Maths M.C.Q (1 Marks)

The given quadric equation is $ax^2+ bx + c = 0,$ and $\sin\alpha$ and $\cos\beta$ are roots of given equation.
And, $a = a, b = b$ and $c = c$
Then, as we know that sum of the roots
$\sin\alpha+\cos\beta=\frac{-\text{b}}{\text{a}}\ ...(\text{i})$
And the product of the roots
$​​\sin\alpha\cdot\cos\beta=\frac{\text{c}}{\text{a}}\ ....(\text{ii})$
Squaring both sides of equation (i) we get
$(\sin\alpha+\cos\beta)^2=\Big(\frac{-\text{b}}{\text{a}}\Big)^2$
$\sin^2\alpha+\cos^2\beta+2\sin\alpha\cos\beta=\frac{\text{b}^2}{\text{a}^2}$
Putting the value of $\sin\alpha+\cos\beta=1$ we get
$1+2\sin\alpha\cos\beta=\frac{\text{b}^2}{\text{a}^2}$
$\text{a}^2(1+2\sin\alpha\cos\beta)=\text{b}^2$
Putting the value of $​​\sin\alpha\cdot\cos\beta=\frac{\text{c}}{\text{a}}$ we get
$\text{a}^2\Big(1+2\frac{\text{c}}{\text{a}}\Big)=\text{b}^2$
$\text{a}^2\Big(\frac{\text{a}+2\text{c}}{\text{a}}\Big)=\text{b}^2$
$\text{a}^2+2\text{ac}=\text{b}^2$
Threfore, the value of $b^2= a^2+ 2ac$
Thus, the correct answer is $(b)$

In the equation $a x^2+2 x+a=0$
$ \Rightarrow D=b^2-4 a c $
$ \Rightarrow D=(2)^2-4 \times a \times a $
$ \Rightarrow D=4-4 a^2$
Roots are real and equal
$ \Rightarrow D=0 $
$ \Rightarrow 4-4 a^2=0 $
$ \Rightarrow 4=4 a^2 $
$ \Rightarrow 1=a^2 $
$ \Rightarrow a^2=1$
$\Rightarrow\text{a}^2=(\pm1)^2$
$\Rightarrow\text{a}=\pm1$

The given quadratic equation $16 \mathrm{x}^2+4 \mathrm{kx}+9=0$, has equal roots.
Here, $a = 16, b = 4k$ and $c = 9$
As we know that $D=b^2-4 a c$
Putting the value of $a = 16, b = 4k$ and $c = 9$
$\Rightarrow D=(4 k)^2-4(16)(9)$
$\Rightarrow D=16 k^2-576$
The given equation will have real and equal roots, if $D = 0$
Thus, $16 \mathrm{k}^2-576=0$
$\Rightarrow \mathrm{k}^2-36=0 $
$ \Rightarrow(\mathrm{k}+6)(\mathrm{k}-6)=0$
$ \Rightarrow \mathrm{k}+6=0 \text { or } \mathrm{k}=6$
Therefore, the value of $k$ is $6, -6$
Hence, the correct option is $(c)$

The given quadric equation is $x^2- ax + 1 = 0,$ and roots are dostinct.
Then fond the value of $a$.
Here, $a = 1, b = a$ and $c = 1$
As we know that $D = b^2- 4ac$
Putting the value of $a = 1, b = a$ and $c = 1$
$= (a)^2- 4 × 1 × 1$
$= a^2- 4$
The given equation will have real and distinct roots, if $D > 0$
$a^2- 4 > 0$
$a^2 > 4$
$\text{a}>\sqrt{4}$
$\text{a}>\pm2$
Therefore, the value of $|a| > 2$
Thus, the correct answer is $(c)$

Quad equation is $ax^2+ bx + c = 0$
Let first root $=\alpha,$ Then
Second root $=3\alpha$
$\therefore$ Sum of root $=\alpha+3\alpha=\frac{-\text{b}}{\text{a}}$
$\Rightarrow4\alpha=\frac{-\text{b}}{\text{a}}$
$\Rightarrow\alpha=\frac{-\text{b}}{4\text{a}}\ ....(\text{i})$
and produt of roots $=\alpha\times3\alpha=\frac{\text{c}}{\text{a}}$
$\Rightarrow3\alpha^2=\frac{\text{c}}{\text{a}}$
$\Rightarrow\alpha^2=\frac{\text{c}}{3\text{a}}$
$\Rightarrow\Big(\frac{-\text{b}}{4\text{a}}\Big)^2=\frac{\text{c}}{3\text{a}}$ [From (i)]
$\Rightarrow\frac{\text{b}^2}{16\text{a}^2}=\frac{\text{c}}{3\text{a}}$ (Dividing by a)
$\Rightarrow\frac{\text{b}^2}{16\text{a}}=\frac{\text{c}}{3}$
$\Rightarrow\frac{\text{b}^2}{\text{ac}}=\frac{16}{3}$
$\Rightarrow\text{b}^2:\text{ac}=16:3$

The given quadric equation is $\left(a^2+b^2\right) x^2-2 b(a+c) x+\left(b^2+c^2\right)=0$, and roots equal.
Here, $a=\left(a^2+b^2\right), b=-2 b(a+c)$ and $c=b^2+c^2$
As we know that $D=b^2-4 a c$
Putting the value of $a=\left(a^2+b^2\right), b=-2 b(a+c)$ and $c=b^2+c^2$
$ =\{-2 b(a+c)\}^2-4 \times\left(a^2+b^2\right) \times\left(b^2+c^2\right) $
$ =4 a^2 b^2+4 b^2 c^2+8 a b^2 c-4\left(a^2 b^2+a^2 c^2+b^4+b^2 c^2\right) $
$ =4 a^2 b^2+4 b^2 c^2+8 a b^2 c-4 a^2 b^2-4 a^2 c^2-4 b^4-4 b^2 c^2 $
$ =+8 a b^2 c-4 a^2 c^2-4 b^4 $
$ =-4\left(a^2 c^2+b^4-2 a b^2 c\right)$
The given equation will have equal roots, if $D=0$
$ -4\left(a^2 c^2+b^4-2 a b^2 c\right)=0 $
$ a^2 c^2+b^4-2 a b^2 c=0 $
$ \left(a c-b^2\right)^2=0 $
$ a c-b^2=0 $
$ a c=b^2$
Thus, the correct answer is $(b)$

$ k x^2+6 x+4 k=0 $
$ \text { Here } a=k, b=6, c=4 k $
$ \Rightarrow D=b^2-4 a c $
$ \Rightarrow D=(6)^2-4 \times k \times 4 k $
$ \Rightarrow D=36-16 k^2$
$\text { Roots are equal }$
$ \Rightarrow D=0 $
$ \Rightarrow 36-16 \mathrm{k}^2=0 $
$ \Rightarrow 16 \mathrm{k}^2=36$
$\Rightarrow\text{k}^2=\frac{36}{16}$
$\Rightarrow\text{k}^2=\Big(\frac{6}{4}\Big)^2$
$\Rightarrow\text{k}=\frac{6}{4}$
$\Rightarrow\text{k}=\frac{3}{2}$

$a$ and $b$ are the roots of the equation $x^2+ ax + b = 0$
Sum of roots $= -a$ and product of roots $= b$
Now $a + b = -a$
and $ab = b$
$\Rightarrow a = 1 ....(i)$
$\Rightarrow 2a + b = 0$
$\Rightarrow 2 \times 1 + b = 0$
$\Rightarrow b = -2$
Now $a + b = 1 - 2 = -1$

The given quadric equation is $x^2-(k+6) x+2(2 k-1)=0$, and roots are equal
Then find the value of $k$.
Let $\alpha$ and $\beta$ be two roots of given equation
And, $a = 1, b = -(k + 6)$ and of the roots
$\alpha+\beta=\frac{-\text{b}}{\text{a}}$
$\alpha+\beta=\frac{-\{-(\text{k}+6)\}}{1}$
$\alpha+\beta=(\text{k}+6)$
And the product of the roots
$\alpha\cdot\beta=\frac{\text{c}}{\text{a}}$
$\alpha\beta=\frac{2(2\text{k}-1)}{1}$
$\alpha\beta=2(2\text{k}-1)$
According to question, sum of the roots $=\frac{1}{2}\times$ product of the roots
$(\text{k}+6)=\frac{1}{2}\times2(2\text{k}-1)$
$\text{k}+6=2\text{k}-1$
$6+1=2\text{k}-\text{k}$
$7=\text{k}$
Therefore, the value of $k = 7$
Thus, the correct answer is $(b)$

$\Rightarrow\text{x}^2-\text{x}=\lambda(2\text{x}-1)$
$\Rightarrow\text{x}^2-\text{x}=2\lambda\text{x}-\lambda$
$\Rightarrow\text{x}^2-\text{x}-2\lambda\text{x}+\lambda=0$
$\Rightarrow\text{x}^2-(1+2\lambda)\text{x}+\lambda=0$
Sum of roots $=\frac{-\text{b}}{\text{a}}$
$=\frac{1+2\lambda}{1}$
$\frac{1+2\lambda}{1}=0$
$\Rightarrow2\lambda=-1$
$\lambda=-\frac{1}{2}$

1. $b = 4$ and $a = 1, 2, 3, 4$
2. $b = 3$ and $a = 1, 2$
3. $b = 2$ and $a = 1$

In the equation
$ \Rightarrow\left(a^2+b^2\right) x^2-2(a c+b d) x+\left(c^2+d^2\right)=0 $
$ \Rightarrow D=B^2-4 A C $
$ \Rightarrow D=[-2(a c+b d)]^2-4\left(a^2+b^2\right)\left(c^2+d^2\right) $
$ \Rightarrow D=4\left[a^2 c^2+b 2 d^2+2 a b c d\right]-4\left[a^2 c^2+a^2 d^2+b^2 c^2+b^2 d^2\right] $
$ \Rightarrow D=4 a^2 c^2+4 b^2 d^2+8 a b c d-4 a^2 c^2-4 a^2 d^2-4 b^2 c^2-4 b^2 d^2 $
$ \Rightarrow D=8 a b c d-4 a^2 d 2-4 b^2 c^2 $
$ \Rightarrow D=-4\left[a^2 d^2+b^2 c^2-2 a b c d\right] $
$ \Rightarrow D=-4(a d-b c)^2$
$\because$ Roots are equal
$\therefore D = 0$
$\Rightarrow -4(ad - bc)^2= 0$
$\Rightarrow ad - bc = 0$
$\Rightarrow ad = bc$

The given quadric equation is $\left(a^2+b^2\right) x^2+2(a b+b d) x+c^2+d^2=0$, and roots are equal. Here, $a=\left(a^2+b^2\right), b=2(a b+b d)$ and, $c=c^2+d^2$
As we know that $D=b^2-4 a c$
Putting the value of $a=\left(a^2+b^2\right), b=2(a b+b d)$ and, $c=c^2+d^2$
$ =\{2(a b+b d)\}^2-4 \times\left(a^2+b^2\right) \times\left(c^2+d^2\right) $
$ =4 a^2 b^2+4 b^2 d^2+8 a b^2 d-4\left(a^2 c^2+a^2 d^2+b^2 c^2+b^2 d^2\right) $
$ =4 a^2 b^2+4 b^2 d^2+8 a b^2 d-4 a^2 c^2-4 a^2 d^2-4 b^2 c^2-4 a^2 d^2 $
$ =4 a^2 b^2+8 a b^2 d-4 a^2 c^2-4 a^2 d^2-4 b^2 c^2 $
$ =4\left(a^2 b^2+2 a b^2 d-a^2 c^2-a^2 d^2-b^2 c^2\right)$
The given equation will have no real roots, if $\mathrm{D}<0$
$ 4\left(a^2 b^2+2 a b^2 d-a^2 c^2-a^2 d^2-b^2 c^2\right)<0 $
$ a^2 b^2+2 a b^2 d-a^2 c^2-a^2 d^2-b^2 c^2<0$
$\text{ad}\neq\text{bc}$
Thus, the correct answer is $(d)$

Let $\alpha$ and $\beta$ be the roots of quadratic equation $2x^2+ kx + 4 = 0$ in such a way that $a = 2$
Here, $a = 2, b = k$ and $c = 4$
Then, according to question sum of the roots
$\alpha+\beta=\frac{-\text{b}}{\text{a}}$
$2+\beta=\frac{-\text{k}}{2}$
$\beta=\frac{-\text{k}}{2}-2$
$\beta=\frac{-\text{k}-4}{2}$
And the product of the roots
$\alpha\cdot\beta=\frac{\text{c}}{\text{a}}$
$\alpha\cdot\beta=\frac{4}{2}$
$\alpha\cdot\beta=2$
Putting the value $\beta=\frac{-\text{k}-4}{2}$ in above
$2\times\frac{(-\text{k}-4)}{2}=2$
$(-\text{k}-4)=2$
$\text{k}=-4-2$
$\text{k}=-6$
Putting the value of k in $\beta=\frac{-\text{k}-4}{2}$
$\beta=\frac{-(-6)-4}{2}$
$\beta=\frac{6-4}{2}$
$\beta=\frac{2}{2}$
$\beta=1$
Therefore, value of other root be $\beta=1$
Thus, the correct answer is $(d)$

The given quadric equation is $x^2+k(4 x+k-1)+2=0$, and roots are equal Then find the value of $k$.
$ x^2+k(4 x+k-1)+2=0 $
$ x^2+4 k x+\left(k^2-k+2\right)=0$
Here, $a=1, b=4 k$ and $c=k^2-k+2$
As we know that $D=b^2-4 a c$
Putting the value of $a=1, b=4 k$ and $c=k^2-k+2$
$ =(4 k)^2-4 \times 1 \times\left(k^2-k+2\right) $
$ =16 k^2-4 k^2+4 k-8 $
$ =12 k^2+4 k-8 $
$ =4\left(3 k^2+k-2\right)$
The given equation will have real and distinct roots, if $D=0$
$ 4\left(3 k^2+k-2\right)=0 $
$ 3 k^2+k-2=0 $
$ 3 k^2+3 k-2 k-2=0 $
$ 3 k(k+1)-2(k+1)=0 $
$ (k+1)(3 k-2)=0 $
$ (k+1)=0 \text { or }(3 k-2)=0$
$k = -1$ or $\text{k}=\frac{2}{3}$
Therefore, the value of $\text{k}=\frac{2}{3},-1$
Thus, the correct answer is $(b)$

Let $\text{x}=\sqrt{6+\sqrt{6+\sqrt{6+}}}...$
$\Rightarrow\text{x}=\sqrt{6+\text{x}}$
$ \Rightarrow x^2=6+x $
$ \Rightarrow x^2-x-6=0 $
$ \Rightarrow x^2-3 x+2 x-6=0 $
$ \Rightarrow x(x-3)+2(x-3)=0 $
$ \Rightarrow(x-3)(x+2)=0$
Either $x - 3 = 0$, then $x = 3$
Or $x + 2 = 0$, then $x = -2$
Now if x = 3, then
$3=\sqrt{6+\sqrt{6+\sqrt{6+}}}...$
$=\sqrt{6+ 3}=\sqrt{9}$
$=3$
If $x = -2$, then
$\Rightarrow\text{x}=\sqrt{6+\text{x}}$
$\Rightarrow-2=\sqrt{6-2}$
$\Rightarrow-2=\sqrt{4}$
$\Rightarrow-2\neq2$
Which is not possible $x = 3$ is correct.

In the equation
$ 9 x^2+6 k x+4=0 $
$ a=9, b=6 k, c=4 \text { then } $
$ \Rightarrow D=b^2-4 a c $
$ \Rightarrow D=(6 k)^2-4 \times 9 \times 4 $
$ \Rightarrow D=36 k^2-144$
$\text { Roots are equal }$
$ \Rightarrow D=0 $
$ \Rightarrow 36 k^2-144=0 $
$ \Rightarrow 36 k^2=144$
$\Rightarrow\text{k}^2=\frac{144}{36}$
$\Rightarrow\text{k}^2=4$
$\Rightarrow\text{k}^2=(\pm2)^2$
$\therefore\text{k}=\pm2$
$\therefore$ Roots are $=\frac{-\text{b}}{2\text{a}}$
$=\frac{\pm2\times6}{2\times9}$
$=\pm\frac{2}{3}$

The given quadric equation is $ax^2+ bx + c = 0,$ and roots are equal
Then find the value of $c$.
Let $\alpha$ and $\beta$ be two roots of given equation $\alpha=\beta$
Then, as we know that sum of the roots
$\alpha+\beta=\frac{-\text{b}}{\text{a}}$
$\alpha+\alpha=\frac{-\text{b}}{\text{a}}$
$2\alpha=\frac{-\text{b}}{\text{a}}$
$\alpha=\frac{-\text{b}}{2\text{a}}$
And the product of the roots
$\alpha\cdot\beta=\frac{\text{c}}{\text{a}}$
$\alpha\alpha=\frac{\text{c}}{\text{a}}$
Putting the value of $\alpha$
$\frac{-\text{b}}{2\text{a}}\times\frac{-\text{b}}{2\text{a}}=\frac{\text{c}}{\text{a}}$
$\frac{\text{b}^2}{4\text{a}}=\text{c}$
Therefore, the value of $\text{c}=\frac{\text{b}^2}{4\text{a}}$
Thus, the correct answer is $(d)$

$ \Rightarrow y=1 $
$ \Rightarrow a y^2+a y+3=0 $
$ \therefore a \times(1)^2+a \cdot 1+3=0 $
$ \Rightarrow a+a+3=0 $
$ \Rightarrow 2 a=-3 $
$ \Rightarrow a=\frac{-3}{2} $
$ \text { and } y^2+y+b=0 $
$ \Rightarrow(1)^2+(1)+b=0 $
$ \Rightarrow 1+1+b=0 $
$ \Rightarrow 2+b=0 $
$ \therefore b=-2$
$\text{ab}=\frac{-3}{2}\times(-2)=3$

$2$ is a root of equation $x^2+a x+12=0$
$ \Rightarrow(2)^2+\mathrm{a} \times 2+12=0 $
$ \Rightarrow 4+2 \mathrm{a}+12=0 $
$ \Rightarrow 2 \mathrm{a}=-(12+4) $
$ \Rightarrow 2 \mathrm{a}=-16 $
$ \Rightarrow \mathrm{a}=\frac{-16}{2} $
$ \Rightarrow \mathrm{a}=-8$
and in quadratic equation roots are equal $x^2+a x+q=0$
$ \Rightarrow b^2-4 a c=0 $
$ \Rightarrow a^2-4 q=0 $
$ \Rightarrow(-8)^2-4 q=0 $
$ \Rightarrow 64-4 q=0 $
$ \Rightarrow 4 q=64$
$\Rightarrow\text{q}=\frac{64}{4}$
$\Rightarrow q = 16$
$\therefore q = 16$