MCQ 11 Mark
If sin $\alpha$ and $\cos\alpha$ are the roots of the equations $ax^2 + bx + c = 0,$ then $b^2 =$
- A
$a^2 - 2ac$
- ✓
$a^2 + 2ac$
- C
$a^2 - ac$
- D
$a2 + ac$
AnswerCorrect option: B. $a^2 + 2ac$
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 $(b)$
View full question & answer→MCQ 21 Mark
If $2 $is a root of the equation $x^2 + bx + 12 = 0$ and the equation $x^2 + bx + q = 0$ has equal roots, then $q =$
Answer$2$ is the common roots given quadric equation are $x^2+b x+12=0$, and $x^2+b x+q=0$ Then find the value of $q$.
Here, $x^2+b x+12=0$
$x^2+b x+q=0$
Putting the value of $x=2$ in equation $(i)$ we get
$2^2+b \times 2+12=0$
$4+2 b+12=0$
$2 b=-16$
$b=-8$
Now, putting the value of $b=-8$ in equation $(ii)$ we get
$x^2-8 x+q=0$
Then,
$a_2=1, b_2=-8 \text { and } c_2=q$
As we know that $D_1=b^2-4 a c$
Putting the value of $a_2=1, b_2=-8$ and $c_2=q$
$=(-8)^2-4 \times 1 \times q$
$=64-4 q$
The given equation will have equal roots, if $D=0$
$64-4 q=0$
$4 q=64$
$q=\frac{64}{4}$
$q=16$
Thus, the correct answer is $(c)$
View full question & answer→MCQ 31 Mark
If the equation $ax^2 + 2x + a = 0$ has two distinct roots, if:
- ✓
$\text{a}=\pm1$
- B
$a = 0$
- C
$a = 0, 1$
- D
$a = -1, 0$
AnswerCorrect option: A. $\text{a}=\pm1$
In the equation a$x^2 + 2x + a = 0$
$\Rightarrow D = b^2 - 4ac$
$\Rightarrow D = (2)^2 - 4 \times a \times a$
$\Rightarrow D = 4 - 4a^2$
Roots are real and equal
$\Rightarrow D = 0$
$\Rightarrow 4 - 4a^2 = 0$
$\Rightarrow 4 = 4a^2$
$\Rightarrow 1 = a^2$
$\Rightarrow a^2 = 1$
$\Rightarrow\text{a}^2=(\pm1)^2$
$\Rightarrow\text{a}=\pm1$
View full question & answer→Question 41 Mark
If the equation $x^2 + 4x + k = 0$ has real and distinct roots, then:
Answer$k < 4$
Solution:
In the equation $x^2 + 4x + k = 0$
$a = 1, b = 4, c = k$
$\Rightarrow D = b^2 - 4ac$
$\Rightarrow D = (4)^2 - 4 \times 1 \times k$
$\Rightarrow D = 16 - 4k$
Roots are real and distinct
$\Rightarrow D > 0$
$\Rightarrow 16 - 4k > 0$
$\Rightarrow 16 > 4k$
$\Rightarrow 4 > k$
$\Rightarrow k < 4$
View full question & answer→MCQ 51 Mark
The value of $c$ for which the equation $ax^2 + 2bx + c = 0$ has equal roots is:
- ✓
$\frac{\text{b}^2}{\text{a}}$
- B
$\frac{\text{b}^2}{4\text{a}}$
- C
$\frac{\text{a}^2}{\text{b}}$
- D
$\frac{\text{a}^2}{4\text{b}}$
AnswerCorrect option: A. $\frac{\text{b}^2}{\text{a}}$
$\Rightarrow ax^2 + 2bx + c = 0$
$\Rightarrow D = b^2 - 4ac$
$\Rightarrow D = (2b)^2 - 4 \times a \times c$
$\Rightarrow D = 4b^2 - 4ac$
Roots are equal
$\Rightarrow D = 0$
$\Rightarrow 4b^2 - 4ac = 0$
$\Rightarrow 4ac = 4b^2$
$\Rightarrow\text{c}=\frac{4\text{b}^2}{4\text{a}}$
$\Rightarrow\text{c}=\frac{\text{b}^2}{\text{a}}$
View full question & answer→MCQ 61 Mark
The positive value of $k$ for which the equation $x^2+ kx + 64 = 0$ and $x^2 - 8x + k = 0$ will both have real roots, is:
AnswerSolution:
The given quadric equation are $x^2+k x+64=0$ and $x^2-8 x+k=0$ roots are real.
Then find the value of a.
Here, $x^2+k x+64=0$
$x^2-8 x+k=0....(ii)$
$a_1=1, b_1=k$ and $c_1=64$
$a_2=1, b_2=-8$ and $c_2=k$
As we know that $D_1=b^2-4 a c$
Putting the value of $a_1=1, b_1=k$ and $c_1=64$
$=(k)^2-4 \times 1 \times 64$
$=k^2-256$
The given equation will have real and distinct roots, if $D>0$
$k^2-256=0$
$k^2=256$
$k=\sqrt{256}$
$k= \pm 16$
Therefore, putting the value of $k=16$ in equation $(ii)$ we get
$x^2-8 x+16=0$
$(x-4)^2=0$
$x-4=0$
$x=4$
The value of $k=16$ satisfying to both equations.
Thus, the correct answer is $(d)$
View full question & answer→MCQ 71 Mark
The values of k for which the quadratic equation $16x^2 + 4kx + 9 = 0$ has real and equal roots are:
AnswerCorrect option: C. $6,-6$
The given quadratic equation $16x^2 + 4kx + 9 = 0$, has equal roots.
Here, $a = 16, b = 4k$ and $c = 9$
As we know that $D = b^2 - 4ac$
Putting the value of $a = 16, b = 4k$ and $c = 9$
$\Rightarrow D = (4k)^2 - 4(16)(9)$
$\Rightarrow D = 16k^2 - 576$
The given equation will have real and equal roots, if $D = 0$
Thus, $16k^2 - 576 = 0$
$\Rightarrow k^2 - 36 = 0$
$\Rightarrow (k + 6)(k - 6) = 0$
$\Rightarrow k + 6 = 0$ or $k = 6$
Therefore, the value of k is $6, -6$
Hence, the correct option is $(c)$
View full question & answer→MCQ 81 Mark
If the equation $x^2 - ax + 1 = 0$ has two distinct roots, then:
- A
$|a| = 2$
- B
$|a| < 2$
- ✓
$|a| >2$
- D
AnswerCorrect option: C. $|a| >2$
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 $c = 1$
$= (a)^2 - 4 \times 1 \times 1$
$= a^2 - 4$
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 $(c)$
View full question & answer→MCQ 91 Mark
If one root of the equation $ax^2 + bx + c = 0$ is three times the other, then $b^2 : ac =$
- A
$3 : 1$
- B
$3 : 16$
- ✓
$16 : 3$
- D
$16 : 1$
AnswerCorrect option: C. $16 : 3$
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$)$
$\Rightarrow\frac{\text{b}^2}{16\text{a}}=\frac{\text{c}}{3}$
$\Rightarrow\frac{\text{b}^2}{\text{ac}}=\frac{16}{3}$
$\Rightarrow\text{b}^2:\text{ac}=16:3$
View full question & answer→MCQ 101 Mark
If the roots of the equations $(a^2 + b^2)x^2 - 2b(a + c)x + (b^2 + c^2) = 0$ are equal, then:
AnswerCorrect option: B. $b^2 = ac$
The given quadric equation is $(a^2 + b^2)x^2 - 2b(a + c)x + (b^2 + c^2) = 0$, and roots equal.
Here, $a = (a^2 + b^2), b = -2b(a + c)$ and $c = b^2 + c^2$
As we know that $D = b^2 - 4ac$
Putting the value of $a = (a^2 + b^2), b = -2b(a + c)$ and $c = b^2 + c^2$
$= {-2b(a + c)}^2 - 4 \times (a^2 + b^2) \times (b^2 + c^2)$
$= 4a^2b^2 + 4b^2c^2 + 8ab^2c - 4(a^2b^2 + a^2c^2 + b^4 + b^2c^2)$
$= 4a^2b^2 + 4b^2c^2 + 8ab^2c - 4a^2b^2 - 4a^2c^2 - 4b^4 - 4b^2c^2$
$= +8ab^2c - 4a^2c^2 - 4b^4$
$= -4(a^2c^2 + b^4 - 2ab^2c)$
The given equation will have equal roots, if $D = 0$
$-4(a^2c^2 + b^4 - 2ab^2c) = 0$
$a^2c^2 + b^4 - 2ab^2c = 0$
$(ac - b^2)^2 = 0$
$ac - b^2 = 0$
$ac = b^2$
Thus, the correct answer is $(b)$
View full question & answer→MCQ 111 Mark
If the sum and product of the roots of the equation $kx^2 + 6x + 4k = 0$ are equal, then $k =$
- A
$-\frac{3}{2}$
- ✓
$\frac{3}{2}$
- C
$\frac{2}{3}$
- D
$-\frac{2}{3}$
AnswerCorrect option: B. $\frac{3}{2}$
$kx^2 + 6x + 4k = 0$
Here $a = k, b = 6, c = 4k$
$\Rightarrow D = b^2 - 4ac$
$\Rightarrow D = (6)^2 - 4 \times k \times 4k$
$\Rightarrow D = 36 - 16k^2$
Roots are equal
$\Rightarrow D = 0$
$\Rightarrow 36 - 16k^2 = 0$
$\Rightarrow 16k^2 = 36$
$\Rightarrow\text{k}^2=\frac{36}{16}$
$\Rightarrow\text{k}^2=\Big(\frac{6}{4}\Big)^2$
$\Rightarrow\text{k}=\frac{6}{4}$
$\Rightarrow\text{k}=\frac{3}{2}$
View full question & answer→MCQ 121 Mark
If $a$ and $b$ are roots of the equation $x^2 + ax + b = 0$, then $a + b =$
Answer$a$ 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 $a + b = 1 - 2 = -1$
View full question & answer→MCQ 131 Mark
If the sum of the roots of the equation $x^2 - (k + 6)x + 2(2k - 1) = 0$ is equal to half of their product, then $k =$
AnswerThe given quadric equation is $x^2 - (k + 6)x + 2(2k - 1) = 0$, 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 $(b)$
View full question & answer→MCQ 141 Mark
If the sum of the roots of the equation $\text{x}^2-\text{x}=\lambda(2\text{x}-1)$ is zero, then $\lambda=$
- A
$-2$
- B
$2$
- ✓
$-\frac{1}{2}$
- D
$\frac{1}{2}$
AnswerCorrect option: C. $-\frac{1}{2}$
$\Rightarrow\text{x}^2-\text{x}=\lambda(2\text{x}-1)$
$\Rightarrow\text{x}^2-\text{x}=2\lambda\text{x}-\lambda$
$\Rightarrow\text{x}^2-\text{x}-2\lambda\text{x}+\lambda=0$
$\Rightarrow\text{x}^2-(1+2\lambda)\text{x}+\lambda=0$
Sum of roots $=\frac{-\text{b}}{\text{a}}$
$=\frac{1+2\lambda}{1}$
$\frac{1+2\lambda}{1}=0$
$\Rightarrow2\lambda=-1$
$\lambda=-\frac{1}{2}$
View full question & answer→MCQ 151 Mark
A quadratic equation whose one root is $2$ and the sum of whose roots is zero, is:
- A
$x^2 + a^4 = 0$
- ✓
$x^2 - 4 = 0$
- C
$4x^2 - 1 = 0$
- D
$x^2 - 2 = 0$
AnswerCorrect option: B. $x^2 - 4 = 0$
Let $\alpha$ and $\beta$ be the roots of quadratic equation in such a way that $\alpha=2$
Then, according to question sum of the roots
$\alpha+\beta=0$
$2+\beta=0$
$\beta=-2$
And the product of the roots
$\alpha\cdot\beta=2\times(-2)$
$\alpha\cdot\beta=-4$
As we know that the quadratic equation
$\text{x}^2-(\alpha+\beta)\text{x}+\alpha\beta=0$
Putting the value of $\alpha$ and $\beta$ in above
Therefore, the require be
$\text{x}^2-0\times\text{x}+(-4)=0$
$\text{x}^2-4=0$
Thus, the correct answer is $(b)$
View full question & answer→MCQ 161 Mark
If one root of the equation $x^2 + ax + 3 = 0$ is $1$, then its other root is:
AnswerThe quad equation is $x^2 + ax + 3 = 0$
One root $ =1$
and product of roots $=\frac{\text{c}}{\text{a}}=\frac{3}{1}=3$
Second root $=\frac{3}{1}=3$
View full question & answer→MCQ 171 Mark
If $a$ and $b$ can take values $1, 2, 3, 4$. Then the number of the equations of the from $ax^2 + bx + 1 = 0$ having real roots is:
AnswerGiven that the equation $ax^2 + bx + 1 = 0$
For given equation to have real roots, discriminant $(\text{D})\geq0$
$\Rightarrow\text{b}^2-4\text{a}\geq0$
$\Rightarrow\text{b}^2\geq4\text{a}$
$\Rightarrow\text{b}\geq2\sqrt{\text{a}}$
Now, it is given that $a$ and $b$ can take values of $1, 2, 3$ and $4$
The above condition $\text{b}\geq2\sqrt{\text{a}}$ can be satisfied when
$i. b = 4$ and $a = 1, 2, 3, 4$
$ii. b = 3$ and $a = 1, 2$
$iii. b = 2$ and $a = 1$
So, there will be a maximum of $7$ equations for the values of $(a, b) = (1, 4), (2, 4), (3, 4), (4, 4), (1, 3), (2, 3)$ and $(1, 2)$
Thus, the correct option is $(b)$
View full question & answer→MCQ 181 Mark
If $p$ and $q$ are the roots of the equation $x^2 - px + q + 0$, then:
- ✓
$p = 1, q = -2$
- B
$p = 0, q = 1$
- C
$p = -2, q = 0$
- D
$p = -2, q = 1$
AnswerCorrect option: A. $p = 1, q = -2$
Given that $p$ and $q$ be the roots of the equation $x^2 - Px + q + 0$
Then find the value of $p$ and $q.$
Here, $a = 1, b = -p$ and $c = q$
p and q be the roots of the given equation
Therefore, sum of the roots
$\text{p}+\text{q}=\frac{-\text{b}}{\text{a}}$
$\text{p}+\text{q}=\frac{-\text{p}}{1}$
$\text{p}+\text{q}=-\text{p}$
$\text{q}=-\text{p}-\text{p}$
$\text{q}=-2\text{p}\ ....(\text{i})$
Product of the roots
$\text{p}\times\text{q}=\frac{\text{q}}{1}$
As we know that
$\text{p}=\frac{\text{q}}{\text{q}}$
$\text{p}=1$
Putting the value of $p = 1$ in equation $(i)$
$q = -2 \times 1$
$q = -2$
Therefore, the value of $p = 1, q = -2$
Thus, the correct answer is $(a)$
View full question & answer→MCQ 191 Mark
If the equation $(a^2 + b^2)x^2 - 2(ac + bd)x + c^2 + d2 = 0$ has equal roots, then:
AnswerCorrect option: B. $\text{ad}=\text{bc}$
In the equation
$\Rightarrow (a^2 + b^2)x^2 - 2(ac + bd)x + (c^2 + d^2) = 0$
$\Rightarrow D = B^2 - 4AC$
$\Rightarrow D = [-2(ac + bd)]^2 - 4(a^2 + b^2)(c^2 + d^2)$
$\Rightarrow D = 4[a^2c^2 + b2d^2 + 2abcd] - 4[a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2]$
$\Rightarrow D = 4a^2c^2 + 4b^2d^2 + 8abcd - 4a^2c^2 - 4a^2d^2 - 4b^2c^2 - 4b^2d^2$
$\Rightarrow D = 8abcd - 4a^2d2 - 4b^2c^2$
$\Rightarrow D = -4[a^2d^2 + b^2c^2 - 2abcd]$
$\Rightarrow D = -4(ad - bc)^2$
$\because$ Roots are equal
$\therefore D = 0$
$\Rightarrow - 4(ad - bc)^2 = 0$
$\Rightarrow ad - bc = 0$
$\Rightarrow ad = bc$
View full question & answer→MCQ 201 Mark
If $(a^2 + b^2)x^2 + 2(ab + bd)x + c^2 + d^2 = 0$ has no real roots, then:
- A
$ab = bc$
- B
$ab = cd$
- C
$ac = bd$
- ✓
$\text{ad}\neq\text{bc}$
AnswerCorrect option: D. $\text{ad}\neq\text{bc}$
The given quadric equation is $(a^2 + b^2)x^2 + 2(ab + bd)x + c^2 + d^2 = 0$, and roots are equal.
Here, $a = (a^2 + b^2), b = 2(ab + bd)$ and, $c = c^2 + d^2$
As we know that $D = b^2 - 4ac$
Putting the value of $a = (a^2 + b^2), b = 2(ab + bd)$ and, $c = c^2 + d^2$
$= {2(ab + bd)}^2 - 4 \times (a^2 + b^2) \times (c^2 + d^2)$
$= 4a^2b^2 + 4b^2d^2 + 8ab^2d - 4(a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2)$
$= 4a^2b^2 + 4b^2d^2 + 8ab^2d - 4a^2c^2 - 4a^2d^2 - 4b^2c^2 - 4a^2d^2$
$= 4a^2b^2 + 8ab^2d - 4a^2c^2 - 4a^2d^2 - 4b^2c^2$
$= 4(a^2b^2 + 2ab^2d - a^2c^2 - a^2d^2 - b^2c^2)$
The given equation will have no real roots, if $D < 0$
$4(a^2b^2 + 2ab^2d - a^2c^2 - a^2d^2 - b^2c^2) < 0$
$a^2b^2 + 2ab^2d - a^2c^2 - a^2d^2 - b^2c^2 < 0$
$\text{ad}\neq\text{bc}$
Thus, the correct answer is $(d)$
View full question & answer→MCQ 211 Mark
If one root the equation $2x^2 + kx + 4 = 0$ is $2$, then the other root is:
AnswerLet $\alpha$ 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 $(d)$
View full question & answer→MCQ 221 Mark
if $x^2 + k(4x + k - 1) + 2 = 0$ has equal rrots, then $k =$
- A
$-\frac{2}{3},1$
- ✓
$\frac{2}{3},-1$
- C
$\frac{3}{2},\frac{1}{3}$
- D
$-\frac{3}{2},-\frac{1}{3}$
AnswerCorrect option: B. $\frac{2}{3},-1$
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 $D=0$
$4\left(3 k^2+k-2\right)=0$
$3 k^2+k-2=0$
$3 k^2+3 k-2 k-2=0$
$3 k(k+1)-2(k+1)=0$
$(k+1)(3 k-2)=0$
$(k+1)=0 \text { or }(3 k-2)=0$
$k=-1 \text { or } k=\frac{2}{3}$
Therefore, the value of $k =\frac{2}{3},-1$
Thus, the correct answer is $(b)$
View full question & answer→MCQ 231 Mark
The value of $\sqrt{6+\sqrt{6+\sqrt{6+}}}...$ is:
View full question & answer→MCQ 241 Mark
If one root of the equation $4\text{x}^2-2\text{x}+(\lambda-4)=0$ be the reciprocal of the other, then $\lambda=$
AnswerLet $\alpha$ and $\beta$ be the roots of quadratic equation $4\text{x}^2-2\text{x}+(\lambda-4)=0$ in such a way
Then, $\alpha=\frac{1}{\beta}$
Here, $\text{a}=4,\text{b}=-2$ and $\text{c}=(\lambda-4)$
Then, according to question sum of the roots
$\alpha+\beta=\frac{-\text{b}}{\text{a}}$
$\frac{1}{\beta}+\beta=\frac{-(-2)}{4}$
$\frac{1+\beta^2}{\beta}=\frac{1}{2}$
$2+2\beta^2=\beta$
$2\beta^2-\beta+2=0$
And the product of the roots
$\alpha\cdot\beta=\frac{\text{c}}{\text{a}}$
$\frac{1}{\beta}\times\beta=\frac{\lambda-4}{4}$
$1=\frac{\lambda-4}{4}$
$\lambda-4=4$
$\lambda=4+4$
$\lambda=8$
Therefore, value of $\lambda=8$
Thus, the correct answer is $(a)$
View full question & answer→MCQ 251 Mark
The number of quadratic equations having real roots and which do not change by squaring their roots is:
AnswerAs we know that the number of quadratic equations having real roots and which do not change by squaring their roots is $2.$
Thus, the correct answer is $(c)$
View full question & answer→MCQ 261 Mark
If the equation $9x^2 + 6kx + 4 = 0$, has equal roots, then the roots are both equal to.
- ✓
$\pm\frac{2}{3}$
- B
$\pm\frac{3}{2}$
- C
$0$
- D
$\pm3$
AnswerCorrect option: A. $\pm\frac{2}{3}$
In the equation
$9x^2 + 6kx + 4 = 0$
$a = 9, b = 6k, c = 4$ then
$\Rightarrow D = b^2 - 4ac$
$\Rightarrow D = (6k)^2 - 4 \times 9 \times 4$
$\Rightarrow D = 36k^2 - 144$
Roots are equal
$\Rightarrow D = 0$
$\Rightarrow 36k^2 - 144 = 0$
$\Rightarrow 36k^2 = 144$
$\Rightarrow\text{k}^2=\frac{144}{36}$
$\Rightarrow\text{k}^2=4$
$\Rightarrow\text{k}^2=(\pm2)^2$
$\therefore\text{k}=\pm2$
$\therefore$ Roots are $=\frac{-\text{b}}{2\text{a}}$
$=\frac{\pm2\times6}{2\times9}$
$=\pm\frac{2}{3}$
View full question & answer→MCQ 271 Mark
If $x = 1$ is a common roots of the equations $ax^2 + ax + 3 = 0$ and $x^2 + x + b = 0$, then $ab =$
Answer$x=1$ is the common roots given quadric equation are $a x^2+a x+3=0$, and $x^2+x+b=0$ Then find the value of $q$.
Here, $a x^2+a x+3=0 \ldots... (i)$
$x^2+x+b=0$
Putting the value of $x=1$ in equation $(i)$ we get
$a \times 1^2+a \times 1+3=0$
$a+a+3=0$
$2 a=-3$
$a=-\frac{3}{2}$
Now, putting the value of $x=1$ in equation $(ii)$ we get
$1^2+1+b=0$
$2+b=0$
$b=-2$
Then,
$ab=\frac{-3}{2} \times(-2)$
$=3$
Thus, the correct answer is $(a)$
View full question & answer→MCQ 281 Mark
If $x = 1$ is a common root of $ax^2 + ax + 2 = 0$ and $x^2 + x + b = 0$, then $ab =$
Answer$x =1$ is the common roots given quadric equation are $a x^2+a x+2=0$, and $x^2+x+b=0$ Then find the value of $a b$.
Here, $a x^2+a x+2=0 \ldots . .. (i)$
$x^2+x+b=0 \ldots . . .(i i)$
Putting the value of $x=1$ in equation $(ii)$ we get
$1^2+1+b=0$
$2+b=0$
$b=-2$
Now, putting the value of $x=1$ in equation $(i)$ we get
$a+a+2=0$
$2 a+2=0$
$a=\frac{-2}{2}$
$a=-1$
$a b=(-1) \times(-2)$
Then, $a b=2$
Thus, the correct answer is $(b)$
View full question & answer→MCQ 291 Mark
If the equation $x^2 - bx + 1 = 0$ does not possess real roots, then:
- A
$-3 < b < 3$
- ✓
$-2 < b < 2$
- C
$b > 2$
- D
$b < -2$
AnswerCorrect option: B. $-2 < b < 2$
In the equation
$x^2 - bx + 1 = 0$
$\Rightarrow D = b^2 - 4ac$
$\Rightarrow D = (-b)^2 - 4 \times 1 \times 1$
$\Rightarrow D = b^2 - 4$
$\because$ The roots are not real
$\therefore$ $D < 0$
$\Rightarrow b^2 - 4 < 0$
$\Rightarrow b^2 < 4$
$\text{b}^2<(\pm2)^2$
$\therefore$ $b < 2$ and $b > -2$ or $-2 < b$
$\therefore$ $-2 < b < 2$
View full question & answer→MCQ 301 Mark
If $ax^2 + bx + c = 0$ has equal roots, then $c =$
- A
$\frac{-\text{b}}{2\text{a}}$
- B
$\frac{\text{b}}{2\text{a}}$
- C
$\frac{-\text{b}^2}{4\text{a}}$
- ✓
$\frac{-\text{b}^2}{4\text{a}}$
AnswerCorrect option: D. $\frac{-\text{b}^2}{4\text{a}}$
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 $(d)$
View full question & answer→MCQ 311 Mark
If $y = 1$ is a common root of the equations $ay^2 + ay + 3 = 0$ and $y^2 + y + b = 0$, then ab equals:
- ✓
$3$
- B
$-\frac{1}{2}$
- C
$6$
- D
$-3$
Answer$\Rightarrow y = 1$
$\Rightarrow ay^2 + ay + 3 = 0$
$\therefore$ $a \times (1)^2 + a.1 + 3 = 0$
$\Rightarrow a + a + 3 = 0$
$\Rightarrow 2a = -3$
$\Rightarrow\text{a}=\frac{-3}{2}$
and $y^2 + y + b = 0$
$\Rightarrow (1)^2 + (1) + b = 0$
$\Rightarrow 1 + 1 + b = 0$
$\Rightarrow 2 + b = 0$
$\therefore$ $b = -2$
$\text{ab}=\frac{-3}{2}\times(-2)=3$
View full question & answer→MCQ 321 Mark
If $2$ is a root of the equation $x^2 + ax + 12 = 0$ and the quadratic equation $x^2 + ax + q = 0$ has equal roots, then $q =$
Answer$2$ is a root of equation $x^2+a x+12=0$
$\Rightarrow(2)^2+a \times 2+12=0$
$\Rightarrow 4+2 a+12=0$
$\Rightarrow 2 a=-(12+4)$
$\Rightarrow 2 a=-16$
$\Rightarrow a=\frac{-16}{2}$
$\Rightarrow a=-8$
and in quadratic equation roots are equal $x^2+a x+q=0$
$\Rightarrow b^2-4 a c=0$
$\Rightarrow a^2-4 q=0$
$\Rightarrow(-8)^2-4 q=0$
$\Rightarrow 64-4 q=0$
$\Rightarrow 4 q=64$
$\Rightarrow q=\frac{64}{4}$
$\Rightarrow q=16$
$\therefore q=16$
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