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

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]

0%

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

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

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

[IMAGE 15]

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.

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

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

[IMAGE 16]

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

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

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

$Integral value of k = 7$

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]

0%

(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]

0%

(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 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]

0%

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]

0%

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]

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Topic Question Set

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]

0%

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]

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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]

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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]