Factorise:
Question: Factorise: $x^{2}+2 x y+y^{2}-a^{2}+2 a b-b^{2}$ Solution: $x^{2}+2 x y+y^{2}-a^{2}+2 a b-b^{2}$ $=\left(x^{2}+2 x y+y^{2}\right)-\left(a^{2}-2 a b+b^{2}\right)$ $=(x+y)^{2}-(a-b)^{2} \quad\left[a^{2}+2 a b+b^{2}=(a+b)^{2}\right.$ and $\left.a^{2}-2 a b+b^{2}=(a-b)^{2}\right]$ $=[(x+y)+(a-b)][(x+y)-(a-b)] \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$ $=(x+y+a-b)(x+y-a+b)$...
Read More →The least value of k which makes the roots of the equation x
Question: The least value of $k$ which makes the roots of the equation $x^{2}+5 x+k=0$ imaginary is (a) 4 (b) 5 (c) 6 (d) 7 Solution: (d) 7 The roots of the quadratic equation $x^{2}+5 x+k=0$ will be imaginary if its discriminant is less than zero. $\therefore 25-4 k0$ $\Rightarrow k\frac{25}{4}$ Thus, the minimum integral value ofkfor which the roots are imaginary is 7....
Read More →The least value of k which makes the roots of the equation x
Question: The least value of $k$ which makes the roots of the equation $x^{2}+5 x+k=0$ imaginary is (a) 4 (b) 5 (c) 6 (d) 7 Solution: (d) 7 The roots of the quadratic equation $x^{2}+5 x+k=0$ will be imaginary if its discriminant is less than zero. $\therefore 25-4 k0$ $\Rightarrow k\frac{25}{4}$ Thus, the minimum integral value ofkfor which the roots are imaginary is 7....
Read More →If α, β are the roots of the equation x
Question: If $\alpha, \beta$ are the roots of the equation $x^{2}-p(x+1)-c=0$, then $(\alpha+1)(\beta+1)=$ (a)c (b)c 1 (c) 1 c (d) none of these Solution: (c) 1 c Given equation: $x^{2}-p(x+1)-c=0$ or $\quad x^{2}-p x-p-c=0$ Also $\alpha$ and $\beta$ are the roots of the equation. Sum of the roots $=\alpha+\beta=p$ Product of the roots $=\alpha \beta=-(c+p)$ Then, $(\alpha+1)(\beta+1)=\alpha \beta+\alpha+\beta+1$ $=-(c+p)+p+1$ $=-c-p+p+1$ $=1-c$...
Read More →If α, β are the roots of the equation x
Question: If $\alpha, \beta$ are the roots of the equation $x^{2}-p(x+1)-c=0$, then $(\alpha+1)(\beta+1)=$ (a)c (b)c 1 (c) 1 c (d) none of these Solution: (c) 1 c Given equation: $x^{2}-p(x+1)-c=0$ or $\quad x^{2}-p x-p-c=0$ Also $\alpha$ and $\beta$ are the roots of the equation. Sum of the roots $=\alpha+\beta=p$ Product of the roots $=\alpha \beta=-(c+p)$ Then, $(\alpha+1)(\beta+1)=\alpha \beta+\alpha+\beta+1$ $=-(c+p)+p+1$ $=-c-p+p+1$ $=1-c$...
Read More →If the difference of the roots of x
Question: If the difference of the roots of $x^{2}-p x+q=0$ is unity, then (a) $p^{2}+4 q=1$ (b) $p^{2}-4 q=1$ (c) $p^{2}+4 q^{2}=(1+2 q)^{2}$ (d) $4 p^{2}+q^{2}=(1+2 p)^{2}$ Solution: (b) $p^{2}-4 q=1$ Given equation: $x^{2}-p x+q=0$ Also $\alpha$ and $\beta$ are the roots of the equation such that $\alpha-\beta=1$. Sum of the roots $=\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}=-\left(\frac{-p}{1}\right)=p$ Product of the roots $=\alpha \beta=\frac{C \text { ...
Read More →If the difference of the roots of x
Question: If the difference of the roots of $x^{2}-p x+q=0$ is unity, then (a) $p^{2}+4 q=1$ (b) $p^{2}-4 q=1$ (c) $p^{2}+4 q^{2}=(1+2 q)^{2}$ (d) $4 p^{2}+q^{2}=(1+2 p)^{2}$ Solution: (b) $p^{2}-4 q=1$ Given equation: $x^{2}-p x+q=0$ Also $\alpha$ and $\beta$ are the roots of the equation such that $\alpha-\beta=1$. Sum of the roots $=\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}=-\left(\frac{-p}{1}\right)=p$ Product of the roots $=\alpha \beta=\frac{C \text { ...
Read More →Factorise:
Question: Factorise: $x^{2}+y^{2}-z^{2}-2 x y$ Solution: $x^{2}+y^{2}-z^{2}-2 x y$ $=\left(x^{2}+y^{2}-2 x y\right)-z^{2}$ $=(x-y)^{2}-z^{2} \quad\left[a^{2}-2 a b+b^{2}=(a-b)^{2}\right]$ $=(x-y+z)(x-y-z) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$...
Read More →If α, β are the roots of the equation x
Question: If $\alpha, \beta$ are the roots of the equation $x^{2}+p x+q=0$ then $-\frac{1}{\alpha}+\frac{1}{\beta}$ are the roots of the equation (a) $x^{2}-p x+q=0$ (b) $x^{2}+p x+q=0$ (c) $q x^{2}+p x+1=0$ (d) $q x^{2}-p x+1=0$ Solution: (d) $q x^{2}-p x+1=0$ Given equation: $x^{2}+p x+q=0$ Also, $\alpha$ and $\beta$ are the roots of the given equation. Then, sum of the roots $=\alpha+\beta=-p$ Product of the roots $=\alpha \beta=q$ Now, for roots $-\frac{1}{\alpha},-\frac{1}{\beta}$, we have:...
Read More →If α, β are the roots of the equation x
Question: If $\alpha, \beta$ are the roots of the equation $x^{2}+p x+q=0$ then $-\frac{1}{\alpha}+\frac{1}{\beta}$ are the roots of the equation (a) $x^{2}-p x+q=0$ (b) $x^{2}+p x+q=0$ (c) $q x^{2}+p x+1=0$ (d) $q x^{2}-p x+1=0$ Solution: (d) $q x^{2}-p x+1=0$ Given equation: $x^{2}+p x+q=0$ Also, $\alpha$ and $\beta$ are the roots of the given equation. Then, sum of the roots $=\alpha+\beta=-p$ Product of the roots $=\alpha \beta=q$ Now, for roots $-\frac{1}{\alpha},-\frac{1}{\beta}$, we have:...
Read More →Factorize:
Question: Factorize: $(a+b)^{3}-a-b$ Solution: $(a+b)^{3}-a-b=(a+b)^{3}-(a+b)$ $=(a+b)\left[(a+b)^{2}-1\right]$ $=(a+b)\left[(a+b)^{2}-1^{2}\right]$ $=(a+b)(a+b-1)(a+b+1)$...
Read More →If α and β are the roots of
Question: If $\alpha$ and $\beta$ are the roots of $4 x^{2}+3 x+7=0$, then the value of $\frac{1}{\alpha}+\frac{1}{\beta}$ is(a) $\frac{4}{7}$ (b) $-\frac{3}{7}$ (c) $\frac{3}{7}$ (d) $-\frac{3}{4}$ Solution: (b) $-\frac{3}{7}$ Given equation: $4 x^{2}+3 x+7=0$ Also, $\alpha$ and $\beta$ are the roots of the equation. Then, sum of the roots $=\alpha+\beta=\frac{-C \text { oefficient of } x}{C \text { oefficient of } x^{2}}=-\frac{3}{4}$ Product of the roots $=\alpha \beta=\frac{C \text { onstant...
Read More →The number of roots of the equation
Question: The number of roots of the equation $\frac{(x+2)(x-5)}{(x-3)(x+6)}=\frac{x-2}{x+4}$ is (a) 0 (b) 1 (c) 2 (d) 3 Solution: (b) 1 $\frac{(x+2)(x-5)}{(x-3)(x+6)}=\frac{(x-2)}{(x+4)}$ $\Rightarrow\left(x^{2}-3 x-10\right)(x+4)=\left(x^{2}+3 x-18\right)(x-2)$ $\Rightarrow x^{3}+4 x^{2}-3 x^{2}-12 x-10 x-40=x^{3}-2 x^{2}+3 x^{2}-6 x-18 x+36$ $\Rightarrow x^{2}-22 x-40=x^{2}-24 x+36$ $\Rightarrow 2 x=76$ $\Rightarrow x=38$ Hence, the equation has only 1 root....
Read More →The number of roots of the equation
Question: The number of roots of the equation $\frac{(x+2)(x-5)}{(x-3)(x+6)}=\frac{x-2}{x+4}$ is (a) 0 (b) 1 (c) 2 (d) 3 Solution: (b) 1 $\frac{(x+2)(x-5)}{(x-3)(x+6)}=\frac{(x-2)}{(x+4)}$ $\Rightarrow\left(x^{2}-3 x-10\right)(x+4)=\left(x^{2}+3 x-18\right)(x-2)$ $\Rightarrow x^{3}+4 x^{2}-3 x^{2}-12 x-10 x-40=x^{3}-2 x^{2}+3 x^{2}-6 x-18 x+36$ $\Rightarrow x^{2}-22 x-40=x^{2}-24 x+36$ $\Rightarrow 2 x=76$ $\Rightarrow x=38$ Hence, the equation has only 1 root....
Read More →Find fog and gof if
Question: (i) $f(x)=e^{x}, g(x)=\log _{e} x$ (ii) $f(x)=x^{2}, g(x)=\cos x$ (iii) $f(x)=|x|, g(x)=\sin x$ (iv) $f(x)=x+1, g(x)=e^{x}$ (v) $f(x)=\sin ^{-1} x, g(x)=x^{2}$ (vi) $f(x)=x+1, g(x)=\sin x$ (vii) $f(x)=x+1, g(x)=2 x+3$ (viii) $f(x)=c, c \in R, g(x)=\sin x^{2}$ (ix) $f(x)=x^{2}+2, g(x)=1-\frac{1}{1-x}$ Solution: (i) $f(x)=e^{x}, g(x)=\log _{e} x$ $f: R \rightarrow(0, \infty) ; g:(0, \infty) \rightarrow R$ Computingfog: Clearly,therangeofgisasubsetofthedomainoff. $f o g:(0, \infty) \right...
Read More →Factorize:
Question: Factorize: $108 a^{2}-3(b-c)^{2}$ Solution: $108 a^{2}-3(b-c)^{2}=3\left[36 a^{2}-(b-c)^{2}\right]$ $=3\left[(6 a)^{2}-(b-c)^{2}\right]$ $=3(6 a-b+c)(6 a+b-c)$...
Read More →The set of all values of m for which both the roots of the equation x
Question: The set of all values of $m$ for which both the roots of the equation $x^{2}-(m+1) x+m+4=0$ are real and negative, is (a) $(-\infty,-3] \cup[5, \infty)$ (b) $[-3,5]$ (c) $(-4,-3]$ (d) $(-3,-1]$ Solution: (c) $(-4,-3]$ The roots of the quadratic equation $x^{2}-(m+1) x+m+4=0$ will be real, if its discriminant is greater than or equal to zero. $\therefore(m+1)^{2}-4(m+4) \geq 0$ $\Rightarrow(m-5)(m+3) \geq 0$ $\Rightarrow m \leq-3$ or $m \geq 5 \quad \ldots(1)$ It is also given that, the...
Read More →Factorize:
Question: Factorize: $a^{2}+2 a b+b^{2}-9 c^{2}$ Solution: $a^{2}+2 a b+b^{2}-9 c^{2}=(a+b)^{2}-(3 c)^{2}$ $=(a+b-3 c)(a+b+3 c)$...
Read More →Factorise:
Question: Factorise: $4 a^{2}-4 b^{2}+4 a+1$ Solution: $4 a^{2}-4 b^{2}+4 a+1$ $=\left(4 a^{2}+4 a+1\right)-4 b^{2}$ $=\left[(2 a)^{2}+2 \times 2 a \times 1+1^{2}\right]-4 b^{2}$ $=(2 a+1)^{2}-(2 b)^{2} \quad\left[a^{2}+2 a b+b^{2}=(a+b)^{2}\right]$ $=[(2 a+1)+2 b][(2 a+1)-2 b] \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$ $=(2 a+1+2 b)(2 a+1-2 b)$ $=(2 a+2 b+1)(2 a-2 b+1)$...
Read More →The value of p and q (p ≠ 0, q ≠ 0) for which p, q are the roots of the equation
Question: The value of $p$ and $q(p \neq 0, q \neq 0)$ for which $p, q$ are the roots of the equation $x^{2}+p x+q=0$ are (a)p= 1,q= 2 (b)p= 1,q= 2 (c)p= 1,q= 2 (d)p= 1,q= 2 Solution: (a)p= 1,q = 2 It is given that, $p$ and $q(p \neq 0, q \neq 0)$ are the roots of the equation $x^{2}+p x+q=0$. $\therefore$ Sum of roots $=p+q=-p$ $\Rightarrow 2 p+q=0 \quad \ldots(1)$ Product of roots $=p q=q$ $\Rightarrow q(p-1)=0$ $\Rightarrow p=1, q=0$ but $q \neq 0$ Now, substitutingp= 1 in (1), we get, $2+q=0...
Read More →The value of p and q (p ≠ 0, q ≠ 0) for which p, q are the roots of the equation
Question: The value of $p$ and $q(p \neq 0, q \neq 0)$ for which $p, q$ are the roots of the equation $x^{2}+p x+q=0$ are (a)p= 1,q= 2 (b)p= 1,q= 2 (c)p= 1,q= 2 (d)p= 1,q= 2 Solution: (a)p= 1,q = 2 It is given that, $p$ and $q(p \neq 0, q \neq 0)$ are the roots of the equation $x^{2}+p x+q=0$. $\therefore$ Sum of roots $=p+q=-p$ $\Rightarrow 2 p+q=0 \quad \ldots(1)$ Product of roots $=p q=q$ $\Rightarrow q(p-1)=0$ $\Rightarrow p=1, q=0$ but $q \neq 0$ Now, substitutingp= 1 in (1), we get, $2+q=0...
Read More →If one root of the equation x
Question: If one root of the equation $x^{2}+p x+12=0$ is 4 , while the equation $x^{2}+p x+q=0$ has equal roots, the value of $q$ is (a) 49/4 (b) 4/49 (c) 4 (d) none of these Solution: (a) 49/4 It is given that, 4 is the root of the equation $x^{2}+p x+12=0$. $\therefore 16+4 p+12=0$ $\Rightarrow p=-7$ It is also given that, the equation $x^{2}+p x+q=0$ has equal roots. So, the discriminant of $x^{2}+p x+q=0$ will be zero. $\therefore p^{2}-4 q=0$ $\Rightarrow 4 q=(-7)^{2}=49$ $\Rightarrow q=\f...
Read More →Factorise:
Question: Factorise: $a^{2}-b^{2}+2 b c-c^{2}$ Solution: $a^{2}-b^{2}+2 b c-c^{2}$ $=a^{2}-\left(b^{2}-2 b c+c^{2}\right)$ $=a^{2}-(b-c)^{2} \quad\left[a^{2}-2 a b+b^{2}=(a-b)^{2}\right]$ $=[a+(b-c)][a-(b-c)] \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$ $=(a+b-c)(a-b+c)$...
Read More →If one root of the equation x
Question: If one root of the equation $x^{2}+p x+12=0$ is 4 , while the equation $x^{2}+p x+q=0$ has equal roots, the value of $q$ is (a) 49/4 (b) 4/49 (c) 4 (d) none of these Solution: (a) 49/4 It is given that, 4 is the root of the equation $x^{2}+p x+12=0$. $\therefore 16+4 p+12=0$ $\Rightarrow p=-7$ It is also given that, the equation $x^{2}+p x+q=0$ has equal roots. So, the discriminant of $x^{2}+p x+q=0$ will be zero. $\therefore p^{2}-4 q=0$ $\Rightarrow 4 q=(-7)^{2}=49$ $\Rightarrow q=\f...
Read More →If the equations x
Question: If the equations $x^{2}+2 x+3 \lambda=0$ and $2 x^{2}+3 x+5 \lambda=0$ have a non-zero common roots, then $\lambda=$ (a) 1 (b) 1 (c) 3 (d) none of these. Solution: (b) 1 Let $\alpha$ be the common roots of the equations, $x^{2}+2 x+3 \lambda=0$ and $2 x^{2}+3 x+5 \lambda=0$ Therefore, $\alpha^{2}+2 \alpha+3 \lambda=0$ ....(1) $2 \alpha^{2}+3 \alpha+5 \lambda=0$ ....(2) Solving (1) and (2) by cross multiplication, we get $\frac{\alpha^{2}}{10 \lambda-9 \lambda}=\frac{\alpha}{6 \lambda-5...
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