3 y)^{2}\right]-(2 x+3 y)$ $=(2 x-3 y)(2 x+3 y)-1(2 x+3 y)$ $=(2 x+3 y)(2 x-3 y-1)$...
Read More →Let f be any real function and let g be a function given by
Question: Let $f$ be any real function and let $g$ be a function given by $g(x)=2 x$. Prove that $g$ of $=f+f$. Solution: Given, $f: R \rightarrow R$ Since $\mathrm{g}(x)=2 x$ is a polynomial, $g: R \rightarrow R$ Clearly, gof $: R \rightarrow R$ and $f+f: R \rightarrow R$ So, domains of gof and $\mathrm{f}+\mathrm{f}$ are the same. $($ gof $)(x)=g(f(x))=2 f(x)$ $(f+f)(x)=f(x)+f(x)=2 f(x)$ $\Rightarrow(g o f)(x)=(f+f)(x), \forall x \in R$ Hence, $g o f=f+f$...
Read More →Let $f, g, h$ be real functions given by $f(x)=sin x, g(x)=2 x$ and $h(x)=cos x$. Prove that fog $=g circ(f h)$.
Question: Let $f, g, h$ be real functions given by $f(x)=\sin x, g(x)=2 x$ and $h(x)=\cos x$. Prove that fog $=g \circ(f h)$. Solution: We know that $f: R \rightarrow[-1,1]$ and $g: R \rightarrow R$ Clearly, the range of $g$ is a subset of the domain of $f$. $f o g: R \rightarrow R$ Now, $(f h)(x)=f(x) h(x)=(\sin x)(\cos x)=\frac{1}{2} \sin (2 x)$ Domain of $f h$ is $R$. Since range of $\sin x$ is $[-1,1]$, $-1 \leq \sin 2 x \leq 1$ $\Rightarrow \frac{-1}{2} \leq \frac{\sin x}{2} \leq \frac{1}{2...
Read More →The value of
Question: The value of $\sqrt{6+\sqrt{6+\sqrt{6+\ldots \ldots \text { to } \infty}}}$ is Solution: Let $\sqrt{6+\sqrt{6+\sqrt{6+-----\infty}}}=x$ $=\sqrt{6+x}=x$ i. e. $6+x=x^{2}$ (squaring both sides) i. e. $x^{2}-x-6=0$ i. e. $x^{2}-3 x+2 x-6=0$ i.e. $(x-3)(x+2)=0$ i.e. $x=3$ or $x=-2$ Since $x=-2$ is not possible (being negative) $\Rightarrow x=3$ i. e. $\sqrt{6+\sqrt{6+\sqrt{6+_{-----\infty}}}}=3$...
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