Quadratic Equations jee mains questions | JEE Main Maths Quadratic Equations Previous Year Question

Q 1 :

Let the sum of the maximum and the minimum values of the function $f (x) = \frac{2 x^{2} - 3 x + 8}{2 x^{2} + 3 x + 8}$ be $\frac{m}{n}$ , where $g c d (m, n) = 1 .$ Then $m + n$ is equal to [2024]

182
217
201
195

1

2

3

4

(3)

Let $f (x) = \frac{2 x^{2} - 3 x + 8}{2 x^{2} + 3 x + 8} = y,$ $2 x^{2} + 3 x + 8 > 0 \forall x \in R$

$\Rightarrow x^{2} (2 y - 2) + x (3 y + 3) + 8 y - 8 = 0$

For real roots, $D \geq 0$

$\Rightarrow {(3 y + 3)}^{2} - 4 (2 y - 2) (8 y - 8) \geq 0$

$\Rightarrow {(3 y + 3)}^{2} - {(8 y - 8)}^{2} \geq 0$

$\Rightarrow (11 y - 5) (- 5 y + 11) \geq 0$

$\Rightarrow (y - \frac{5}{11}) (y - \frac{11}{5}) \leq 0 \Rightarrow y \in [\frac{5}{11}, \frac{11}{5}]$

Sum of maximum and minimum values

$y_{max} + y_{min} = \frac{5}{11} + \frac{11}{5} = \frac{146}{55} = \frac{m}{n} \Rightarrow m + n = 201$

Q 2 :

The coefficients $a, b, c$ in the quadratic equation $a x^{2} + b x + c = 0$ are chosen from the set {1, 2, 3, 4, 5, 6, 7, 8}. The probability of this equation having repeated roots is: [2024]

$\frac{3}{256}$
$\frac{1}{128}$
$\frac{1}{64}$
$\frac{3}{128}$

1

2

3

4

(3)

For equation $a x^{2} + b x + c = 0$ to have repeated roots we have

$D = b^{2} - 4 a c = 0$

$\Rightarrow b^{2} = 4 a c = 2^{2} (a c)$

$\Rightarrow a c$ must be a perfect square.

$\Rightarrow (a, c) = {(1, 1), (1, 4), (2, 2), (2, 8), (3, 3), (4, 1), (4, 4), (5, 5), (6, 6), (7, 7), (8, 2), (8, 8)}$

$∴ (a, b, c) = {(1, 2, 1), (1, 4, 4), (2, 4, 2), (2, 8, 8), (3, 6, 3), (4, 4, 1), (4, 8, 4), (8, 8, 2)}$

Required probability $= \frac{8}{8^{3}} = \frac{1}{64}$

Q 3 :

Let $α, β$ be the distinct roots of the equation $x^{2} - (t^{2} - 5 t + 6) x + 1 = 0, t \in R$ and $a_{n} = α^{n} + β^{n}$ . Then the minimum value of $\frac{a_{2023} + a_{2025}}{a_{2024}}$ is [2024]

$- 1 / 2$
$1 / 2$
$1 / 4$
$- 1 / 4$

1

2

3

4

(4)

Newton's Theorem says that for a quadratic equation $a x^{2} + b x + c = 0$ if $a$ and $b$ are its roots and $a_{n} = α^{n} \pm β^{n},$ then $a a_{n + 1} + b a_{n} + c a_{n - 1} = 0 .$

So, by Newton's theorem, we have

$a_{2025} - (t^{2} - 5 t + 6) a_{2024} + a_{2023} = 0$

$\Rightarrow t^{2} - 5 t + 6 = \frac{a_{2025} + a_{2023}}{a_{2024}}$

Now, $t^{2} - 5 t + 6 = {(t - \frac{5}{2})}^{2} - \frac{1}{4}$

Minimum value $= - \frac{1}{4}$

Q 4 :

Let $S$ be the set of positive integral values of $a$ for which $\frac{a x^{2} + 2 (a + 1) x + 9 a + 4}{x^{2} - 8 x + 32} < 0, \forall x \in R .$ Then, the number of elements is $S$ is [2024]

$\infty$
$0$
$1$
$3$

1

2

3

4

(2)

Given $\frac{a x^{2} + 2 (a + 1) x + 9 a + 4}{x^{2} - 8 x + 32} < 0, \forall x \in R$

For $x^{2} - 8 x + 32,$

$D = b^{2} - 4 a c, D = {(- 8)}^{2} - 4 (1) (32), D = 64 - 128,$

$D = - 64 < 0, a > 0$

So, $x^{2} - 8 x + 32 > 0,$ when

$∴$ $a x^{2} + 2 (a + 1) x + 9 a + 4 < 0, \forall x \in R$

$a < 0 and D < 0$

Here, $S$ is the set of positive integral values of $x,$ which is not possible.

So, the number of elements in $S = 0$

Q 5 :

The coefficients $a, b, c$ in the quadratic equation $a x^{2} + b x + c = 0$ are from the set {1, 2, 3, 4, 5, 6}. If the probability of this equation having one real root bigger than the other is $p$ , then $216 p$ equals: [2024]

76
38
57
19

1

2

3

4

(2)

We need to find the probability that the given equation $a x^{2} + b x + c = 0$ has real and distinct roots.

[ $∵$ One root will be bigger if roots are distinct]

$∴$ $D > 0$

$\Rightarrow b^{2} - 4 a c > 0$

$b \neq 1, 2$ $∵$ if $b = 1$ or $2$

then $a c < \frac{1}{4}$ or $a c < 1$ which is not possible as

$a, b, c \in {1, 2, 3, 4, 5, 6} .$

If $b = 3$

$\Rightarrow a c < \frac{9}{4} \Rightarrow (a, c) = {(1, 1), (1, 2), (2, 1)}$

i.e., 3 ways

If $b = 4$

$\Rightarrow a c < 4 \Rightarrow (a, c) = {(1, 1), (1, 2), (2, 1), (3, 1), (1, 3)} .$

i.e., 5 ways

If $b = 5$

$\Rightarrow a c < \frac{25}{4} \Rightarrow (a, c) = {(1, 1), (1, 2), (2, 1), (3, 1), (1, 3), (2, 2), (1, 4), (1, 5), (2, 3), (3, 2), (4, 1), (5, 1), (6, 1), (1, 6)} .$

i.e., 14 ways

If $b = 6$

$\Rightarrow a c < 9 \Rightarrow (a, c) = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (4, 1), (4, 2), (5, 1), (6, 1)}$

i.e., 16 ways

$∴$ Total number of ways = 3 + 5 + 14 + 16 = 38

$\Rightarrow$ Required probability $= \frac{38}{6^{3}} = p$

So, $216 p = \frac{38}{216} \times 216 = 38$

Q 6 :

The number of distinct real roots of the equation $| x | | x + 2 | - 5 | x + 1 | - 1 = 0$ is _________ [2024]

(3)

$f (x) = | x | | x + 2 | - 5 | x + 1 | - 1 = 0$

Case 1 : $x \geq 0$

$∴$ $f (x) = x (x + 2) - 5 (x + 1) - 1 = 0$

$\Rightarrow x^{2} + 2 x - 5 x - 6 = 0 \Rightarrow x^{2} - 3 x - 6 = 0$

$\Rightarrow x = \frac{3 \pm \sqrt{9 + 24}}{2}$

$\Rightarrow x = \frac{3 \pm \sqrt{33}}{2},$ One positive root [ $∵$ $x \geq 0$ ]

Case 2: $- 1 \leq x < 0$

$\Rightarrow f (x) = - x (x + 2) - 5 (x + 1) - 1 = 0$

$\Rightarrow - x^{2} - 7 x - 6 = 0 \Rightarrow x^{2} + 7 x + 6 = 0$

$\Rightarrow (x + 6) (x + 1) = 0$

$\Rightarrow x = - 6, - 1$ One root i.e., $x = - 1$

Case 3 : $- 2 \leq x < - 1$

$f (x) = - x (x + 2) + 5 (x + 1) - 1 = 0$

$\Rightarrow - x^{2} - 2 x + 5 x + 4 = 0$

$\Rightarrow - x^{2} + 3 x + 4 = 0 \Rightarrow x^{2} - 3 x - 4 = 0$

$\Rightarrow (x + 1) (x - 4) = 0 \Rightarrow x = - 1, 4$

No root possible in given range of $x$ .

Case 4 : $x < - 2$

$f (x) = - x (- x - 2) + 5 (x + 1) - 1 = 0$

$\Rightarrow x^{2} + 2 x + 5 x + 4 = 0$

$\Rightarrow x^{2} + 7 x + 4 = 0$

$\Rightarrow x = \frac{- 7 \pm \sqrt{49 - 16}}{2} = \frac{- 7 \pm \sqrt{33}}{2}$

One root in the range.

$∴$ We have 3 distinct real roots

Q 7 :

The number of real solutions of the equation $x | x + 5 | + 2 | x + 7 | - 2 = 0$ is _______. [2024]

(3)

Let $f (x) = x | x + 5 | + 2 | x + 7 | - 2$ and $f (x) = 0$

Case 1 : $x \geq - 5$

$\Rightarrow x^{2} + 5 x + 2 x + 12 = 0$

$\Rightarrow x^{2} + 7 x + 12 = 0 \Rightarrow x = - 3 or x = - 4$

Case 2 : $- 7 < x < - 5$

$\Rightarrow - x^{2} - 5 x + 2 x + 12 = 0$

$\Rightarrow - x^{2} - 3 x + 12 = 0 \Rightarrow x^{2} + 3 x - 12 = 0$

$\Rightarrow x = \frac{- 3 \pm \sqrt{57}}{2}$

$\Rightarrow x \neq \frac{- 3 + \sqrt{57}}{2}$ $[∵$ $- 7 < x < - 5]$

Case 3 : $x \leq - 7$

$\Rightarrow - x^{2} - 7 x - 16 = 0 \Rightarrow x^{2} + 7 x + 16 = 0$

No real solution

So Number of real solutions = 3

Q 8 :

The number of distinct real roots of the equation $| x + 1 | | x + 3 | - 4 | x + 2 | + 5 = 0$ is ________. [2024]

(2)

$| x + 1 | | x + 3 | - 4 | x + 2 | + 5 = 0$

$(I) If x < - 3, x^{2} + 4 x + 3 + 4 x + 8 + 5 = 0$

$\Rightarrow x^{2} + 8 x + 16 = 0 \Rightarrow x = - 4 (one solution)$

$(II) If - 3 \leq x < - 2, - x^{2} - 4 x - 3 + 4 x + 8 + 5 = 0$

$\Rightarrow x^{2} - 10 = 0 \Rightarrow x = \pm \sqrt{10}$

$which do not satisfy - 3 \leq x < - 2$

$(III) If - 2 \leq x < - 1, - x^{2} - 4 x - 3 - 4 x - 8 + 5 = 0$

$\Rightarrow x^{2} + 8 x + 6 = 0 \Rightarrow {(x + 4)}^{2} = 10$

$\Rightarrow x = - 4 \pm \sqrt{10}$ which do not satisfy $- 2 \leq x < - 1$

$(IV) If x \geq - 1, x^{2} + 4 x + 3 - 4 x - 8 + 5 = 0$

$\Rightarrow x^{2} = 0 \Rightarrow x = 0 (one solution)$

$Hence, the number of distinct real roots are two.$

Q 9 :

Let the set of all values of $p \in R$ , for which both the roots of the equation $x^{2} - (p + 2) x + (2 p + 9) = 0$ are negative real numbers, be the interval $(α, β]$ . Then $β - 2 α$ is equal to [2025]

5
0
20
9

1

2

3

4

(1)

Given, $x^{2} - (p + 2) x + (2 p + 9) = 0$

$D \geq 0$

$\Rightarrow {(p + 2)}^{2} - 4 (2 p + 9) \geq 0$

$\Rightarrow p^{2} + 4 p + 4 - 8 p - 36 \geq 0$

$\Rightarrow p^{2} - 4 p - 32 \geq 0$

$\Rightarrow (p - 8) (p + 4) \geq 0$

$∴ p \in (- \infty, - 4] \cup [8, \infty)$ ... (i)

So, sum of roots : p + 2 < 0

$\Rightarrow$ p < –2 ... (ii)

and product of roots : 2p + 9 > 0

$\Rightarrow p > \frac{- 9}{2}$ ... (iii)

Using (i), (ii) and (iii), we get

$p \in (\frac{- 9}{2}, - 4]$

$∴ β - 2 α = - 4 - 2 (\frac{- 9}{2}) = 5$ .

Q 10 :

The number of real roots of the equation $x$ $| x - 2 |$ $+ 3$ $| x - 3 |$ $+ 1 = 0$ [2025]

3
2
4
1

1

2

3

4

(4)

(I) When x < 2, we have

$- x^{2} + 2 x - 3 x + 9 + 1 = 0$

$\Rightarrow x^{2} + x - 10 = 0$

$\Rightarrow x = \frac{- 1 + \sqrt{41}}{2}, \frac{- 1 - \sqrt{41}}{2}; x = \frac{- 1 - \sqrt{41}}{2}$ ( $∵$ x < 2)

(II) When $2 \leq x < 3$ , we have

$x^{2} - 2 x - 3 x + 9 + 1 = 0$

$\Rightarrow x^{2} - 5 x + 10 = 0$

As D < 0 $\Rightarrow$ No real roots.

(III) When $x \geq 3$ , we have

$x^{2} - 2 x + 3 x - 9 + 1 = 0$

$\Rightarrow x^{2} + x - 8 = 0$

$\Rightarrow x = \frac{- 1 + \sqrt{33}}{2}, \frac{- 1 - \sqrt{33}}{2}$ (rejected)

Thus, only one real root exists.

Q 11 :

The sum of the squares of the roots of $| x - 2 |^{2} + | x - 2 | - 2 = 0$ and the squares of the roots of $x^{2} - 2 | x - 3 | - 5 = 0$ is [2025]

24
36
30
26

1

2

3

4

(2)

We have, $| x - 2 |^{2} + | x - 2 | - 2 = 0$

$\Rightarrow | x - 2 |^{2} + 2 | x - 2 | - | x - 2 | - 2 = 0$

$\Rightarrow (| x - 2 | + 2) (| x - 2 | - 1) = 0$

$\Rightarrow | (x - 2) | = 1$ [ $∵ | x - 2 | \neq - 2$ ]

$\Rightarrow x = 2 \pm 1 \Rightarrow x = 3, 1$

Sum of square of roots = 9 + 1 = 10

Now, we have $x^{2} - 2 | x - 3 | - 5 = 0$

Case I : When x – 3 > 0

$\Rightarrow x^{2} - 2 x + 1 = 0 \Rightarrow {(x - 1)}^{2} = 0 \Rightarrow x = 1$

but x > 3 $\Rightarrow x \neq 1$

Case II : When x – 3 < 0

$\Rightarrow x^{2} + 2 x - 11 = 0$

Discriminant, D = 4 + 44 = 48 > 0

$x = \frac{- 2 \pm \sqrt{48}}{2} = \frac{- 2 \pm 4 \sqrt{3}}{2} = - 1 \pm 2 \sqrt{3}$

SInce, x < 3, so both roots are valid.

Sum of squares of roots = ${(- 1 + 2 \sqrt{3})}^{2} + {(- 1 - 2 \sqrt{3})}^{2}$

$= 1 + 12 - 4 \sqrt{3} + 1 + 12 + 4 \sqrt{3} = 26$

$∴$ Required sum = 10 + 26 = 36.

Q 12 :

If the set of all $a \in R - {1}$ , for which the roots of the equation $(1 - a) x^{2} + 2 (a - 3) x + 9 = 0$ are positive is $(- \infty, - α] \cup [β, γ)$ , then $2 α + β + γ$ is equal to __________. [2025]

(7)

Let $x_{1}$ and $x_{2}$ be the roots of $(1 - a) x^{2} + 2 (a - 3) x + 9 = 0$ .

Since, $x_{1}, x_{2} > 0$

$∴ x_{1} + x_{2} > 0 and x_{1} x_{2} > 0$

$\Rightarrow \frac{- 2 (a - 3)}{1 - a} > 0 \Rightarrow \frac{a - 3}{1 - a} < 0$

$\Rightarrow a \in (- \infty, 1) \cup (3, \infty)$

Also, $x_{1} x_{2} > 0 \Rightarrow \frac{9}{1 - a} > 0 \Rightarrow 1 - a > 0 \Rightarrow a < 1$

On combining both conditions, we get $a \in (- \infty, 1)$

Now, for real roots, discriminant must be non negative.

i.e., $(2 {(a - 3))}^{2} - 4 (1 - a) 9 \geq 0$

$\Rightarrow 4 (a^{2} - 6 a + 9) - 36 + 36 a \geq 0$

$\Rightarrow 4 a^{2} - 24 a + 36 - 36 + 36 a \geq 0$

$\Rightarrow 4 a^{2} + 12 a \geq 0 \Rightarrow a (a + 3) \geq 0$

$\Rightarrow a \in (- \infty, - 3] \cup [0, \infty)$

Combining all the conditions, we get $a \in (- \infty, - 3] \cup [0, 1)$

$∴ α = 3, β = 0 and γ = 1$

$∴ 2 α + β + γ = 2 \times 3 + 0 + 1 = 7$

Q 13 :

The set of all $a \in ℝ$ for which the equation $x$ $| x - 1 |$ $+$ $| x + 2 |$ $+ a = 0$ has exactly one real root, is [2023]

$(- \infty, \infty)$
$(- 6, \infty)$
$(- \infty, - 3)$
$(- 6, - 3)$

1

2

3

4

(1)

Let $f (x) = x$ $| x - 1 |$ $+$ $| x + 2 |$

Given, $x$ $| x - 1 |$ $+$ $| x + 2 |$ $+ a = 0$

$\Rightarrow f (x) + a = 0 \Rightarrow - a = f (x)$

$f (x) =$ ${\begin{matrix} - x^{2} - 2, & - \infty < x < - 2 \\ - x^{2} + 2 x + 2, & - 2 \leq x < 1 \\ x^{2} + 2, & 1 \leq x < \infty \end{matrix}$

All values are increasing.

Q 14 :

The number of real roots of the equation $x$ $| x |$ $- 5$ $| x + 2 |$ $+ 6 = 0,$ is [2023]

5
3
4
6

1

2

3

4

(2)

We have, $x$ $| x |$ $- 5$ $| x + 2 |$ $+ 6 = 0$

Case I: $x \leq - 2$

$- x^{2} + 5 x + 10 + 6 = 0$

$\Rightarrow - x^{2} + 5 x + 16 = 0 \Rightarrow x^{2} - 5 x - 16 = 0$

$\Rightarrow x = \frac{5 \pm \sqrt{{(- 5)}^{2} + 16 \times 4}}{2} = \frac{5 \pm \sqrt{25 + 64}}{2} = \frac{5 \pm \sqrt{89}}{2}$

Only $x = \frac{5 - \sqrt{89}}{2}$ belongs to $(- \infty, - 2]$

So, one solution exists in this case.

Case II: $- 2 < x \leq 0$

$- x^{2} - 5 x - 10 + 6 = 0$

$\Rightarrow - x^{2} - 5 x - 4 = 0 \Rightarrow x^{2} + 5 x + 4 = 0$

$\Rightarrow (x + 1) (x + 4) = 0 \Rightarrow x = - 1 or x = - 4$

Since $- 2 < x \leq 0$ , $x = - 4$ is not a solution.

So, $x = - 1$ is a solution in this case.

Case III: $x > 0$

$x^{2} - 5 x - 4 = 0 \Rightarrow x = \frac{5 \pm \sqrt{25 + 16}}{2} = \frac{5 \pm \sqrt{41}}{2}$

But $x > 0$ . So, $x = \frac{5 + \sqrt{41}}{2}$ is the only solution in this case.

Therefore, the equation has three solutions.

Q 15 :

The number of integral values of $k$ , for which one root of the equation $2 x^{2} - 8 x + k = 0$ lies in the interval (1, 2) and its other root lies in the interval (2, 3), is [2023]

2
0
1
3

1

2

3

4

(3)

$2 x^{2} - 8 x + k = 0$ represents an upward parabola.

$f (1) \cdot f (2) < 0 and f (2) \cdot f (3) < 0$

$(2 {(1)}^{2} - 8 (1) + k) \cdot (2 {(2)}^{2} - 8 (2) + k) < 0$

$and (2 {(2)}^{2} - 8 (2) + k) (2 {(3)}^{2} - 8 (3) + k) < 0$

$(k - 6) (k - 8) < 0 and (k - 8) (k - 6) < 0$

$k \in (6, 8) and k \in (6, 8)$

$Integral value of k = 7$

Q 16 :

Let $m$ and $n$ be the numbers of real roots of the quadratic equations $x^{2} - 12 x + [x] + 31 = 0$ and $x^{2} - 5$ $| x + 2 |$ $- 4 = 0$ respectively, where $[x]$ denotes the greatest integer $\leq x$ . Then $m^{2} + m n + n^{2}$ is equal to __________ . [2023]

(9)

$x^{2} - 12 x + [x] + 31 = 0$

$\Rightarrow {x} = x^{2} - 11 x + 31 \Rightarrow 0 \leq x^{2} - 11 x + 31 < 1$

$\Rightarrow x^{2} - 11 x + 30 < 0 \Rightarrow x \in (5, 6)$

$So, [x] = 5$

$Now, x^{2} - 12 x + 5 + 31 = 0$

$\Rightarrow x^{2} - 12 x + 36 = 0 \Rightarrow x = 6 but x \in (5, 6)$

$m = 0$

$Now, for x^{2} - 5$ $| x + 2 |$ $- 4 = 0$

When

$x \geq - 2;$ $x < - 2$

$x^{2} - 5 x - 14 = 0;$ $x^{2} + 5 x + 6 = 0$

$(x - 7) (x + 2) = 0;$ $(x + 3) (x + 2) = 0$

$x = 7, - 2;$ $x = - 3, - 2$

$Now, x^{2} - 5$ $| x + 2 |$ $- 4 = 0 \Rightarrow x = {7, - 2, - 3}$ $∴ n = 3$

$So, m^{2} + m n + n^{2} = 9$

Q 17 :

Let $S = {α : \log_{2} (9^{2 α - 4} + 13) - \log_{2} (\frac{5}{3} \cdot 3^{2 α - 4} + 1) = 2} .$ Then the maximum value of $β$ for which the equation

$x^{2} - 2 {(\sum_{α \in S} α)}^{2} x + \sum_{α \in S} {(α + 1)}^{2} β = 0$ has real roots, is ________ . [2023]

(25)

$Given \log_{2} (9^{2 α - 4} + 13) - \log_{2} (\frac{5}{2} \cdot 3^{2 α - 4} + 1) = 2$

$\Rightarrow \log_{2} (\frac{9^{2 α - 4} + 13}{\frac{5}{2} \cdot 3^{2 α - 4} + 1}) = 2 \Rightarrow \frac{9^{2 α - 4} + 13}{\frac{5}{2} \cdot 3^{2 α - 4} + 1} = 4$

$\Rightarrow 9^{2 α - 4} + 13 = 10 \cdot 3^{2 α - 4} + 4 \Rightarrow 10 \cdot 3^{2 α - 4} - 9^{2 α - 4} = 9$

$Put α = 1, 10 \cdot 3^{- 2} - 9^{- 2} = \frac{10}{9} - \frac{1}{81} = \frac{89}{81} \neq 9 (not satisfy)$

$Put α = 2, 10 \cdot 3^{0} - 9^{0} = 10 - 1 = 9 (satisfy)$

$Put α = 3, 10 \cdot 3^{2} - 9^{2} = 90 - 81 = 9 (satisfy)$

$Hence, α = 2 or 3 .$

Now,

$\sum_{α \in S} α = 2 + 3 = 5$ and $\sum_{α \in S} {(α + 1)}^{2}$ $= \sum_{α \in S} α^{2} + 2 \sum_{α \in S} α + \sum_{α \in S} 1$

$= 4 + 9 + 2 (5) + 2 = 25$

So, $x^{2} - 2 {(5)}^{2} x + 25 β = 0$ $= x^{2} - 50 x + 25 β = 0$

For real roots, $D \geq 0$

$\Rightarrow 2500 - 4 (25 β) \geq 0 \Rightarrow 100 β \leq 2500 \Rightarrow β \leq 25$

$Hence, β_{\max} = 25$

Topic Question Set

Q 1 :

Let the sum of the maximum and the minimum values of the function $f (x) = \frac{2 x^{2} - 3 x + 8}{2 x^{2} + 3 x + 8}$ be $\frac{m}{n}$ , where $g c d (m, n) = 1 .$ Then $m + n$ is equal to [2024]

Q 2 :

The coefficients $a, b, c$ in the quadratic equation $a x^{2} + b x + c = 0$ are chosen from the set {1, 2, 3, 4, 5, 6, 7, 8}. The probability of this equation having repeated roots is: [2024]

Q 3 :

Let $α, β$ be the distinct roots of the equation $x^{2} - (t^{2} - 5 t + 6) x + 1 = 0, t \in R$ and $a_{n} = α^{n} + β^{n}$ . Then the minimum value of $\frac{a_{2023} + a_{2025}}{a_{2024}}$ is [2024]

Q 4 :

Let $S$ be the set of positive integral values of $a$ for which $\frac{a x^{2} + 2 (a + 1) x + 9 a + 4}{x^{2} - 8 x + 32} < 0, \forall x \in R .$ Then, the number of elements is $S$ is [2024]

Q 5 :

The coefficients $a, b, c$ in the quadratic equation $a x^{2} + b x + c = 0$ are from the set {1, 2, 3, 4, 5, 6}. If the probability of this equation having one real root bigger than the other is $p$ , then $216 p$ equals: [2024]

Q 6 :

The number of distinct real roots of the equation $| x | | x + 2 | - 5 | x + 1 | - 1 = 0$ is _________ [2024]

Q 7 :

The number of real solutions of the equation $x | x + 5 | + 2 | x + 7 | - 2 = 0$ is _______. [2024]

Q 8 :

The number of distinct real roots of the equation $| x + 1 | | x + 3 | - 4 | x + 2 | + 5 = 0$ is ________. [2024]

Q 9 :

Let the set of all values of $p \in R$ , for which both the roots of the equation $x^{2} - (p + 2) x + (2 p + 9) = 0$ are negative real numbers, be the interval $(α, β]$ . Then $β - 2 α$ is equal to [2025]

Q 10 :

The number of real roots of the equation $x$ $| x - 2 |$ $+ 3$ $| x - 3 |$ $+ 1 = 0$ [2025]

Q 11 :

The sum of the squares of the roots of $| x - 2 |^{2} + | x - 2 | - 2 = 0$ and the squares of the roots of $x^{2} - 2 | x - 3 | - 5 = 0$ is [2025]

Q 12 :

If the set of all $a \in R - {1}$ , for which the roots of the equation $(1 - a) x^{2} + 2 (a - 3) x + 9 = 0$ are positive is $(- \infty, - α] \cup [β, γ)$ , then $2 α + β + γ$ is equal to __________. [2025]

Q 13 :

The set of all $a \in ℝ$ for which the equation $x$ $| x - 1 |$ $+$ $| x + 2 |$ $+ a = 0$ has exactly one real root, is [2023]

Q 14 :

The number of real roots of the equation $x$ $| x |$ $- 5$ $| x + 2 |$ $+ 6 = 0,$ is [2023]

Q 15 :

The number of integral values of $k$ , for which one root of the equation $2 x^{2} - 8 x + k = 0$ lies in the interval (1, 2) and its other root lies in the interval (2, 3), is [2023]

Q 16 :

Let $m$ and $n$ be the numbers of real roots of the quadratic equations $x^{2} - 12 x + [x] + 31 = 0$ and $x^{2} - 5$ $| x + 2 |$ $- 4 = 0$ respectively, where $[x]$ denotes the greatest integer $\leq x$ . Then $m^{2} + m n + n^{2}$ is equal to __________ . [2023]

Q 17 :

Let $S = {α : \log_{2} (9^{2 α - 4} + 13) - \log_{2} (\frac{5}{3} \cdot 3^{2 α - 4} + 1) = 2} .$ Then the maximum value of $β$ for which the equation

$x^{2} - 2 {(\sum_{α \in S} α)}^{2} x + \sum_{α \in S} {(α + 1)}^{2} β = 0$ has real roots, is ________ . [2023]

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 1 : Let the sum of the maximum and the minimum values of the function f(x)=2x2-3x+82x2+3x+8 be mn, where gcd(m,n)=1. Then m+n is equal to [2024]

Q 2 : The coefficients a,b,c in the quadratic equation ax2+bx+c=0 are chosen from the set {1, 2, 3, 4, 5, 6, 7, 8}. The probability of this equation having repeated roots is: [2024]

Q 3 : Let α,β be the distinct roots of the equation x2-(t2-5t+6)x+1=0, t∈R and an=αn+βn. Then the minimum value of a2023+a2025a2024 is [2024]

Q 4 : Let S be the set of positive integral values of a for which ax2+2(a+1)x+9a+4x2-8x+32<0, ∀ x∈R. Then, the number of elements is S is [2024]

Q 5 : The coefficients a,b,c in the quadratic equation ax2+bx+c=0 are from the set {1, 2, 3, 4, 5, 6}. If the probability of this equation having one real root bigger than the other is p, then 216p equals: [2024]

Q 6 : The number of distinct real roots of the equation |x| |x+2|-5|x+1|-1=0 is _________ [2024]

Q 7 : The number of real solutions of the equation x|x+5|+2|x+7|-2=0 is _______. [2024]

Q 8 : The number of distinct real roots of the equation |x+1||x+3|-4|x+2|+5=0 is ________. [2024]

Q 9 : Let the set of all values of p∈R, for which both the roots of the equation x2–(p+2)x+(2p+9)=0 are negative real numbers, be the interval (α,β]. Then β–2α is equal to [2025]

Q 10 : The number of real roots of the equation x|x–2|+3|x–3|+1=0 [2025]

Q 11 : The sum of the squares of the roots of |x–2|2+|x–2|–2=0 and the squares of the roots of x2–2|x–3|–5=0 is [2025]

Q 12 : If the set of all a∈R–{1}, for which the roots of the equation (1–a)x2+2(a–3)x+9=0 are positive is (–∞,–α]∪[β,γ), then 2α+β+γ is equal to __________. [2025]

Q 13 : The set of all a∈ℝ for which the equation x|x-1|+|x+2|+a=0 has exactly one real root, is [2023]

Q 14 : The number of real roots of the equation x|x|-5|x+2|+6=0, is [2023]

Q 15 : The number of integral values of k, for which one root of the equation 2x2-8x+k=0 lies in the interval (1, 2) and its other root lies in the interval (2, 3), is [2023]

Q 16 : Let m and n be the numbers of real roots of the quadratic equations x2-12x+[x]+31=0 and x2-5|x+2|-4=0 respectively, where [x] denotes the greatest integer ≤x. Then m2+mn+n2 is equal to __________ . [2023]

Q 17 : Let S={α:log2(92α-4+13)-log2(53·32α-4+1)=2}. Then the maximum value of β for which the equation x2-2(∑α∈Sα)2x+∑α∈S(α+1)2β=0 has real roots, is ________ . [2023]

Want Us to Email you About Special Offers & Updates?

Q 1 :

Let the sum of the maximum and the minimum values of the function $f (x) = \frac{2 x^{2} - 3 x + 8}{2 x^{2} + 3 x + 8}$ be $\frac{m}{n}$ , where $g c d (m, n) = 1 .$ Then $m + n$ is equal to [2024]

Q 2 :

The coefficients $a, b, c$ in the quadratic equation $a x^{2} + b x + c = 0$ are chosen from the set {1, 2, 3, 4, 5, 6, 7, 8}. The probability of this equation having repeated roots is: [2024]

Q 3 :

Let $α, β$ be the distinct roots of the equation $x^{2} - (t^{2} - 5 t + 6) x + 1 = 0, t \in R$ and $a_{n} = α^{n} + β^{n}$ . Then the minimum value of $\frac{a_{2023} + a_{2025}}{a_{2024}}$ is [2024]

Q 4 :

Let $S$ be the set of positive integral values of $a$ for which $\frac{a x^{2} + 2 (a + 1) x + 9 a + 4}{x^{2} - 8 x + 32} < 0, \forall x \in R .$ Then, the number of elements is $S$ is [2024]

Q 5 :

The coefficients $a, b, c$ in the quadratic equation $a x^{2} + b x + c = 0$ are from the set {1, 2, 3, 4, 5, 6}. If the probability of this equation having one real root bigger than the other is $p$ , then $216 p$ equals: [2024]

Q 6 :

The number of distinct real roots of the equation $| x | | x + 2 | - 5 | x + 1 | - 1 = 0$ is _________ [2024]

Q 7 :

The number of real solutions of the equation $x | x + 5 | + 2 | x + 7 | - 2 = 0$ is _______. [2024]

Q 8 :

The number of distinct real roots of the equation $| x + 1 | | x + 3 | - 4 | x + 2 | + 5 = 0$ is ________. [2024]

Q 9 :

Let the set of all values of $p \in R$ , for which both the roots of the equation $x^{2} - (p + 2) x + (2 p + 9) = 0$ are negative real numbers, be the interval $(α, β]$ . Then $β - 2 α$ is equal to [2025]

Q 10 :

The number of real roots of the equation $x$ $| x - 2 |$ $+ 3$ $| x - 3 |$ $+ 1 = 0$ [2025]

Q 11 :

The sum of the squares of the roots of $| x - 2 |^{2} + | x - 2 | - 2 = 0$ and the squares of the roots of $x^{2} - 2 | x - 3 | - 5 = 0$ is [2025]

Q 12 :

If the set of all $a \in R - {1}$ , for which the roots of the equation $(1 - a) x^{2} + 2 (a - 3) x + 9 = 0$ are positive is $(- \infty, - α] \cup [β, γ)$ , then $2 α + β + γ$ is equal to __________. [2025]

Q 13 :

The set of all $a \in ℝ$ for which the equation $x$ $| x - 1 |$ $+$ $| x + 2 |$ $+ a = 0$ has exactly one real root, is [2023]

Q 14 :

The number of real roots of the equation $x$ $| x |$ $- 5$ $| x + 2 |$ $+ 6 = 0,$ is [2023]

Q 15 :

The number of integral values of $k$ , for which one root of the equation $2 x^{2} - 8 x + k = 0$ lies in the interval (1, 2) and its other root lies in the interval (2, 3), is [2023]

Q 16 :

Let $m$ and $n$ be the numbers of real roots of the quadratic equations $x^{2} - 12 x + [x] + 31 = 0$ and $x^{2} - 5$ $| x + 2 |$ $- 4 = 0$ respectively, where $[x]$ denotes the greatest integer $\leq x$ . Then $m^{2} + m n + n^{2}$ is equal to __________ . [2023]

Q 17 :

Let $S = {α : \log_{2} (9^{2 α - 4} + 13) - \log_{2} (\frac{5}{3} \cdot 3^{2 α - 4} + 1) = 2} .$ Then the maximum value of $β$ for which the equation

$x^{2} - 2 {(\sum_{α \in S} α)}^{2} x + \sum_{α \in S} {(α + 1)}^{2} β = 0$ has real roots, is ________ . [2023]