Factorize:
Question: Factorize: $x^{2}-y^{2}+6 y-9$ Solution: $x^{2}-y^{2}+6 y-9=x^{2}-\left(y^{2}-6 y+9\right)$ $=x^{2}-\left(y^{2}-2 \times y \times 3+3^{2}\right)$ $=x^{2}-(y-3)^{2}$ $=[x+(y-3)][x-(y-3)]$ $=(x+y-3)(x-y+3)$...
Read More →Factorise:
Question: Factorise: $9 a^{2}+6 a+1-36 b^{2}$ Solution: $9 a^{2}+6 a+1-36 b^{2}$ $=\left[(3 a)^{2}+2 \times 3 a \times 1+1^{2}\right]-(6 b)^{2}$ $=(3 a+1)^{2}-(6 b)^{2} \quad\left[a^{2}+2 a b+b^{2}=(a+b)^{2}\right]$ $=(3 a+1-6 b)(3 a+1+6 b) \quad\left[a^{2}-b^{2}=(a-b)(a+b)\right]$ $=(3 a-6 b+1)(3 a+6 b+1)$...
Read More →If α, β are roots of the equation x
Question: If , are roots of the equationx2+x+ 1 = 0, then the equation whose roots are 19and 7is ____________. Solution: Given : $\alpha$ and $\beta$ are roots of $x^{2}+x+1=0$ i. e. $\alpha^{2}+\alpha+1=0$ and $\beta^{2}+\beta+1=0$ Since $x^{3}-1=(x-1)\left(x^{2}+x+1\right)$ $\Rightarrow \alpha^{3}-1=(\alpha-1)\left(\alpha^{2}+\alpha+1\right)$ and $\beta^{3}-1=(\beta-1)\left(\beta^{2}+\beta+1\right)$ $\Rightarrow \alpha^{3}-1=0$ and $\beta^{3}-1=0$ i.e. $\alpha^{3}=1$ and $\beta^{3}=1$ Now, $\a...
Read More →If α, β are roots of the equation x
Question: If , are roots of the equationx2+x+ 1 = 0, then the equation whose roots are 19and 7is ____________. Solution: Given : $\alpha$ and $\beta$ are roots of $x^{2}+x+1=0$ i. e. $\alpha^{2}+\alpha+1=0$ and $\beta^{2}+\beta+1=0$ Since $x^{3}-1=(x-1)\left(x^{2}+x+1\right)$ $\Rightarrow \alpha^{3}-1=(\alpha-1)\left(\alpha^{2}+\alpha+1\right)$ and $\beta^{3}-1=(\beta-1)\left(\beta^{2}+\beta+1\right)$ $\Rightarrow \alpha^{3}-1=0$ and $\beta^{3}-1=0$ i.e. $\alpha^{3}=1$ and $\beta^{3}=1$ Now, $\a...
Read More →If f(x) = sin x and g(x) = 2x be two real functions, then describe gof and fog. Are these equal functions?
Question: Iff(x) = sinxandg(x) = 2xbe two real functions, then describegofandfog.Are these equal functions? Solution: We know that $f: R \rightarrow[-1,1]$ and $g: R \rightarrow R$ Clearly, the range of $f$ is a subset of the domain of $g$. gof $: R \rightarrow R$ $(g o f)(x)=g(f(x))$ $=g(\sin x)$ $=2 \sin x$ Clearly, the range of $g$ is a subset of the domain of $f$. $f o g: R \rightarrow R$ So, $(f o g)(x)=f(g(x))$ $=f(2 x)$ $=\sin (2 x)$ Clearly,foggof...
Read More →Factorise:
Question: Factorise: $1+2 a b-\left(a^{2}+b^{2}\right)$ Solution: $1+2 a b-\left(a^{2}+b^{2}\right)$ $=1+2 a b-a^{2}-b^{2}$ $=1-a^{2}+2 a b-b^{2}$ $=1^{2}-\left(a^{2}-2 a b+b^{2}\right)$ $=1^{2}-(a-b)^{2} \quad\left[a^{2}-2 a b+b^{2}=(a-b)^{2}\right]$ $=[1+(a-b)][1-(a-b)] \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$ $=(1+a-b)(1-a+b)$...
Read More →If one root of the equation
Question: If one root of the equationx2+px+ 12 = 0 is 4, then the sum of the roots is ____________. Solution: One root of $x^{2}+p x+12=0$ is 4 Let us suppose the other solution is $z$. $\therefore$ Sum of roots $=\frac{-p}{1}=-p$ i. e. $z+4=-p$ and product of roots is 12 i. e. $4 z=12$ i. e. $z=3$ $\Rightarrow$ sum of roots is $4+3=7$...
Read More →If one root of the equation
Question: If one root of the equationx2+px+ 12 = 0 is 4, then the sum of the roots is ____________. Solution: One root of $x^{2}+p x+12=0$ is 4 Let us suppose the other solution is $z$. $\therefore$ Sum of roots $=\frac{-p}{1}=-p$ i. e. $z+4=-p$ and product of roots is 12 i. e. $4 z=12$ i. e. $z=3$ $\Rightarrow$ sum of roots is $4+3=7$...
Read More →Factorize:
Question: Factorize: $x^{3}-5 x^{2}-x+5$ Solution: $x^{3}-5 x^{2}-x+5=x^{2}(x-5)-1(x-5)$ $=(x-5)\left(x^{2}-1\right)$ $=(x-5)\left(x^{2}-1^{2}\right)$ $=(x-5)(x-1)(x+1)$...
Read More →The number of real roots of the equation
Question: The number of real roots of the equationx2+ 5 |x| + 4 = 0 is _________. Solution: $x^{2}+5|x|+4=0$ for $|x|$, we have 2 cases case (i) if $x0$ Quadratic equation is $x^{2}+5 x+4=0$ i. e. $x^{2}+4 x+x+4=0$ i. e. $(x+4)(x+1)=0$ i.e. $x=-1$ or $-4$ which is not possible since $x0$ case (ii) for $x0$, Quadratic equation is $x^{2}-5 x+4=0$ i. e. $x^{2}-4 x-x+4=0$ i.e. $(x-4)(x-1)=0$ i.e. $x=1$ or $x=4$ which is not possible, since $x$ is negative in this case Hence, no real solution exist f...
Read More →The number of real roots of the equation
Question: The number of real roots of the equationx2+ 5 |x| + 4 = 0 is _________. Solution: $x^{2}+5|x|+4=0$ for $|x|$, we have 2 cases case (i) if $x0$ Quadratic equation is $x^{2}+5 x+4=0$ i. e. $x^{2}+4 x+x+4=0$ i. e. $(x+4)(x+1)=0$ i.e. $x=-1$ or $-4$ which is not possible since $x0$ case (ii) for $x0$, Quadratic equation is $x^{2}-5 x+4=0$ i. e. $x^{2}-4 x-x+4=0$ i.e. $(x-4)(x-1)=0$ i.e. $x=1$ or $x=4$ which is not possible, since $x$ is negative in this case Hence, no real solution exist f...
Read More →Factorize:
Question: Factorize: $9-a^{2}+2 a b-b^{2}$ Solution: $9-a^{2}+2 a b-b^{2}=9-\left(a^{2}-2 a b+b^{2}\right)$ $=3^{2}-(a-b)^{2}$ $=[3-(a-b)][3+(a-b)]$ $=(3-a+b)(3+a-b)$...
Read More →If f(x) = 2x + 5 and
Question: If $f(x)=2 x+5$ and $g(x)=x^{2}+1$ be two real functions, then describe each of the following functions: (i)fog(ii)gof(iii)fof(iv)f2 Also, show that $f \circ f \neq f^{2}$ Solution: f(x)andg(x)are polynomials. $\Rightarrow f: R \rightarrow R$ and $g: R \rightarrow R$. So, $f \circ g: R \rightarrow R$ and $g \circ f: R \rightarrow R$. (i) $(f \circ g)(x)=f(g(x))$ $=f\left(x^{2}+1\right)$ $=2\left(x^{2}+1\right)+5$ $=2 x^{2}+2+5$ $=2 x^{2}+7$ (ii) $(g \circ f)(x)=g(f(x))$ $=g(2 x+5)$ $=(...
Read More →If the equation
Question: If the equation 2x2kx+x+ 8 = 0 has real and equal roots, thenk= __________. Solution: for $2 x^{2}-k x+x+8=0$ given; roots one real and equal i.e. $D=0$ (Discriminat) i. e. $b^{2}-4 a c=0$ Here $b=-k+1$ $a=2$ $c=8$ with reference to standard equations $a x^{2}+b x+c=0$ $\Rightarrow(1-K)^{2}-4(2) 8=0$ i.e. $K^{2}+1-2 K-64=0$ i.e. $K^{2}-2 K-63=0$ i. e. $K^{2}-9 K+7 K-63=0$ i.e. $(K-9)(k+7)=0$ i.e. $K=9$ or $-7$...
Read More →If the equation
Question: If the equation 2x2kx+x+ 8 = 0 has real and equal roots, thenk= __________. Solution: for $2 x^{2}-k x+x+8=0$ given; roots one real and equal i.e. $D=0$ (Discriminat) i. e. $b^{2}-4 a c=0$ Here $b=-k+1$ $a=2$ $c=8$ with reference to standard equations $a x^{2}+b x+c=0$ $\Rightarrow(1-K)^{2}-4(2) 8=0$ i.e. $K^{2}+1-2 K-64=0$ i.e. $K^{2}-2 K-63=0$ i. e. $K^{2}-9 K+7 K-63=0$ i.e. $(K-9)(k+7)=0$ i.e. $K=9$ or $-7$...
Read More →Factorize:
Question: Factorize: $a^{2}-b^{2}-4 a c+4 c^{2}$ Solution: $a^{2}-b^{2}-4 a c+4 c^{2}=\left(a^{2}-4 a c+4 c^{2}\right)-b^{2}$ $=a^{2}-2 \times 2 a \times c+(2 c)^{2}-b^{2}$ $=(a-2 c)^{2}-b^{2}$ $=(a-2 c+b)(a-2 c-b)$...
Read More →If the difference of the roots of the equation x
Question: If the difference of the roots of the equationx2Px+ 8 = 0 is 2, thenP=___________. Solution: For $x^{2}-P x+8=0$ Let us suppose two roots are $z_{1}$ and $z_{2}$ given : $z_{1}-z_{2}=2$ also $z_{1} z_{2}=8$n (product of roots) i. e. $\left(z_{2}+2\right) Z_{2}=8$ i. e. $z_{2}^{2}+2 z_{2}=8$ i.e. $z_{2}^{2}+2 z_{2}-8=0$ i. e. $z_{2}^{2}+4 z_{2}-2 z_{2}-8=0$ i. e. $\left(z_{2}+4\right)\left(z_{2}-2\right)=0$ i.e. $z_{2}=-4$ or $z_{2}=2$ $\Rightarrow z_{1}=-2$ or $z_{2}=4$ Since $P=z_{1}+...
Read More →If the difference of the roots of the equation x
Question: If the difference of the roots of the equationx2Px+ 8 = 0 is 2, thenP=___________. Solution: For $x^{2}-P x+8=0$ Let us suppose two roots are $z_{1}$ and $z_{2}$ given : $z_{1}-z_{2}=2$ also $z_{1} z_{2}=8$n (product of roots) i. e. $\left(z_{2}+2\right) Z_{2}=8$ i. e. $z_{2}^{2}+2 z_{2}=8$ i.e. $z_{2}^{2}+2 z_{2}-8=0$ i. e. $z_{2}^{2}+4 z_{2}-2 z_{2}-8=0$ i. e. $\left(z_{2}+4\right)\left(z_{2}-2\right)=0$ i.e. $z_{2}=-4$ or $z_{2}=2$ $\Rightarrow z_{1}=-2$ or $z_{2}=4$ Since $P=z_{1}+...
Read More →Factorize:
Question: Factorize: $a-b-a^{2}+b^{2}$ Solution: $a-b-a^{2}+b^{2}=(a-b)-\left(a^{2}-b^{2}\right)$ $=(a-b)-(a-b)(a+b)$ $=(a-b)[1-(a+b)]$ $=(a-b)(1-a-b)$...
Read More →If 1 – i is a root of the equation x
Question: If 1 iis a root of the equationx2+ax+b= 0, wherea,bR, then the values ofaandbare ____________. Solution: for equation $x^{2}+a x+b=0$ one root is $1-i$. Let us suppose other root is $p+i q$ when $p, q \in \mathbb{R}$ Sum of roots of quadratic equations is $=-a$ i.e. $(1-i)+(p+i q)=-a$ i. e. $1+p+i(-1+q)=-a$ since $a$ is real (given) $\Rightarrow-1+q=0$ (i. e. imaginary part is zero) i.e. $q=1$ also product of roots $=b$ $\Rightarrow(1-i)(p+i q)=b$ i.e. $p-i^{2} q-i p+i q=b$ i. e. $p+q+...
Read More →If f(x) = |x|, prove that fof = f.
Question: If $f(x)=|x|$, prove that $f \circ f=f$ Solution: Domains offandfofare same asR. $(f o f)(x)=f(f(x))=f(|x|)=|| x||=|x|=f(x)$ So, $(f o f)(x)=f(x), \forall x \in R$ Hence, $f o f=f$...
Read More →If 1 – i is a root of the equation x
Question: If 1 iis a root of the equationx2+ax+b= 0, wherea,bR, then the values ofaandbare ____________. Solution: for equation $x^{2}+a x+b=0$ one root is $1-i$. Let us suppose other root is $p+i q$ when $p, q \in \mathbb{R}$ Sum of roots of quadratic equations is $=-a$ i.e. $(1-i)+(p+i q)=-a$ i. e. $1+p+i(-1+q)=-a$ since $a$ is real (given) $\Rightarrow-1+q=0$ (i. e. imaginary part is zero) i.e. $q=1$ also product of roots $=b$ $\Rightarrow(1-i)(p+i q)=b$ i.e. $p-i^{2} q-i p+i q=b$ i. e. $p+q+...
Read More →Factorise:
Question: Factorise: $25 x^{2}-10 x+1-36 y^{2}$ Solution: $25 x^{2}-10 x+1-36 y^{2}$ $=\left[(5 x)^{2}-2 \times 5 x \times 1+1^{2}\right]-(6 y)^{2}$ $=(5 x-1)^{2}-(6 y)^{2} \quad\left[a^{2}-2 a b+b^{2}=(a-b)^{2}\right]$ $=[(5 x-1+6 y)][(5 x-1-6 y)] \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$ $=(5 x+6 y-1)(5 x-6 y-1)$...
Read More →The equation of the smallest degree with real coefficients having 1 + i as one of the roots is
Question: The equation of the smallest degree with real coefficients having 1 +ias one of the roots is (a) $x^{2}+x+1=0$ (b) $x^{2}-2 x+2=0$ (c) $x^{2}+2 x+2=0$ (d) $x^{2}+2 x-2=0$ Solution: (b) $x^{2}-2 x+2=0$ We know that, imaginary roots of a quadratic equation occur in conjugate pair. It is given that, $1+i$ is one of the roots. So, the other root will be $1-i$. Thus, the quadratic equation having roots $1+i$ and $1-i$ is, $x^{2}-(1+i+1-i) x+(1+i)(1-i)=0$ $\Rightarrow x^{2}-2 x+2=0$...
Read More →The equation of the smallest degree with real coefficients having 1 + i as one of the roots is
Question: The equation of the smallest degree with real coefficients having 1 +ias one of the roots is (a) $x^{2}+x+1=0$ (b) $x^{2}-2 x+2=0$ (c) $x^{2}+2 x+2=0$ (d) $x^{2}+2 x-2=0$ Solution: (b) $x^{2}-2 x+2=0$ We know that, imaginary roots of a quadratic equation occur in conjugate pair. It is given that, $1+i$ is one of the roots. So, the other root will be $1-i$. Thus, the quadratic equation having roots $1+i$ and $1-i$ is, $x^{2}-(1+i+1-i) x+(1+i)(1-i)=0$ $\Rightarrow x^{2}-2 x+2=0$...
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