MCQ 11 Mark
The roots of the quadratic equation $2x^2- x - 6 = 0$ are :
- A
$-2,\frac{3}{2}$
- ✓
$2,\frac{-3}{2}$
- C
$-2,\frac{-3}{2}$
- D
$2,\frac{3}{2}$
AnswerCorrect option: B. $2,\frac{-3}{2}$
Given that, $2 x^2-x-6=0 $
$ \Rightarrow 2 x^2-(4 x-3 x)-6=0 $
$ \Rightarrow 2 x^2-4 x+3 x-6=0 $
$ \Rightarrow 2 x(x-2)+3(x-2)=0 $
$ \Rightarrow(x-2)(2 x+3)=0 $
$\Rightarrow\text{x}=2,\frac{-3}{2}$
View full question & answer→MCQ 21 Mark
The roots of the equation $x^2- 3x - m(m + 3) = 0,$ where $m$ is a constant, are :
- A
$m, m + 3$
- ✓
$-m, m + 3$
- C
$m, -(m + 3)$
- D
$-m, -(m: 3)$
AnswerCorrect option: B. $-m, m + 3$
The given quadratic equation is $x^2- 3x - m(m + 3) = 0, $
where $m$ is a constant.
$ x^2-3 x-m(m+3)=0 $
$ \therefore x^2-[(m+3)-m] x-m(m+3)=0 $
$ \Rightarrow x^2-(m+3) x+m x-m(m+3)=0 $
$\Rightarrow x [x - (m +3)] + m[x - (m + 3)] = 0$
$\Rightarrow [x - (m +3)] (x + m) = 0$
$\Rightarrow x - (m +3) = 0$ or $x + m = 0$
$\Rightarrow x = m + 3$ or $x = -m$
Thus, the roots of the given quadratic equation are $m + 3$ and $-m$.
View full question & answer→MCQ 31 Mark
If $1$ is a root of the equations $ay^2+ ay + 3 = 0$ and $y^2+ y + b = 0$, then $ab$ equals :
- ✓
$3$
- B
$-\frac{7}{2}$
- C
$6$
- D
$-3$
AnswerGiven :
$1$ is the root.
So, $ay^2+ ay + 3 = 0$
$(\because\ \text{root is 1})$
$a(1)^2+ a.1 + 3 = 0$
$\Rightarrow a+ a + 3 = 0$
$\Rightarrow 2a + 3 = 0$
$\Rightarrow\text{a}=\frac{-3}{2}$
Again,
$y^2+ y + b = 0$
$\Rightarrow 1 + 1 + b = 0$
$\Rightarrow 2 + b = 0$
$\Rightarrow b = -2$
Hence , required product is
$\text{a}=\frac{-3}{2},\text{b}=-2$
$(\text{ab})=\frac{-3}{2}(-2)=3$
View full question & answer→MCQ 41 Mark
The roots of the equation $x^2+ x - p(p + 1) = 0$, where $p$ is a constant, are :
- A
$p, p + 1$
- B
$-p, p + 1$
- ✓
$p, - (p + 1)$
- D
$−p, - (p + 1)$
AnswerCorrect option: C. $p, - (p + 1)$
$ x^2+x-p(p+1)=0 $
$x^2+(p+1) x-p x-p(p+1)=0 $
$ x(x+p+1)-p(x+p+1)=0 $
$ (x+p+1)(x-p)=0 $
$ x=-p-1, p $
View full question & answer→MCQ 51 Mark
The value $(s)$ of $k$ for which the quadratic equation $2x^2+ kx + 2 = 0$ has equal roots, is :
AnswerCorrect option: B. $\pm4$
Given equation is $2 x^2+k x+2=0$
On comparing with $a x^2+b x+c=0$, we get
$a = 2, b = k$ and $c = 2$
For equal roots, the discriminant must be zero.
$D=b^2-4 a c=0$
$k^2-4 \times 2 \times 2=0$
$k^2-16=0$
$\text{k}=\pm4$
View full question & answer→MCQ 61 Mark
The equation $x^2-8 x+k=0$ has real and distinct roots if :
- A
$k = 16$
- B
$k > 16$
- C
$k = 8$
- ✓
$k < 16$
AnswerCorrect option: D. $k < 16$
Since the given equation has real roots,
i.e., $D=b^2-4 a c=0$
Here $, a = 1, b = -8, c = k$
$\therefore(-8)^2-4(1)(\text{k})\geq0$
Or, $64-4\text{k}\geq0$
$4\text{k}\leq64$
OR
$\text{k}\leq\frac{64}{4}$
Or, $\text{k}\leq16$
View full question & answer→MCQ 71 Mark
The quadratic equation $x^2– 4x + k = 0$ has distinct real roots if :
- A
$k = 4$
- B
$k > 4$
- C
$k = 16$
- ✓
$k < 4$
AnswerCorrect option: D. $k < 4$
The given quadric equation is $x^2+ 4x + k = 0,$ and roots are real and distinct.
Then find the value of $k$.
Here $, a = 1, b = 4$ and $c = k$
As we know that $D = b^2- 4ac$
Putting the value of $a = 1, b = 4$ and $, c = k$
Putting the value of $a = 1, b = 4$ and $, c = k$
$(4)^2- 4 \times 1 \times k$
$= 16 - 4k$
The given equation will have real and distinct roots, if $D > 0$
$16 - 4k > 0$
$4k < 16$
$\text{ k} < \frac{16}{4}$
$k < 4$
$\therefore$ The value of $k < 4$.
View full question & answer→MCQ 81 Mark
If $\triangle\text{ABC} \sim \triangle\text{DEF}$ such that $AB = 1.2\ cm$ and $DE = 1.4\ cm,$ the ratio of the areas of $\triangle\text{ABC}$ and $\triangle\text{DEF}$ is :
- A
$49 : 36$
- B
$6 : 7$
- C
$7 : 6$
- ✓
$36 : 49$
AnswerCorrect option: D. $36 : 49$
We have,
$\triangle\text{ABC} \sim \triangle\text{DEF}$
$AB = 1.2\ cm$ and $DF = 1.4\ cm$
By area of similar triangle theorem
$\frac{\text{Area}(\triangle\text{ABC})}{\text{Area}(\triangle\text{DEF})}=\frac{\text{AB}^2}{\text{DE}^2}$
$=\frac{1.2^2}{1.4^2}$
$=\frac{1.44}{1.96}$
$=\frac{36}{49}$
View full question & answer→MCQ 91 Mark
The common difference of an $AP,$ whose $n^{th}$ term is $an = (3n + 7),$ is :
Answer$ a_n=3 n+7 $
$ a_1=3 \times 1+7=10 $
$ a_2=3 \times 2+7=13 $
$ d=a_2-a_1 $
$d = 13 - 10$
$d = 3$
common difference $= 3$
View full question & answer→MCQ 101 Mark
The roots of the quadratic equation $x^2-0.04=0$ are :
- ✓
$\pm0.2$
- B
$\pm00.2$
- C
$0.4$
- D
$2$
AnswerCorrect option: A. $\pm0.2$
$\text{x}^2-0.04=0$
$\Rightarrow\text{x}^2=0.04$
$\Rightarrow \text{x}= \sqrt{0.04}$
$\Rightarrow\text{x}= \pm 0.2$
View full question & answer→MCQ 111 Mark
$5x^2+ 8x + 4 = 2x^2+ 4x + 6$ is $a$ :
AnswerGiven : $ 5 x^2+8 x+4=2 x^2+4 x+6 $
$ \Rightarrow 5 x^2-2 x^2+8 x-4 x+4-6 $
$ \Rightarrow 3 x^2+4 x-2=0 $
Here, the degree is $2,$
$\therefore$ it is a quadratic equation.
View full question & answer→MCQ 121 Mark
The ratio of the sum and product of the roots of the equation $7x^2- 12x + 18 = 0$ is :
- A
$7 : 12$
- B
$7 : 18$
- C
$3 : 2$
- ✓
$2 : 3$
AnswerCorrect option: D. $2 : 3$
$7x^2- 12x + 18 = 0$
Comparing with $ax^2+ bx + c = 0$, we have
$a = 7, b = -12, c = 18$
Sum of the roots $=-\frac{\text{b}}{\text{a}}=-\frac{(-12)}{7}=\frac{12}{7}$
Product of the roots $=\frac{\text{c}}{\text{a}}=\frac{18}{7}$
Now $, \frac{\text{Sum of the roots}}{\text{Product of the roots}}=\frac{\frac{12}{7}}{\frac{18}{7}}=\frac{12}{18}$
$=\frac{2}{3}=2:3$
View full question & answer→MCQ 131 Mark
If the equation $x^2- kx + 1 = 0$ has no real roots, then :
- A
$k < -2$
- B
$k > 2$
- ✓
$-2 < k < 2$
- D
AnswerCorrect option: C. $-2 < k < 2$
Since the equation $x^2+ 5kx + 16 = 0$ has no real roots,
$\Rightarrow D < 0$
$\Rightarrow b^2- 4ac > 0$
$\Rightarrow (-k)^2- 4 \times 1 \times 1 < 0$
$\Rightarrow k^2- 4 < 0$
$\Rightarrow k^2 < 4$
$\Rightarrow\text{k}<\sqrt{4}$ or $\text{k}>-\sqrt{4}$
$\Rightarrow k < 2$ or $k > -2$
$\Rightarrow -2 < k < 2$
View full question & answer→MCQ 141 Mark
If $\sin \alpha$ and $\cos\alpha$ are the roots of the equations $a x^2+b x+c=0$, then $b^2=$
- A
$ a^2-2 a c $
- ✓
$ a^2+2 a c $
- C
$ a^2-a c $
- D
$ a 2+a c $
AnswerCorrect option: B. $ a^2+2 a c $
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 151 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 $x^2+ bx + 12 = 0$, and $x^2+ bx + q = 0$
Then find the value of $q$.
Here, $x^2+ bx + 12 = 0 ....(i)$
$x^2+ bx + q = 0 ....(ii)$
Putting the value of $x = 2$ in equation $(i)$ we get
$2^2+ b \times 2 + 12 = 0$
$4 + 2b + 12 = 0$
$2b = -16$
$b = -8$
Now, putting the value of $b = -8$ in equation $(ii)$ we get
$x^2- 8x + q = 0$
Then,
$a_2= 1, b_2= -8$ and $c_2= q$
As we know that $D_1= b^2- 4ac$
Putting the value of $a_2= 1, b_2= -8$ and $c_2= q$
$= (-8)^2- 4 \times 1 \times q$
$= 64 - 4q$
The given equation will have equal roots, if $D = 0$
$64 - 4q = 0$
$4q = 64$
$\text{q}=\frac{64}{4}$
$q = 16$
Thus, the correct answer is $(c)$
View full question & answer→MCQ 161 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 $ax^2+ 2x + a = 0$
$\Rightarrow D = b^2- 4ac$
$\Rightarrow D = (2)^2- 4 \times a \times a$
$\Rightarrow D = 4 - 4a^2$
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→MCQ 171 Mark
The number of quadratic equations having real roots and which do not change by squaring their roots is :
AnswerWe are given that quadratic equations have real roots and the quadratic equation does not change by squaring their roots.
We have to find the number of quadratic equations.
The possible roots $(1, 1) (1, 0) (0, 0)$
The general formula of quadratic equation is.
$x^2- ($sum of roots$)\ x \ +$ product of roots
So, we have:
Case $-I:$ When roots are $1$ and $1$
$x^2- (1 + 1) x + 1 = 0$
$x^2- 2x + 1 = 0$
Case $-II:$ When roots are $1$ and $0$
$x^2- x = 0$
Case $-III:$ When roots are $0$ and $0$
Then, $x^2= 0$
$\therefore 3$ possible quadratic equation.
View full question & answer→MCQ 181 Mark
Choose the correct answer from the given four options in the following questions : $(x^2+ 1)^2- x^2= 0$ has :
AnswerGiven equation is $\left(x^2+1\right)^2-x^2=0$
$ \Rightarrow x^4+1+2 x-x^2=0 $
$ {\left[\because(a+b)^2=a^2+b^2+2 a b\right]} $
$ \Rightarrow x^4+x^2+1=0$
Let $x^2=y$
$ \therefore\left(x^2\right)^2+x^2+1=0$
$ y^2+y+1=0$
On comparing with $a y^2+b y+c=0$, we get
$a = 1, b = 1$ and $c= 1$
Discrinimant, $ D=b^2-4 a c $
$ =(1)^2-4(1)(1) $
$ =1-4=-3 $
Since $, D<0 $
$ \therefore y^2+y+1=0$
i.e.$ , x^4+x^2+1=0 $ or
$ \left(x^2+1\right)^2-x^2=0$ has no real roots.
View full question & answer→MCQ 191 Mark
The product of two consecutive integers is $240$. The quadratic representation of the above situation is :
- A
$x^2+(x+1)=240$
- B
$x + (x + 1) = 240$
- C
$x(x+1)^2=240$
- ✓
$x(x + 1) = 240$
AnswerCorrect option: D. $x(x + 1) = 240$
Let one of the two consecutive integers be $x$
Then the other consecutive integer will be $(x + 1)$
$\therefore$ According to question $, (x) \times (x + 1) = 240$
$\Rightarrow x(x + 1) = 240$
View full question & answer→MCQ 201 Mark
If the equation $x^2+ 4x + k = 0$ has real and distinct roots, then :
- ✓
$\text{k}<4$
- B
$\text{k}>4$
- C
$\text{k}\geq4$
- D
$\text{k}\leq4$
AnswerCorrect option: A. $\text{k}<4$
In the equation $x^2+4 x+k=0$
$ a=1, b=4, c=k $
$ \Rightarrow D=b^2-4 a c $
$ \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 211 Mark
A quadratic equation $ax^2+ bx + c = 0$ has non $-$ real roots, if :
- A
$ b^2-4 a c=0 $
- ✓
$ b^2-4 a c > 0 $
- C
$ b^2-4 a c < 0 $
- D
$ b^2-a c=0 $
AnswerCorrect option: B. $ b^2-4 a c > 0 $
The roots of the quadratic equation $a x^2+b x+c=0,$ In this formula the term $b^2-4 a c$ is called the discriminant. If $b^2-4 a c=0,$
so the equation has a single repeated root. If $b^2-4 a c>0,$ the equation has two real roots.
If $b^2-4 a c<0,$ the equation has two complex roots.
View full question & answer→MCQ 221 Mark
Let $b = a + c$. Then the equation $ax^2+ bx + c = 0$ has equal roots if :
- A
$a = -c$
- B
$a = 2c$
- ✓
$a = c$
- D
$a = -2c$
AnswerCorrect option: C. $a = c$
Since, If $a x^2+b x+c=0$ has equal roots,
then $ b^2-4 a c=0 $
$ \Rightarrow(a+c)^2-4 a c=0\ [$Given : $b=a+c]$
$ \Rightarrow a^2+c^2+2 a c-4 a c=0 $
$ \Rightarrow a^2+c^2-2 a c=0 $
$ \Rightarrow(a-c)^2=0$
$\Rightarrow a - c = 0$
$\Rightarrow a = c$
View full question & answer→MCQ 231 Mark
If $x = 3$ is a solution of the equation $3x^2+ (k - 1)x + 9 = 0$ then $k = ?$
AnswerSince $x=3$ is a solution of the equation $3 x^2+(k-1) x+9=0$,
we have $3(3)^2+(k-1) 3+9=0$
$\Rightarrow 27 + 3k - 3 + 9 = 0$
$\Rightarrow 3k + 33 = 0$
$\Rightarrow 3k = -33$
$\Rightarrow k = -11$
View full question & answer→MCQ 241 Mark
Directions : In the following questions, the Assertions $(A)$ and Reason $(s) \ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following :
Assertion : $(2 x-1)^2-4 x^2+5=0$ is not a quadratic equation.
Reason : $x = 0,3$ are the roots of the equation $2 x^2-6 x=0$
- A
If both assertion and reason are true and reason is the correct explanation of assertion.
- ✓
If both assertion and reason are true but reason is not the correct explanation of assertion.
- C
If assertion is true but reason is false.
- D
If both assertion and reason are false.
AnswerCorrect option: B. If both assertion and reason are true but reason is not the correct explanation of assertion.
Assertion $(2x - 1)2 - 4x^2 + 5 = 0$
$\Rightarrow -4x + 6 = 0$
Reason $2x^2 - 6x = 0$
$\Rightarrow 2x (x - 3) = 0 $
$\Rightarrow x = 0$
and $x = 3$
View full question & answer→MCQ 251 Mark
The angry Arjun carried some arrows for fighting with Bheeshma. With half the arrows, he cut down the arrows thrown by Bheeshma on him and with six other arrows he killed the rath driver of Bheeshma. With one arrow each, he knocked down respectively the rath, flag and bow of Bheeshma. Finally, with one more than four times the square root of arrows, he laid Bheeshma unconscious on an arrow bed. The total number of arrows that Arjun had, is:
AnswerLet Arjun had $x$ arrows.
According to question,
$\frac{\text{x}}{2}+6+3+{4}\sqrt{\text{x}}+1=\text{x}$
$\Rightarrow10+4\sqrt{\text{x}} = \frac{\text{x}}{2}$
$\Rightarrow{20+8}\sqrt{\text{x}}=\text{x}$
${8}\sqrt{\text{x}}=\text{x}-20$
$ \Rightarrow 64 x=x^2-40 x+400 $
$ \Rightarrow x^2-104 x+400=0 $
$ \Rightarrow x^2-100 x-4 x+400=0 $
$\Rightarrow x(x - 100) -4(x - 100) = 0$
$\Rightarrow (x - 100) (x - 4) = 0$
$\Rightarrow x - 100 = 0 $ and $x - 4 = 0$
$\Rightarrow x = 100$ and $x = 4\ [$which is not possible$]$
$\therefore$ Arjun had $100$ arrows.
View full question & answer→MCQ 261 Mark
If $x = 3$ is a solution of the equation $3x^2+ (k - 1)x + 9 = 0$ then $k = $?
Answer$3x^2+ (k - 1)x + 9 = 0$
$x = 3$ is a solution of the equation means it satisfies the equation
Put $x = 3,$ we get
$3(3)^2+ (k - 1) 3 + 9 = 0$
$27 + 3k - 3 + 9 = 0$
$27 + 3k + 6 = 0$
$3k = -33$
$k = -11$
View full question & answer→MCQ 271 Mark
A quadratic equation $ax^2+ bx + c = 0,$ has coincident roots, if :
- A
$ b^2-4 a c<0 $
- B
$ b^2-4 a c>0 $
- ✓
$ b^2-4 a c=0 $
- D
$ b^2-a c=0 $
AnswerCorrect option: C. $ b^2-4 a c=0 $
The roots of the quadratic equation $a x^2+b x+c=0$, In this formula the term $b^2-4 a c$ is called the discriminant.
If $b^2-4 a c=0$ so the equation has a single repeated root.
If $b^2-4 a c>0$, the equation has two real roots.
If $b^2-4 a c<0,$ the equation has two complex roots.
View full question & answer→MCQ 281 Mark
Directions : In the following questions, the Assertions $(A)$ and Reason $(s) \ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion : $(2x - 1)2 - 4x^2+ 5 = 0$ is not a quadratic equation.
Reason : $x = 0, 3$ are the roots of the equation $2x^2- 6x = 0.$
- A
Both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ isthe correct explanation of assertion $(A$).
- ✓
Both assertion $(A)$ and reason $(R)$ are true but reason $(R)$ is not the correct explanation of assertion $(A)$.
- C
Assertion $(A)$ is true but reason $(R)$ is false.
- D
Assertion $(A)$ is false but reason $(R)$ is true
AnswerCorrect option: B. Both assertion $(A)$ and reason $(R)$ are true but reason $(R)$ is not the correct explanation of assertion $(A)$.
Assertion and Reason both are true statements.
But Reason is not the correct explanation.
Assertion $(2 x-1)^2-4 x^2+5=0$
$=-4 x+6=0$
Reason $2 x^2-6 x=0$
$2x(x - 3) = 0 x = 0$
and $x = 3$
View full question & answer→MCQ 291 Mark
A train travels $360\ km$ at a uniform speed. If the speed had been $5\ km/ hr$ more, it would have taken $1$ hour less for the same journey, then the actual speed of the train is :
- A
$48km/ hr$
- ✓
$40km/ hr$
- C
$36km/ hr$
- D
$45km/ hr$
AnswerCorrect option: B. $40km/ hr$
Let the actual speed of the train be $x \ km/ hr$
Time taken to cover $360\ km$ at this speed $=\frac{360}{\text{x}}\text{hr}$
Time taken to cover $360\ km$ at the increased speed $=\frac{360}{\text{x + 5}}\text{hr}$
According to condition, $\frac{360}{\text{x}}-\frac{360}{\text{x}+5} = 1$
$\Rightarrow 360\big[\frac{1}{\text{x}}-\frac{1}{\text{x}+5}\big] = 1$
$\Rightarrow{360}\big[\frac{\text{x}+5\text{-x}}{{\text{x}}(\text{x}+5)}\big] = 1$
$\Rightarrow{360}\big[\frac5{{\text{x}}(\text{x}+5)}\big] = 1$
$ \Rightarrow x^2+5 x-1800=0 $
$ \Rightarrow x^2+45 x-40 x-1800 $
$\Rightarrow x(x + 45) -40(x + 45) = 0$
$\Rightarrow (x - 40) (x - 45) = 0$
$\Rightarrow x - 40 = 0$ and $x + 45 = 0$
$\Rightarrow x = 40\ km/ hr$ and $x = -45\ km/ hr\ [$But $x = -45$ is not possible$]$
$\therefore$ The actual speed of the train is $40\ km/ hr$
View full question & answer→MCQ 301 Mark
Directions : In the following questions, the Assertions $(A)$ and Reason $(s) \ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion : Both the roots of the equation $x^2- x + 1 = 0$ are real.
Reason : The roots of the equation $ax^2+ bx + c = 0$ are real if and only if $b^2 - 4ac \geq 0.$
- A
If both assertion and reason are true and reason is the correct explanation of assertion.
- B
If both assertion and reason are true but reason is not the correct explanation of assertion.
- C
If assertion is true but reason is false.
- ✓
If both assertion and reason are false.
AnswerCorrect option: D. If both assertion and reason are false.
Clearly, Reason is Correct.
Now, given quadratic equation is $x^2-x+1=0$
$ D=b^2-4 a c $
$ \Rightarrow D=(-1)^2-4(1)(1)$
$\Rightarrow D = -3 < 0$
View full question & answer→MCQ 311 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 a x^2+2 b x+c=0 $
$ \Rightarrow D=b^2-4 a c $
$ \Rightarrow D=(2 b)^2-4 \times a \times c $
$ \Rightarrow D=4 b^2-4 a c$
Roots are equal
$ \Rightarrow D=0 $
$ \Rightarrow 4 b^2-4 a c=0 $
$ \Rightarrow 4 a c=4 b^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 321 Mark
The perimeter of a right triangle is $70\ cm$ and its hypotenuse is $29\ cm$. The area of the triangle is :
- A
$200 \text{sq. cm}$
- B
$250 \text{sq. cm}$
- ✓
$210 \text{sq. cm}$
- D
$180 \text{sq. cm}$
AnswerCorrect option: C. $210 \text{sq. cm}$
Let base of the right triangle be $x \ cm$.
Given: Perpendicular $= x + 29 = 70$
$\Rightarrow $ Perpendicular $= (41- x) (41- x)\ cm$
Now, using Pythagoras theorem,
$ (29)^2=x^2+(41-x)^2 $
$ \Rightarrow 841=1681+x^2-82 x+x^2 $
$ \Rightarrow 2 x^2-82 x+840=0 $
$ \Rightarrow x^2-41 x+420=0 $
$ \Rightarrow x^2-20 x-21 x+420=0 $
$\Rightarrow x(x - 20) -21(x - 20) = 0$
$\Rightarrow (x - 20) (x - 21) = 0$
$\Rightarrow x - 20 = 0$ and $x - 21 = 0$
$\Rightarrow x = 20$ and $x = 21$
$\therefore$ The two sides other than hypotenuse are of $20\ cm$ and $21\ cm.$
View full question & answer→MCQ 331 Mark
If the product of the roots of the equation $x^2- 3x + k = 10$ is $-2$ then the value of $k$ is :
Answer$ x^2-3 x+k=10 $
$ \Rightarrow x^2-3 x+(k-10)=0$
Comparing with $ {ax}^2+ {bx}+ {c}=0,$ we have
$a = 1, b = -3, c = k - 10$
Product of the roots $= -2$
$ \Rightarrow\frac{\text{c}}{\text{a}}=-2$
$\Rightarrow\text{k}-10=-2$
$\Rightarrow\text{k}=8$
View full question & answer→MCQ 341 Mark
If the sum of a number and its reciprocal is $2\frac{1}{2}$ then the number are :
- A
$3$ and $\frac{1}{3}$
- ✓
$2$ and $\frac{1}{2}$
- C
$1$ and $\frac{3}{2}$
- D
AnswerCorrect option: B. $2$ and $\frac{1}{2}$
Let the one number be $x$ then its reciprocal will be $\frac{1}{\text{x}}$ According to question,
$\Rightarrow\frac{{\text{x}}^{2}+{1}}{\text{x}}=\frac{5}{2}$
$ \Rightarrow 2 x^2+2=5 x $
$ \Rightarrow 2 x^2-5 x+2=0$
using factorisation method
$ \Rightarrow 2 x^2-4 x-x+2=0$
$\Rightarrow 2x(x - 2) -1(x - 2) = 0$
$\Rightarrow (x - 2) (2x - 1) = 0$
$\Rightarrow x - 2 = 0$ and $2x - 1 = 0$
$\Rightarrow x = 2$ and $\text{x}=\frac{1}{2}$
$\therefore$ The number are $2$ and $\frac{1}{2}$
View full question & answer→MCQ 351 Mark
If the equation $x^2+ 5kx + 16 = 0$ has no real roots then :
AnswerCorrect option: C. $\frac{-8}{5} < \text{k} < \frac{8}{5}$
Since the equation $x^2+ 5kx + 16 = 0$ has no real roots,
$\Rightarrow D < 0$
$ \Rightarrow b^2-4 a c>0 $
$ \Rightarrow(5 k)^2-4 \times 16<0 $
$ \Rightarrow 25 k^2-64<0 $
$ \Rightarrow 25 k^2<64 $
$\Rightarrow\text{k}^2<\frac{64}{25}$
$\Rightarrow\text{k}<\sqrt{\frac{64}{25}}$ or $\text{k}>-\sqrt{\frac{64}{25}}$
$\Rightarrow\text{k}<\frac{8}{5}$ or $\text{k}>-\frac{8}{5}$
$\Rightarrow-\frac{8}{5}<\text{k}<\frac{8}{5}$
View full question & answer→MCQ 361 Mark
$2x^2- 3x + 2 = 0$ have :
Answer$D = b^2- 4ac$
$D = (-3)^2- 4 \times 2 \times 2$
$D = 9 - 16$
$D = - 7$
$D < 0$.
No Real roots.
View full question & answer→MCQ 371 Mark
Rohan’s mother is $26$ years older than him. The product of their ages $3$ years from now will be $360,$ then Rohan’s present age is :
- A
$10$ years
- B
$6$ years
- ✓
$7$ years
- D
$8$ years
AnswerCorrect option: C. $7$ years
Let Rohan’s present age be $x$ years.
Then Rohan’s mother age will be $(x + 26)$ years.
And after $3$ years their ages will be $(x + 3)$ and $(x + 29)$ years.
According to question,
$(x + 3) (x + 29) = 360$
$ \Rightarrow x^2+29 x+3 x+87=360 $
$ \Rightarrow x^2+32 x-273=0 $
$ \Rightarrow x^2+39 x+7 x-273=0 $
$\Rightarrow x(x + 39) -7(x + 39) = 0$
$\Rightarrow (x - 7) (x + 39) = 0$
$\Rightarrow (x - 7) = 0 $ and $x + 39 = 0$
$\Rightarrow x = 7$ and $x = -39 \ [x = -39$ is not possible$]$
$\therefore$ Rohan’s present is $7$ years.
View full question & answer→MCQ 381 Mark
The sum of the roots of the equation $x^2-6 x+2=0$ is :
AnswerSum of the roots of the equation of $a x^2+b x+c=0$ is $\frac{-\text{b}}{\text{a}}$
Here $, a = 1, b = -6, c = 2$
By substitution of values we get
$\frac{-\text{b}}{\text{a}}=\frac{-(-6)}{1} = 6$
View full question & answer→MCQ 391 Mark
A quadratic equation $ax^2+ bx + c = 0$ has real and equal roots, if :
- A
$ b^2-4 a c>0 $
- ✓
$ b^2-4 a c=0 $
- C
$ b^2-4 a c<0 $
- D
AnswerCorrect option: B. $ b^2-4 a c=0 $
A quadratic equation $ax^2+ bx + c = 0$ has real and equal roots, if $ b^2-4 a c=0 $
View full question & answer→MCQ 401 Mark
If the equation $9x^2+ 6kx + 4 = 0$ has equal roots then $k = ?$
- A
$2$ or $0$
- B
$-2$ or $0$
- ✓
$2$ or $-2$
- D
$0$ only
AnswerCorrect option: C. $2$ or $-2$
Since the roots of the equation $9x^2+ 6kx + 4 = 0$ are equal,
$D = 0$
$ \Rightarrow b^2-4 a c=0 $
$ \Rightarrow(6 k)^2-4 \times 9 \times 4=0 $
$ \Rightarrow 36 k^2-144=0 $
$ \Rightarrow 36 k^2=144 $
$ \Rightarrow k^2=4 $
$\Rightarrow\text{k}=\pm2$
View full question & answer→MCQ 411 Mark
The ratio of sum and the product of the roots of $7 x^2-12 x+18=0$ is :
- A
$3 : 2$
- B
$7 : 12$
- C
$7 : 18$
- ✓
$2 : 3$
AnswerCorrect option: D. $2 : 3$
Ratio of sum and product of the roots $7 x^2-12 x+18=0$ is $\frac{\alpha+\beta}{\alpha\beta}$
$\Rightarrow\frac{-\text{b}}{\text{c}}$
$\Rightarrow \frac{12}{18}=\frac{2}{3} = 2:3$
View full question & answer→MCQ 421 Mark
The sum of two number is $17$ and the sum of their reciprocals is $\frac{17}{36}$ The quadratic representation of the above situation is :
- A
$\frac{1}{\text{x}} - \frac{1}{17-\text{ x}} = \frac{17}{62}$
- B
$\frac{1}{\text{x}(17 - \text{x)}}=\frac{17}{62}$
- ✓
$\frac{1}{\text{x}} + \frac{1}{17-\text{ x}} = \frac{17}{62}$
- D
$\frac{1}{\text{x}} + \frac{1}{\text{x+17}} = \frac{17}{62}$
AnswerCorrect option: C. $\frac{1}{\text{x}} + \frac{1}{17-\text{ x}} = \frac{17}{62}$
Let one number be $x,$ as the sum of the number is $17,$ then the other number will be $(17 - x) $
their reciprocals will be $\frac{1}{\text{x}}$ and $\frac{1}{17-\text{x}}$
$\therefore$ According to question, $\frac{1}{\text{x}} + \frac{1}{17-\text{ x}} = \frac{17}{62}$
View full question & answer→MCQ 431 Mark
If one root of the equation $2x^2+ ax + 6 = 0$ is $2$ then $a =$ ?
- A
$\frac{-7}{2}$
- B
$\frac{7}{2}$
- ✓
$-7$
- D
$7$
AnswerOne root of the equation $2 x^2+a x+6=0$ is $2$
i.e. it satisfies the equation
$2(2)^2+2 a+6=0$
$8 + 2a + 6 = 0$
$2a = -14$
$a = -7$
View full question & answer→MCQ 441 Mark
If one root of $5 x^2+13 x+k=0$ be the reciprocal of the other root, then the value of $k$ is :
AnswerLet one root of the given equation be $\alpha.$
Then, its root will be $\frac{1}{\alpha}.$
Given equation is $5 x^2+13 x+k=0$
Comparing with $a x^2+b x+c=0,$
we have $a = 5, b = 13, c = k$
Now,
Product of the roots $=\frac{\text{c}}{\text{a}}$
$\Rightarrow\alpha\times\frac{1}{\alpha}=\frac{\text{k}}5{}$
$\Rightarrow\text{k}=5$
View full question & answer→MCQ 451 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 :
AnswerThe 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 \ldots \text { (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 461 Mark
$......$ is called the Discriminant of the quadratic equation $ax^2+ bx + c = 0:$
- A
$ a^2-4 b c $
- ✓
$ b^2-4 a c $
- C
$ c^2-4 a b $
- D
AnswerCorrect option: B. $ b^2-4 a c $
Discriminant of the quadratic equation $ax^2+ bx + c = 0$ is $D = b^2- 4ac$.
View full question & answer→MCQ 471 Mark
$2$ One of the roots of the quadratic equation $ a^2 x^2-2 a b x+2 b^2=0$ is :
- A
$\frac{\text{-2b}}{\text{a}}$
- B
$\frac{\text{-2a}}{\text{b}}$
- ✓
$\frac{\text{2b}}{\text{a}}$
- D
$\frac{\text{2a}}{\text{b}}$
AnswerCorrect option: C. $\frac{\text{2b}}{\text{a}}$
$\Rightarrow a^2 x^2-2 a b x+2 b^2=0$
$\Rightarrow ax(ax - 2b) -b(ax - 2b) = 0$
$\Rightarrow (ax - b) (ax - 2b) = 0$
$\Rightarrow ax - b = 0$ and $ax - 2b = 0$
$\Rightarrow\text{x} = \frac{\text{b}}{\text{a}}$ and $\text{x} = \frac{\text{2b}}{\text{a}}$
View full question & answer→MCQ 481 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 $16 x^2+4 k x+9=0,$ 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, $16 k^2-576=0$
$\Rightarrow \mathrm{k}^2-36=0$
$\Rightarrow (k + 6)(k - 6) = 0$
$\Rightarrow k + 6 = 0$ or $k = 6$
Therefore, the value of $k$ is $6, -6$
Hence, the correct option is $(c)$
View full question & answer→MCQ 491 Mark
Directions : In the following questions, the Assertions $(A)$ and Reason $(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion : The roots of the quadratic equation $x^2+ 2x + 2 = 0$ are imaginary.
Reason : If discriminant $D = b^2- 4ac < 0$ then the roots of quadratic equation $ax^2+ bx + c = 0$ are imaginary.
- A
If both assertion and reason are true and reason is the correct explanation of assertion.
- ✓
If both assertion and reason are true but reason is not the correct explanation of assertion.
- C
If assertion is true but reason is false.
- D
If both assertion and reason are false.
AnswerCorrect option: B. If both assertion and reason are true but reason is not the correct explanation of assertion.
$x^2+2 x+2=0$
Discriminant, $D=b^2-4 a c$
$\Rightarrow D=(2)^2-4(1)(c)$
$\Rightarrow D = 4 - 8 = -4 < 0$
Roots are imaginary.
View full question & answer→MCQ 501 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 $x^2-a x+1=0$, and roots are dostinct.
Then fond the value of a.
Here $, a = 1, b = a$ and $c = 1$
As we know that $D = D=b^2-4 a c$
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→