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

(C)

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

(C)

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

(D)

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

(B)

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

(B)

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]

0%

(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]

0%

(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

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]

0%

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

0%

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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]

0%

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

0%

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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]