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

Q 1 :

The sum of all the solutions of the equation ${(8)}^{2 x} - 16 \cdot {(8)}^{x} + 48 = 0$ is : [2024]

$1 + \log_{8} (8)$
$\log_{8} (6)$
$1 + \log_{8} (6)$
$\log_{8} (4)$

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(3)

We have, $8^{2 x} - 16 \cdot 8^{x} + 48 = 0$

Let $8^{x} = t$

$\Rightarrow t^{2} - 16 t + 48 = 0 \Rightarrow t^{2} - 12 t - 4 t + 48 = 0$

$\Rightarrow (t - 12) (t - 4) = 0 \Rightarrow t = 4 or t = 12$

$\Rightarrow 8^{x} = 4 or 8^{x} = 12$

$\Rightarrow x = \log_{8} 4 or x = \log_{8} 12$

Sum of solutions $= \log_{8} 4 + \log_{8} 12 = \log_{8} 48 = \log_{8} (6 \cdot 8)$

$= \log_{8} 8 + \log_{8} 6 = 1 + \log_{8} 6$

Q 2 :

Let $α, β$ be the roots of the equation $x^{2} + 2 \sqrt{2} x - 1 = 0 .$ The quadratic equation, whose roots are $α^{4} + β^{4}$ and $\frac{1}{10} (α^{6} + β^{6}),$ is: [2024]

$x^{2} - 195 x + 9466 = 0$
$x^{2} - 180 x + 9506 = 0$
$x^{2} - 195 x + 9506 = 0$
$x^{2} - 190 x + 9466 = 0$

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(3)

We have, $x^{2} + 2 \sqrt{2 x} - 1 = 0$

$∴$ $α + β = - 2 \sqrt{2}$ and $α β = - 1$

Now, $α^{2} + β^{2} = {(α + β)}^{2} - 2 α β = {(- 2 \sqrt{2})}^{2} - 2 (- 1)$

$= 8 + 2 = 10$

Also, $α^{4} + β^{4} = {(α^{2} + β^{2})}^{2} - 2 {(α β)}^{2} = (10)$ ²-2(1)

$= 100 - 2 = 98$

and $α^{6} + β^{6} = {(α^{2} + β^{2})}^{3} - 3 α^{2} β^{2} (α^{2} + β^{2}) = {(10)}^{3} - 3 (1) (10)$

$= 1000 - 30 = 970$

So, $\frac{1}{10} (α^{6} + β^{6}) = 97$

Hence, equation whose roots are $α^{4} + β^{4}$ and $\frac{1}{10} (α^{6} + β^{6})$ is $x^{2} - (98 + 97) x + 98 \times 97 = 0$ i.e., $x^{2} - 195 x + 9506 = 0$

Q 3 :

Let $α, β : α > β,$ be the roots of the equation $x^{2} - \sqrt{2} x - \sqrt{3} = 0 .$ Let $P_{n} = α^{n} - β^{n}, n \in N .$ Then $(11 \sqrt{3} - 10 \sqrt{2}) P_{10} + (11 \sqrt{2} + 10) P_{11} - 11 P_{12}$ is equal to [2024]

$11 \sqrt{2} P_{9}$
$10 \sqrt{2} P_{9}$
$10 \sqrt{3} P_{9}$
$11 \sqrt{3} P_{9}$

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(3)

We have, $x^{2} - \sqrt{2} x - \sqrt{3} = 0$ has two roots $α$ and $β$

So for $S_{n} = α^{n} \neq β^{n},$ by Newton's theorem, we have $S_{n + 2} - \sqrt{2} S_{n + 1} - \sqrt{3} S_{n} = 0$

$\Rightarrow (α^{n + 2} - β^{n + 2}) - \sqrt{2} (α^{n + 1} - β^{n + 1}) - \sqrt{3} (α^{n} - β^{n}) = 0$

$\Rightarrow P_{n + 2} - \sqrt{2} P_{n + 1} - \sqrt{3} P_{n} = 0 [∵ P_{n} = α^{n} - β^{n}]$

For $n = 9,$ we have

$P_{12} - \sqrt{2} P_{11} - \sqrt{3} P_{10} = 0$ ...(i)

$\Rightarrow P_{12} = \sqrt{2} P_{11} + \sqrt{3} P_{10}$

For $n = 9,$ we have

$P_{11} - \sqrt{2} P_{10} - \sqrt{3} P_{9} = 0$

$\Rightarrow P_{11} = \sqrt{2} P_{10} + \sqrt{3} P_{9}$

So, $(11 \sqrt{3} - 10 \sqrt{2}) P_{10} + (11 \sqrt{2} + 10) P_{11} - 11 P_{12}$

$= (11 \sqrt{3} - 10 \sqrt{2}) P_{10} + (11 \sqrt{2} + 10) (\sqrt{2} P_{10} + \sqrt{3} P_{9}) - 11 (\sqrt{2} P_{11} + \sqrt{3} P_{10})$

$= 11 \sqrt{3} P_{10} - 10 \sqrt{2} P_{10} + 22 P_{10} + 10 \sqrt{2} P_{10} - 11 \sqrt{3} P_{10} + 11 \sqrt{6} P_{9} + 10 \sqrt{3} P_{9} - 11 \sqrt{2} (\sqrt{2} P_{10} + \sqrt{3} P_{9})$

$= 22 P_{10} + 11 \sqrt{6} P_{9} + 10 \sqrt{3} P_{9} - 22 P_{10} - 11 \sqrt{6} P_{9} = 10 \sqrt{3} P_{9}$

Q 4 :

Let $α, β$ are the roots of the equation, $x^{2} - x - 1 = 0$ and $S_{n} = 2023 α^{n} + 2024 β^{n},$ then : [2024]

$2 S_{12} = S_{11} + S_{10}$
$2 S_{11} = S_{12} + S_{10}$
$S_{11} = S_{10} + S_{12}$
$S_{12} = S_{11} + S_{10}$

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(4)

Given, $α$ and $β$ are the roots of $x^{2} - x - 1 = 0$

$\Rightarrow α^{2} - α - 1 = 0$ and $β^{2} - β - 1 = 0$

$\Rightarrow α^{2} = 1 + α$ and $β^{2} = 1 + β$ (i)

Now, $S_{10} = 2023 α^{10} + 2024 β^{10} [∵ S_{n} = 2023 α^{n} + 2024 β^{n}]$

$S_{11} = 2023 α^{11} + 2024 β^{11}$

$\Rightarrow S_{10} + S_{11} = 2023 α^{10} (1 + α) + 2024 β^{10} (1 + β)$

$= 2023 α^{10} \cdot α^{2} + 2024 β^{10} \cdot β^{2} [From (i)]$

$= 2023 α^{12} + 2024 β^{12} = S_{12}$

$∴ S_{12} = S_{10} + S_{11}$

Q 5 :

Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the solutions of the equation $4 x^{4} + 8 x^{3} - 17 x^{2} - 12 x + 9 = 0$ and $(4 + x_{1}^{2}) (4 + x_{2}^{2}) (4 + x_{3}^{2}) (4 + x_{4}^{2}) = \frac{125}{16} m .$ Then the value of $m$ is _______ . [2024]

(221)

We have, $x_{1}, x_{2}, x_{3}$ and $x_{4}$ are solution of $4 x^{4} + 8 x^{3} - 17 x^{2} - 12 x + 9 = 0$

$\Rightarrow 4 x^{4} + 8 x^{3} - 17 x^{2} - 12 x + 9$

$= 4 (x - x_{1}) (x - x_{2}) (x - x_{3}) (x - x_{4})$ ...(i)

Let us substitute $x = 2 i$ in (i)

$∴ 64 - 64 i + 68 - 24 i + 9$

$= 4 (2 i - x_{1}) (2 i - x_{2}) (2 i - x_{3}) (2 i - x_{4})$

$\Rightarrow 4 (2 i - x_{1}) (2 i - x_{2}) (2 i - x_{3}) (2 i - x_{4}) = 141 - 88 i$ ...(ii)

Substitute $x = - 2 i$ in (i), we get

$4 (2 i + x_{1}) (2 i + x_{2}) (2 i + x_{3}) (2 i + x_{4}) = 141 + 88 i$ ...(iii)

Multiplying (ii) and (iii), we get

$(4 + x_{1}^{2}) (4 + x_{2}^{2}) (4 + x_{3}^{2}) (4 + x_{4}^{2}) = \frac{{(141)}^{2} + 88^{2}}{16}$

$\Rightarrow \frac{125 m}{16} = \frac{27625}{16} \Rightarrow m = 221$

Q 6 :

If $α$ satisfies the equation $x^{2} + x + 1 = 0$ and ${(1 + α)}^{7} = A + B α + C α^{2}, A, B, C \geq 0,$ then $5 (3 A - 2 B - C)$ is equal to _______ . [2024]

(5)

As $α$ satisfies $x^{2} + x + 1 = 0 \Rightarrow α = ω or ω^{2}$

Now, ${(1 + α)}^{7} = {(1 + ω)}^{7} = A + B ω + C ω^{2}$

$\Rightarrow - ω^{2} = A + B ω + C ω^{2} [∵ 1 + ω + ω^{2} = 0 and ω^{3} = 1]$

$\Rightarrow A + B ω + (C + 1) ω^{2} = 0$

As $ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ & $ω^{2} = - \frac{1}{2} - \frac{\sqrt{3}}{2} i$

$\Rightarrow A + B (- \frac{1}{2} + \frac{\sqrt{3}}{2} i) + (C + 1) (- \frac{1}{2} - \frac{\sqrt{3}}{2} i) = 0$

$\Rightarrow A - \frac{B}{2} - \frac{C + 1}{2} = 0 and \frac{\sqrt{3}}{2} B - \frac{\sqrt{3}}{2} (C + 1) = 0$

$\Rightarrow 2 A - B - C - 1 = 0 and B - C = 1$

On solving, we get A = B and B - C = 1

$∴$ $5 (3 A - 2 B - C) = 5 (3 B - 2 B - C) = 5 (B - C) = 5 \times 1 = 5$

Q 7 :

Let $α, β$ be the roots of the equation $x^{2} - x + 2 = 0$ with $Im (α) > Im (β) .$ Then $α^{6} + α^{4} + β^{4} - 5 α^{2}$ is equal to ____ . [2024]

(13)

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

As $α, β$ are the roots of equation so, it satisfy the equation.

$α^{2} - α + 2 = 0 \Rightarrow α^{2} = α - 2$

$\Rightarrow α^{4} = {(α - 2)}^{2}$ [Squaring both sides]

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

and $α^{6} = α^{2} \cdot α^{4} = (α - 2) (2 - 3 α) = 2 α - 3 α^{2} - 4 + 6 α$

$= 8 α - 4 - 3 (α - 2) = 8 α - 4 - 3 α + 6 = 5 α + 2$

Similarly, $β^{4} = 2 - 3 β$

Now, $α^{6} + α^{4} + β^{4} - 5 α^{2}$

$= 5 α + 2 + 2 - 3 α + 2 - 3 β - 5 (α - 2)$

$= 16 - 3 (α + β) = 16 - 3 (1) = 13$

Q 8 :

Let $α, β$ be the roots of the equation $x^{2} - \sqrt{6} x + 3 = 0$ such that $Im (α) > Im (β) .$ Let $a, b$ be integers not divisible by 3 and $n$ be a natural number such that $\frac{α^{99}}{β} + α^{98} = 3^{n} (a + i b), i = \sqrt{- 1} .$ Then $n + a + b$ is equal to ______ . [2024]

(49)

Roots of the equation $x^{2} - \sqrt{6} x + 3 = 0$ are

$x = \frac{\sqrt{6} \pm \sqrt{6 - 12}}{2} \Rightarrow x = \frac{\sqrt{6} \pm \sqrt{6 i}}{2}$

$∵$ $Im (α) > Im (β)$

$∴$ $α = \frac{\sqrt{6} + \sqrt{6 i}}{2}, β = \frac{\sqrt{6} - \sqrt{6 i}}{2}$

$\Rightarrow α^{2} = \frac{2 \cdot \sqrt{6} \cdot \sqrt{6} \cdot i}{4}, β^{2} = \frac{- 2 \cdot \sqrt{6} \cdot \sqrt{6 i}}{4}$

$α^{2} = 3 i, β^{2} = - 3 i \Rightarrow \frac{α}{β} = i$

Now, $\frac{α^{99}}{β} + α^{98} = 3^{n} (a + i b)$

$\Rightarrow α^{98} (\frac{α}{β} + 1) = 3^{n} (a + i b) \Rightarrow {(3 i)}^{49} (i + 1) = 3^{n} (a + i b)$

$\Rightarrow 3^{49} \cdot i (i + 1) = 3^{n} (a + i b) \Rightarrow 3^{49} \cdot (- 1 + i) = 3^{n} (a + i b)$

$\Rightarrow n = 49, a = - 1, b = 1$ $∴$ $n + a + b = 49$

Q 9 :

Let $a, b, c$ be the lengths of three sides of a triangle satisfying the condition $(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0 .$ If the set of all possible values of $x$ is the interval, $(α, β)$ then $12 (α^{2} + β^{2})$ is equal to _____ . [2024]

(36)

We have, $(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0$

$\Rightarrow a^{2} x^{2} + b^{2} x^{2} - 2 a b x - 2 b c x + b^{2} + c^{2} = 0$

$\Rightarrow a^{2} x^{2} - 2 a b x + b^{2} + b^{2} x^{2} - 2 b c x + c^{2} = 0$

$\Rightarrow {(a x - b)}^{2} + {(b x - c)}^{2} = 0$

$\Rightarrow a x - b = 0, b x - c = 0$

$\Rightarrow a x = b, b x = c \Rightarrow a x^{2} = b x, b x^{2} = c x$

Since, $a + b > c, b + c > a and c + a > b$

$a + a x > b x, a x + b x > a and b x + a > a x$

$\Rightarrow a + a x > a x^{2}, a x + a x^{2} > a and a x^{2} + a > a x$

$\Rightarrow x^{2} - x - 1 < 0, x^{2} + x - 1 > 0 and x^{2} - x + 1 > 0$

$\Rightarrow \frac{1 - \sqrt{5}}{2} < x < \frac{1 + \sqrt{5}}{2} \Rightarrow x < \frac{- 1 - \sqrt{5}}{2} or x > \frac{- 1 + \sqrt{5}}{2}$

$\Rightarrow \frac{\sqrt{5} - 1}{2} < x < \frac{\sqrt{5} + 1}{2} \Rightarrow α = \frac{\sqrt{5} - 1}{2}, β = \frac{\sqrt{5} + 1}{2}$

$∴$ $12 (α^{2} + β^{2}) = 36$

Q 10 :

The number of real solutions of the equation $x (x^{2} + 3 | x | + 5 | x - 1 | + 6 | x - 2 |) = 0$ is ______________. [2024]

(1)

We have, equation $x (x^{2} + 3 | x | + 5 | x - 1 | + 6 | x - 2 |) = 0$

$\Rightarrow x = 0$ is one solution.

Q 11 :

Consider the equation $x^{2} + 4 x - n = 0$ , where n $\in$ [20, 100] is a natural number. Then the number of all distinct values of n, for which the given equation has integral roots, is equal to [2025]

5
7
6
8

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(3)

we have, $x^{2} + 4 x - n = 0$

$\Rightarrow x^{2} + 4 x + 4 = n + 4$

$\Rightarrow {(x + 2)}^{2} = n + 4$

$\Rightarrow x = - 2 \pm \sqrt{n + 4}$

But $20 \leq n \leq 100$

$\Rightarrow 24 \leq n + 4 \leq 104 \Rightarrow \sqrt{24} \leq \sqrt{n + 4} \leq \sqrt{104}$

$\Rightarrow \sqrt{n + 4} \in {5, 6, 7, 8, 9, 10}$ .

$∴$ 6 integral values of 'n' are possible.

Q 12 :

The product of all solutions of the equation $e^{5 {(\log_{e} x)}^{2} + 3} = x^{8}, x > 0$ , is: [2025]

$e^{2}$
e
$e^{8 / 5}$
$e^{6 / 5}$

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(3)

We have, $e^{5 {(\log_{e} x)}^{2} + 3} = x^{8}, x > 0$

Taking log on both sides, we get

$5 {(\ln x)}^{2} + 3 = 8 \ln x \Rightarrow 5 {(\ln x)}^{2} + 3 - 8 \ln x = 0$

Put ln x = t

$\Rightarrow 5 t^{2} - 8 t + 3 = 0 \Rightarrow t_{1} + t_{2} = \frac{8}{5}$

$∴ \ln x_{1} + \ln x_{2} = 8 / 5 \Rightarrow \ln (x_{1} x_{2}) = 8 / 5 \Rightarrow x_{1} x_{2} = e^{8 / 5}$ .

Q 13 :

The product of all the rational roots of the equation ${(x^{2} - 9 x + 11)}^{2} - (x - 4) (x - 5) = 3$ , is equal to [2025]

28
21
7
14

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(4)

We have, ${(x^{2} - 9 x + 11)}^{2} - (x - 4) (x - 5) = 3$

$\Rightarrow {(x^{2} - 9 x + 11)}^{2} - (x - 9 x + 20) = 3$

Let $x^{2} - 9 x = t$ , then

${(t + 11)}^{2} - (t + 20) = 3$

$\Rightarrow t^{2} + 121 + 22 t - t - 20 - 3 = 0$

$\Rightarrow t^{2} + 21 t + 98 = 0 \Rightarrow (t + 14) (t + 7) = 0$

$\Rightarrow t = - 7, - 14$

$\Rightarrow x^{2} - 9 x + 7 = 0 and x^{2} - 9 x + 14 = 0$

$\Rightarrow x = \frac{9 \pm \sqrt{53}}{2} and x = \frac{9 \pm 5}{2}$

$\Rightarrow$ Product of all rational roots = $(\frac{9 + 5}{2}) (\frac{9 - 5}{2}) = 7 \times 2 = 14$ .

Q 14 :

The number of real solution(s) of the equation $x^{2} + 3 x + 2 = \min {| x - 3 |, | x + 2 |}$ is : [2025]

2
3
1
0

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(1)

$In (- \infty, \frac{1}{2})$ , $| x + 2 |$ is minimum and in $(\frac{1}{2}, \infty)$ , $| x - 3 |$ is minimum.

From the graph, only two solutions.

Q 15 :

The sum, of the squares of all the roots of the equation $x^{2} + | 2 x - 3 | - 4 = 0$ , is [2025]

$6 (2 - \sqrt{2})$
$3 (3 - \sqrt{2})$
$6 (3 - \sqrt{2})$
$3 (2 - \sqrt{2})$

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4

(1)

Given, $x^{2} + | 2 x - 3 | - 4 = 0$

Case I: $x^{2} + (2 x - 3) - 4 = 0 when x \geq \frac{3}{2}$

$\Rightarrow x^{2} + 2 x - 7 = 0 \Rightarrow x = 2 \sqrt{2} - 1$ $[∵ x \geq \frac{3}{2}]$

Case II : $x^{2} - (2 x - 3) - 4 = 0, x < \frac{3}{2}$

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

$\Rightarrow x = \frac{2 \pm 2 \sqrt{2}}{2} = 1 \pm \sqrt{2} \Rightarrow x = 1 - \sqrt{2}$ $[∵ x < \frac{3}{2}]$

$∴$ Required sum = ${(2 \sqrt{2} - 1)}^{2} + {(1 - \sqrt{2})}^{2}$

$= 12 - 6 \sqrt{2} = 6 (2 - \sqrt{2})$ .

Q 16 :

If $α + i β$ and $γ + i δ$ are the roots of $x^{2} - (3 - 2 i) x - (2 i - 2) = 0, i = \sqrt{- 1}$ , then $α γ + β δ$ is equal to : [2025]

–6
6
–2
2

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4

(4)

We have, $x^{2} - (3 - 2 i) x - (2 i - 2) = 0$

$x = \frac{(3 - 2 i) \pm \sqrt{{(3 - 2 i)}^{2} + 4 (1) (2 i - 2)}}{2}$

$= \frac{(3 - 2 i) \pm \sqrt{9 - 4 - 12 i + 8 i - 8}}{2} = \frac{(3 - 2 i) \pm \sqrt{- 3 - 4 i}}{2}$

$= \frac{(3 - 2 i) \pm \sqrt{{{(1)}^{2} + (2 i)}^{2} - 2 (1) (2 i)}}{2}$

$= \frac{(3 - 2 i) \pm \sqrt{{(1 - 2 i)}^{2}}}{2} = \frac{(3 - 2 i) \pm (1 - 2 i)}{2}$

$∴ x = \frac{3 - 2 i + 1 - 2 i}{2} or \frac{3 - 2 i - 1 + 2 i}{2}$

$∴ x = 2 - 2 i or 1 + 0 i$

So, $α γ + β δ = 2 (1) + (- 2) (0) = 2$ .

Q 17 :

The number of solutions of the equation $(\frac{9}{x} - \frac{9}{\sqrt{x}} + 2) (\frac{2}{x} - \frac{7}{\sqrt{x}} + 3) = 0$ is : [2025]

1
3
4
2

1

2

3

4

(3)

We have, $(\frac{9}{x} - \frac{9}{\sqrt{x}} + 2) (\frac{2}{x} - \frac{7}{\sqrt{x}} + 3) = 0$ ... (i)

Put $\frac{1}{\sqrt{x}} = y$ in (i), we get

$(9 y^{2} - 9 y + 2) (2 y^{2} - 7 y + 3) = 0$

$\Rightarrow (3 y - 2) (3 y - 1) (2 y - 1) (y - 3) = 0$

$\Rightarrow y = \frac{2}{3}, \frac{1}{3}, \frac{1}{2}, 3 \Rightarrow x = \frac{9}{4}, 9, 4, \frac{1}{9}$

$∴$ Number of solutions = 4.

Q 18 :

If the set of all $a \in R$ , for which the equation $2 x^{2} + (a - 5) x + 15 = 3 a$ has no real root, is the interval $(α, β)$ , and $X = {x \in Z : α < x < β}$ , then $\sum_{x \in X} x^{2}$ is equal to : [2025]

2129
2119
2109
2139

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2

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4

(4)

Since, the given equation has no real root

$∴ b^{2} - 4 a c < 0 \Rightarrow {(a - 5)}^{2} - 4 (2) (15 - 3 a) < 0$

$\Rightarrow {(a - 5)}^{2} - 8 (15 - 3 a) < 0$

$\Rightarrow a^{2} - 10 a + 25 - 120 + 24 a < 0$

$\Rightarrow a^{2} - 14 a - 95 < 0 \Rightarrow a \in (- 5, 19)$

$∴ \sum_{x \in X} x^{2} = (1^{2} + 2^{2} + 3^{2} + 4^{2}) + 0^{2} + {(1^{2} + 2^{2} + . . . + 18)}^{2}$

$= \frac{4 (5) (9)}{6} + \frac{18 (19) (37)}{6} = \frac{12834}{6} = 2139$ .

Q 19 :

Let $α, β$ be the roots of the equation $x^{2} - a x - b = 0$ with $I m (α) < I m (β)$ . Let $P_{n} = α^{n} - β^{n}$ . If $P_{3} = - 5 \sqrt{7} i, P_{4} = - 3 \sqrt{7} i, P_{5} = 11 \sqrt{7} i and P_{6} = 45 \sqrt{7} i$ , then $| α^{4} + β^{4} |$ is equal to __________. [2025]

(31)

We have, $α + β = a and α β = - b$

$∵ P_{6} = a P_{5} + b P_{4}$

$\Rightarrow 45 \sqrt{7} i = a \times 11 \sqrt{7} i + b (- 3 \sqrt{7}) i$

$\Rightarrow 45 = 11 a - 3 b$ ... (i)

and $P_{5} = a P_{4} + b P_{3}$

$\Rightarrow 11 \sqrt{7} i = a (- 3 \sqrt{7} i) + b (- 5 \sqrt{7} i)$

$\Rightarrow 11 = - 3 a - 5 b$ ... (ii)

On solving equations (i) and (ii), we get a = 3, b = –4

$∴$ $| α^{4} + β^{4} |$ $= \sqrt{{(α^{4} - β^{4})}^{2} + 4 α^{4} β^{4}}$

$= \sqrt{- 63 + 4 . 4^{4}} = \sqrt{- 63 + 1024} = \sqrt{961} = 31$ .

Q 20 :

Let $α, β, γ$ be the three roots of the equation $x^{3} + b x + c = 0 .$ If $β γ = 1 = - α$ , then the value of $b^{3} + 2 c^{3} - 3 α^{3} - 6 β^{3} - 8 γ^{3}$ is equal to [2023]

21
$\frac{169}{8}$
$\frac{155}{8}$
19

1

2

3

4

(4)

Q 21 :

Let $α, β$ be the roots of the quadratic equation $x^{2} + \sqrt{6} x + 3 = 0 .$ Then $\frac{α^{23} + β^{23} + α^{14} + β^{14}}{α^{15} + β^{15} + α^{10} + β^{10}}$ is equal to [2023]

729
9
81
72

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2

3

4

(3)

$Given, x^{2} + \sqrt{6} x + 3 = 0 ...(i)$

On solving eqn. (i), we get its roots

$α, β = \frac{- \sqrt{6} \pm \sqrt{6 - 12}}{2} = \frac{- \sqrt{6} \pm \sqrt{6} i}{2}$

$α, β = \sqrt{3} e^{\pm 3 π i / 4}$

So, required expression becomes:

$= \frac{{(\sqrt{3})}^{23} (2 \cos \frac{69 π}{4}) + {(\sqrt{3})}^{14} (2 \cos \frac{42 π}{4})}{{(\sqrt{3})}^{15} (2 \cos \frac{45 π}{4}) + {(\sqrt{3})}^{10} (2 \cos \frac{30 π}{4})}$

$= \frac{{(\sqrt{3})}^{23} \times 2 (- 1) + 0}{{(\sqrt{3})}^{15} \times 2 (- 1) + 0}$

$= {(\sqrt{3})}^{8} = 81$

Q 22 :

Let $α, β$ be the roots of the equation $x^{2} - \sqrt{2} x + 2 = 0 .$ Then $α^{14} + β^{14}$ is equal to [2023]

$- 64 \sqrt{2}$
$- 128$
$- 128 \sqrt{2}$
$- 64$

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(2)

$Given, x^{2} - \sqrt{2} x + 2 = 0$

$x = \frac{\sqrt{2} \pm \sqrt{{(- \sqrt{2})}^{2} - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$

$x = \frac{\sqrt{2} \pm \sqrt{- 6}}{2};$ $x = \frac{1 \pm \sqrt{3} i}{\sqrt{2}}$ $\Rightarrow α = - \sqrt{2} ω, β = - \sqrt{2} ω^{2}$

$\Rightarrow α^{14} + β^{14} = {(- \sqrt{2} ω)}^{14} + {(- \sqrt{2} ω^{2})}^{14}$

$= 2^{7} \cdot ω^{14} + 2^{7} ω^{28} = 2^{7} (ω + ω^{2})$

$= - 2^{7}$ $(∵ ω^{2} + ω + 1 = 0)$

$= - 128$

Q 23 :

Let $S = {x : x \in R, {(\sqrt{3} + \sqrt{2})}^{x^{2} - 4} + {(\sqrt{3} - \sqrt{2})}^{x^{2} - 4} = 10} .$ Then $n (S)$ is equal to [2023]

2
4
0
6

1

2

3

4

(2)

$S = {x : x \in R and {(\sqrt{3} + \sqrt{2})}^{x^{2} - 4} + {(\sqrt{3} - \sqrt{2})}^{x^{2} - 4} = 10}$

Let ${(\sqrt{3} + \sqrt{2})}^{x^{2} - 4} = t and {(\sqrt{3} - \sqrt{2})}^{x^{2} - 4} = \frac{1}{t}$

$∴ t + \frac{1}{t} = 10 \Rightarrow t^{2} - 10 t + 1 = 0 \Rightarrow t = \frac{10 \pm \sqrt{96}}{2}$

$\Rightarrow t = 5 \pm 2 \sqrt{6} = {(\sqrt{3} \pm \sqrt{2})}^{2}$

$\Rightarrow {(\sqrt{3} + \sqrt{2})}^{x^{2} - 4} = {(\sqrt{3} + \sqrt{2})}^{2}$

$and {(\sqrt{3} + \sqrt{2})}^{x^{2} - 4} = {(\sqrt{3} - \sqrt{2})}^{2} = {(\sqrt{3} + \sqrt{2})}^{- 2}$

$\Rightarrow x^{2} - 4 = 2 and x^{2} - 4 = - 2 \Rightarrow x^{2} = 6 and x^{2} = 2$

$\Rightarrow x = \pm \sqrt{6}$ and $x = \pm \sqrt{2} \Rightarrow n (S) = 4$

Q 24 :

The number of real solutions of the equation $3 (x^{2} + \frac{1}{x^{2}}) - 2 (x + \frac{1}{x}) + 5 = 0$ , is [2023]

4
3
2
0

1

2

3

4

(4)

$Given, 3 (x^{2} + \frac{1}{x^{2}}) - 2 (x + \frac{1}{x}) + 5 = 0$

$\Rightarrow 3 [{(x + \frac{1}{x})}^{2} - 2] - 2 (x + \frac{1}{x}) + 5 = 0$

$\Rightarrow 3 [y^{2} - 2] - 2 y + 5 = 0$ $(Let y = x + \frac{1}{x})$

$\Rightarrow 3 y^{2} - 6 - 2 y + 5 = 0 \Rightarrow 3 y^{2} - 2 y - 1 = 0$

$\Rightarrow y = 1, \frac{- 1}{3}$ $But y \in (- \infty, - 2] \cup [2, \infty)$

$Hence, no real solution.$

Q 25 :

The number of real roots of the equation $\sqrt{x^{2} - 4 x + 3} + \sqrt{x^{2} - 9} = \sqrt{4 x^{2} - 14 x + 6},$ is [2023]

2
3
1
0

1

2

3

4

(3)

$\sqrt{x^{2} - 4 x + 3} + \sqrt{x^{2} - 9} = \sqrt{4 x^{2} - 14 x + 6}$

$\Rightarrow \sqrt{(x - 3) (x - 1)} + \sqrt{(x - 3) (x + 3)} = \sqrt{2 (x - 3) (2 x - 1)}$

$\Rightarrow \sqrt{(x - 3)} [\sqrt{x - 1} + \sqrt{x + 3} - \sqrt{2} \sqrt{2 x - 1}] = 0$

$\Rightarrow x - 3 = 0 \Rightarrow x = 3$

$and \sqrt{x - 1} + \sqrt{x + 3} = \sqrt{2} \sqrt{2 x - 1}$

$\Rightarrow (x - 1) + (x + 3) + 2 \sqrt{(x - 1) (x + 3)} = 2 (2 x - 1)$

$\Rightarrow 2 x + 2 + 2 \sqrt{(x - 1) (x + 3)} = 4 x - 2$

$\Rightarrow x + 1 + \sqrt{(x - 1) (x + 3)} = 2 x - 1$

$\Rightarrow \sqrt{(x - 1) (x + 3)} = x - 2 \Rightarrow (x - 1) (x + 3) = x^{2} - 4 x + 4$

$\Rightarrow x^{2} + 2 x - 3 = x^{2} - 4 x + 4 \Rightarrow 6 x = 7 \Rightarrow x = \frac{7}{6}$

$Since, x^{2} - 4 x + 3 \geq 0 \Rightarrow (x - 3) (x - 1) \geq 0$

$\Rightarrow x \in (- \infty, 1] \cup [3, \infty)$ $...(i)$

$and x^{2} - 9 \geq 0 \Rightarrow (x - 3) (x + 3) \geq 0$

$\Rightarrow x \in (- \infty, - 3] \cup [3, \infty)$ $...(ii)$

$and 4 x^{2} - 14 x + 6 \geq 0 \Rightarrow (x - 3) (2 x - 1) \geq 0$

$\Rightarrow x \in (- \infty, \frac{1}{2}] \cup [3, \infty)$ $...(iii)$

$From (i), (ii), and (iii), x \in (- \infty, - 3] \cup [3, \infty)$

$Since x = \frac{7}{6} does not belong to x \in (- \infty, - 3] \cup [3, \infty) .$

$Therefore, x = 3 is the only solution.$

Q 26 :

The equation $e^{4 x} + 8 e^{3 x} + 13 e^{2 x} - 8 e^{x} + 1 = 0, x \in R$ has [2023]

four solutions two of which are negative
two solutions and both are negative
no solution
two solutions and only one of them is negative

1

2

3

4

(2)

$Given, e^{4 x} + 8 e^{3 x} + 13 e^{2 x} - 8 e^{x} + 1 = 0$

$Let e^{x} = t$

Now, $t^{4} + 8 t^{3} + 13 t^{2} - 8 t + 1 = 0 ...(i)$

$Dividing equation (i) by t^{2}, we get$

$\Rightarrow t^{2} + 8 t + 13 - \frac{8}{t} + \frac{1}{t^{2}} = 0 \Rightarrow t^{2} + \frac{1}{t^{2}} + 8 (t - \frac{1}{t}) + 13 = 0$

$\Rightarrow {(t - \frac{1}{t})}^{2} + 2 + 8 (t - \frac{1}{t}) + 13 = 0$

$Let t - \frac{1}{t} = z$

$\Rightarrow z^{2} + 8 z + 15 = 0 \Rightarrow (z + 3) (z + 5) = 0$

$\Rightarrow z = - 3 or z = - 5$ $So, t - \frac{1}{t} = - 3 or t - \frac{1}{t} = - 5$

$\Rightarrow t^{2} + 3 t - 1 = 0 or t^{2} + 5 t - 1 = 0;$ $\Rightarrow t = \frac{- 3 \pm \sqrt{13}}{2} or t = \frac{- 5 \pm \sqrt{29}}{2}$

$As t = e^{x}, it must be positive.$ $\Rightarrow t = \frac{\sqrt{13} - 3}{2} or t = \frac{\sqrt{29} - 5}{2}$

$\Rightarrow x = \ln (\frac{\sqrt{13} - 3}{2}) or x = \ln (\frac{\sqrt{29} - 5}{2})$

$Hence, two solutions are possible and both are negative.$

Q 27 :

If $a$ and $b$ are the roots of the equation $x^{2} - 7 x - 1 = 0$ , then the value of $\frac{a^{21} + b^{21} + a^{17} + b^{17}}{a^{19} + b^{19}}$ is equal to ____ . [2023]

(51)

$Given: a and b are the roots of x^{2} - 7 x - 1 = 0$

$By Newton's theorem,$

$S_{n + 2} - 7 S_{n + 1} - S_{n} = 0$

$S_{21} - 7 S_{20} - S_{19} = 0 ...(i)$

$S_{20} - 7 S_{19} - S_{18} = 0 ...(ii)$

$S_{19} - 7 S_{18} - S_{17} = 0 ...(iii)$

$\frac{S_{21} + S_{17}}{S_{19}} = \frac{S_{21} + (S_{19} - 7 S_{18})}{S_{19}} [From (iii)]$

$= \frac{S_{21} + S_{19} - 7 (S_{20} - 7 S_{19})}{S_{19}} [From (ii)]$

$= \frac{50 S_{19} + (S_{21} - 7 S_{20})}{S_{19}} = \frac{50 S_{19} + S_{19}}{S_{19}} [From (i)]$

$= 51 \cdot \frac{S_{19}}{S_{19}} = 51$

$∴$ $\frac{a^{21} + b^{21} + a^{17} + b^{17}}{a^{19} + b^{19}} = 51$

Q 28 :

The number of points, where the curve $f (x) = e^{8 x} - e^{6 x} - 3 e^{4 x} - e^{2 x} + 1, x \in ℝ$ cuts the $x$ -axis, is equal to ______ . [2023]

(2)

$Given: f (x) = e^{8 x} - e^{6 x} - 3 e^{4 x} - e^{2 x} + 1$

$= (e^{4 x} + \frac{1}{e^{4 x}}) - (e^{2 x} + \frac{1}{e^{2 x}}) = 3$

$= {(e^{2 x} + \frac{1}{e^{2 x}})}^{2} - (e^{2 x} + \frac{1}{e^{2 x}}) = 5$

$Let e^{2 x} + \frac{1}{e^{2 x}} = t$

$\Rightarrow t^{2} - t - 5 = 0 \Rightarrow t = \frac{1 \pm \sqrt{1 + 20}}{2} = \frac{1 \pm \sqrt{21}}{2}$

$\Rightarrow t = \frac{1 + \sqrt{21}}{2} and t = \frac{1 - \sqrt{21}}{2} (rejected)$

$\Rightarrow e^{2 x} + \frac{1}{e^{2 x}} = \frac{1 + \sqrt{21}}{2}$

$Let e^{2 x} = t, then t + \frac{1}{t} = \frac{1 + \sqrt{21}}{2}$

$\Rightarrow t^{2} + 1 = (\frac{1 + \sqrt{21}}{2}) t$

$It is a quadratic equation.$

$∴$ $2 values of e^{2 x} are possible.$

$Hence, 2 real solutions exist.$

Q 29 :

If $α$ and $β (α < β)$ are the roots of the equation $(- 2 + \sqrt{3}) (| \sqrt{x} - 3 |) + (x - 6 \sqrt{x}) + (9 - 2 \sqrt{3}) = 0, x \geq 0,$ then $\sqrt{\frac{β}{α}} + \sqrt{α β}$ is equal to: [2026]

8
10
9
11

1

2

3

4

(2)

$(x - 6 \sqrt{x} + 9) - (2 - \sqrt{3})$ $| \sqrt{x} - 3 |$ $- 2 \sqrt{3} = 0$

$\Rightarrow$ $| \sqrt{x} - 3 |^{2}$ $- (2 - \sqrt{3})$ $| \sqrt{x} - 3 |$ $- 2 \sqrt{3} = 0$

$\Rightarrow | \sqrt{x} - 3 | = 2 or | \sqrt{x} - 3 | = - \sqrt{3} (not possible)$

$\Rightarrow \sqrt{x} = 1 or 5$

$\Rightarrow x = 1 or 25$

$\Rightarrow α = 1 and β = 25$

Aliter:

Let $x \geq 9$ , let $\sqrt{x} = t \Rightarrow t \geq 3$

$(\sqrt{3} - 2) (t - 3) + {(t - 3)}^{2} - 2 \sqrt{3} = 0$

Let $t - 3 = u$

$u^{2} + (\sqrt{3} - 2) u - 2 \sqrt{3} = 0$

$\Rightarrow u = 2 or u = - \sqrt{3}$

$\Rightarrow t - 3 = 2 or t - 3 = - \sqrt{3}$

$\Rightarrow t = 5 or t = 3 - \sqrt{3} (rejected)$

$\Rightarrow x = 25$

Now let $0 < x < 9$

$- (\sqrt{3} - 2) (t - 3) + {(t - 3)}^{2} - 2 \sqrt{3} = 0$

Let $t - 3 = u$

$u^{2} - (\sqrt{3} - 2) u - 2 \sqrt{3} = 0$

$\Rightarrow u = \sqrt{3} or u = - 2$

$\Rightarrow t = 3 + \sqrt{3} (rejected) or t - 3 = - 2$

$\Rightarrow t = 1 \Rightarrow x = 1$

$α = 1, β = 25$

$Now \sqrt{\frac{β}{α}} + \sqrt{α β} = \sqrt{25} + \sqrt{25} = 10$

Q 30 :

The sum of all the roots of the equation ${(x - 1)}^{2} - 5$ $| x - 1 |$ $+ 6 = 0$ is [2026]

1
3
5
4

1

2

3

4

(4)

$Let | x - 1 | = t$

$t^{2} - 5 t + 6 = 0$

$t = 2 and t = 3$

$| x - 1 |$ $= 2 and$ $| x - 1 |$ $= 3$

$x - 1 \pm 2 and x - 1 = \pm 3$

$x = 1 \pm 2 and x = 1 \pm 3$

$∴$ $Roots are 3, - 1, 4, - 2$

$∴$ $Sum of roots = 3 + (- 1) + 4 + (- 2) = 4$

Topic Question Set

Q 1 :

The sum of all the solutions of the equation ${(8)}^{2 x} - 16 \cdot {(8)}^{x} + 48 = 0$ is : [2024]

Q 2 :

Let $α, β$ be the roots of the equation $x^{2} + 2 \sqrt{2} x - 1 = 0 .$ The quadratic equation, whose roots are $α^{4} + β^{4}$ and $\frac{1}{10} (α^{6} + β^{6}),$ is: [2024]

Q 3 :

Let $α, β : α > β,$ be the roots of the equation $x^{2} - \sqrt{2} x - \sqrt{3} = 0 .$ Let $P_{n} = α^{n} - β^{n}, n \in N .$ Then $(11 \sqrt{3} - 10 \sqrt{2}) P_{10} + (11 \sqrt{2} + 10) P_{11} - 11 P_{12}$ is equal to [2024]

Q 4 :

Let $α, β$ are the roots of the equation, $x^{2} - x - 1 = 0$ and $S_{n} = 2023 α^{n} + 2024 β^{n},$ then : [2024]

Q 5 :

Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the solutions of the equation $4 x^{4} + 8 x^{3} - 17 x^{2} - 12 x + 9 = 0$ and $(4 + x_{1}^{2}) (4 + x_{2}^{2}) (4 + x_{3}^{2}) (4 + x_{4}^{2}) = \frac{125}{16} m .$ Then the value of $m$ is _______ . [2024]

Q 6 :

If $α$ satisfies the equation $x^{2} + x + 1 = 0$ and ${(1 + α)}^{7} = A + B α + C α^{2}, A, B, C \geq 0,$ then $5 (3 A - 2 B - C)$ is equal to _______ . [2024]

Q 7 :

Let $α, β$ be the roots of the equation $x^{2} - x + 2 = 0$ with $Im (α) > Im (β) .$ Then $α^{6} + α^{4} + β^{4} - 5 α^{2}$ is equal to ____ . [2024]

Q 9 :

Let $a, b, c$ be the lengths of three sides of a triangle satisfying the condition $(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0 .$ If the set of all possible values of $x$ is the interval, $(α, β)$ then $12 (α^{2} + β^{2})$ is equal to _____ . [2024]

Q 10 :

The number of real solutions of the equation $x (x^{2} + 3 | x | + 5 | x - 1 | + 6 | x - 2 |) = 0$ is ______________. [2024]

Q 11 :

Consider the equation $x^{2} + 4 x - n = 0$ , where n $\in$ [20, 100] is a natural number. Then the number of all distinct values of n, for which the given equation has integral roots, is equal to [2025]

Q 12 :

The product of all solutions of the equation $e^{5 {(\log_{e} x)}^{2} + 3} = x^{8}, x > 0$ , is: [2025]

Q 13 :

The product of all the rational roots of the equation ${(x^{2} - 9 x + 11)}^{2} - (x - 4) (x - 5) = 3$ , is equal to [2025]

Q 14 :

The number of real solution(s) of the equation $x^{2} + 3 x + 2 = \min {| x - 3 |, | x + 2 |}$ is : [2025]

Q 15 :

The sum, of the squares of all the roots of the equation $x^{2} + | 2 x - 3 | - 4 = 0$ , is [2025]

Q 16 :

If $α + i β$ and $γ + i δ$ are the roots of $x^{2} - (3 - 2 i) x - (2 i - 2) = 0, i = \sqrt{- 1}$ , then $α γ + β δ$ is equal to : [2025]

Q 17 :

The number of solutions of the equation $(\frac{9}{x} - \frac{9}{\sqrt{x}} + 2) (\frac{2}{x} - \frac{7}{\sqrt{x}} + 3) = 0$ is : [2025]

Q 18 :

If the set of all $a \in R$ , for which the equation $2 x^{2} + (a - 5) x + 15 = 3 a$ has no real root, is the interval $(α, β)$ , and $X = {x \in Z : α < x < β}$ , then $\sum_{x \in X} x^{2}$ is equal to : [2025]

Q 19 :

Let $α, β$ be the roots of the equation $x^{2} - a x - b = 0$ with $I m (α) < I m (β)$ . Let $P_{n} = α^{n} - β^{n}$ . If $P_{3} = - 5 \sqrt{7} i, P_{4} = - 3 \sqrt{7} i, P_{5} = 11 \sqrt{7} i and P_{6} = 45 \sqrt{7} i$ , then $| α^{4} + β^{4} |$ is equal to __________. [2025]

Q 20 :

Let $α, β, γ$ be the three roots of the equation $x^{3} + b x + c = 0 .$ If $β γ = 1 = - α$ , then the value of $b^{3} + 2 c^{3} - 3 α^{3} - 6 β^{3} - 8 γ^{3}$ is equal to [2023]

(4)

Q 21 :

Let $α, β$ be the roots of the quadratic equation $x^{2} + \sqrt{6} x + 3 = 0 .$ Then $\frac{α^{23} + β^{23} + α^{14} + β^{14}}{α^{15} + β^{15} + α^{10} + β^{10}}$ is equal to [2023]

Q 22 :

Let $α, β$ be the roots of the equation $x^{2} - \sqrt{2} x + 2 = 0 .$ Then $α^{14} + β^{14}$ is equal to [2023]

Q 23 :

Let $S = {x : x \in R, {(\sqrt{3} + \sqrt{2})}^{x^{2} - 4} + {(\sqrt{3} - \sqrt{2})}^{x^{2} - 4} = 10} .$ Then $n (S)$ is equal to [2023]

Q 24 :

The number of real solutions of the equation $3 (x^{2} + \frac{1}{x^{2}}) - 2 (x + \frac{1}{x}) + 5 = 0$ , is [2023]

Q 25 :

The number of real roots of the equation $\sqrt{x^{2} - 4 x + 3} + \sqrt{x^{2} - 9} = \sqrt{4 x^{2} - 14 x + 6},$ is [2023]

Q 26 :

The equation $e^{4 x} + 8 e^{3 x} + 13 e^{2 x} - 8 e^{x} + 1 = 0, x \in R$ has [2023]

Q 27 :

If $a$ and $b$ are the roots of the equation $x^{2} - 7 x - 1 = 0$ , then the value of $\frac{a^{21} + b^{21} + a^{17} + b^{17}}{a^{19} + b^{19}}$ is equal to ____ . [2023]

Q 28 :

The number of points, where the curve $f (x) = e^{8 x} - e^{6 x} - 3 e^{4 x} - e^{2 x} + 1, x \in ℝ$ cuts the $x$ -axis, is equal to ______ . [2023]

Q 29 :

If $α$ and $β (α < β)$ are the roots of the equation $(- 2 + \sqrt{3}) (| \sqrt{x} - 3 |) + (x - 6 \sqrt{x}) + (9 - 2 \sqrt{3}) = 0, x \geq 0,$ then $\sqrt{\frac{β}{α}} + \sqrt{α β}$ is equal to: [2026]

Q 30 :

The sum of all the roots of the equation ${(x - 1)}^{2} - 5$ $| x - 1 |$ $+ 6 = 0$ is [2026]

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Topic Question Set

Q 1 : The sum of all the solutions of the equation (8)2x-16·(8)x+48=0 is : [2024]

Q 2 : Let α,β be the roots of the equation x2+22x-1=0. The quadratic equation, whose roots are α4+β4 and 110(α6+β6), is: [2024]

Q 3 : Let α,β:α>β, be the roots of the equation x2-2x-3=0. Let Pn=αn-βn, n∈N. Then (113-102)P10+(112+10)P11-11P12 is equal to [2024]

Q 4 : Let α,β are the roots of the equation, x2-x-1=0 and Sn=2023αn+2024βn, then : [2024]

Q 5 : Let x1,x2,x3,x4 be the solutions of the equation 4x4+8x3-17x2-12x+9=0 and (4+x12)(4+x22)(4+x32)(4+x42)=12516m. Then the value of m is _______ . [2024]

Q 6 : If α satisfies the equation x2+x+1=0 and (1+α)7=A+Bα+Cα2, A,B,C≥0, then 5(3A-2B-C) is equal to _______ . [2024]

Q 7 : Let α,β be the roots of the equation x2-x+2=0 with Im(α)>Im(β). Then α6+α4+β4-5α2 is equal to ____ . [2024]

Q 8 : Let α,β be the roots of the equation x2-6x+3=0 such that Im(α)>Im(β). Let a,b be integers not divisible by 3 and n be a natural number such that α99β+α98=3n(a+ib), i=-1. Then n+a+b is equal to ______ . [2024]

Q 9 : Let a,b,c be the lengths of three sides of a triangle satisfying the condition (a2+b2)x2-2b(a+c)x+(b2+c2)=0. If the set of all possible values of x is the interval, (α,β) then 12(α2+β2) is equal to _____ . [2024]

Q 10 : The number of real solutions of the equation x(x2+3|x|+5|x-1|+6|x-2|)=0 is ______________. [2024]

Q 11 : Consider the equation x2+4x–n=0, where n ∈ [20, 100] is a natural number. Then the number of all distinct values of n, for which the given equation has integral roots, is equal to [2025]

Q 12 : The product of all solutions of the equation e5(logex)2+3=x8, x>0, is: [2025]

Q 13 : The product of all the rational roots of the equation (x2–9x+11)2–(x–4)(x–5)=3, is equal to [2025]

Q 14 : The number of real solution(s) of the equation x2+3x+2=min{|x–3|,|x+2|} is : [2025]

Q 15 : The sum, of the squares of all the roots of the equation x2+|2x–3|–4=0, is [2025]

Q 16 : If α+iβ and γ+iδ are the roots of x2–(3–2i)x–(2i–2)=0, i=–1, then αγ+βδ is equal to : [2025]

Q 17 : The number of solutions of the equation (9x–9x+2)(2x–7x+3)=0 is : [2025]

Q 18 : If the set of all a∈R, for which the equation 2x2+(a–5)x+15=3a has no real root, is the interval (α,β), and X={x∈Z : α<x<β}, then ∑x∈Xx2 is equal to : [2025]

Q 19 : Let α, β be the roots of the equation x2–ax–b=0 with Im(α)<Im(β). Let Pn=αn–βn. If P3=–57i, P4=–37i, P5=117i and P6=457i, then |α4+β4| is equal to __________. [2025]

Q 20 : Let α,β,γ be the three roots of the equation x3+bx+c=0. If βγ=1=-α, then the value of b3+2c3-3α3-6β3-8γ3 is equal to [2023]

(4)

Q 21 : Let α,β be the roots of the quadratic equation x2+6x+3=0. Then α23+β23+α14+β14α15+β15+α10+β10 is equal to [2023]

Q 22 : Let α,β be the roots of the equation x2-2x+2=0. Then α14+β14 is equal to [2023]

Q 23 : Let S={x:x∈R, (3+2)x2-4+(3-2)x2-4=10}. Then n(S) is equal to [2023]

Q 24 : The number of real solutions of the equation 3(x2+1x2)-2(x+1x)+5=0, is [2023]

Q 25 : The number of real roots of the equation x2-4x+3+x2-9=4x2-14x+6, is [2023]

Q 26 : The equation e4x+8e3x+13e2x-8ex+1=0, x∈R has [2023]

Q 27 : If a and b are the roots of the equation x2-7x-1=0, then the value of a21+b21+a17+b17a19+b19 is equal to ____ . [2023]

Q 28 : The number of points, where the curve f(x)=e8x-e6x-3e4x-e2x+1, x∈ℝ cuts the x-axis, is equal to ______ . [2023]

Q 29 : If α and β(α<β) are the roots of the equation (-2+3)(|x-3|)+(x-6x)+(9-23)=0, x≥0, then βα+αβ is equal to: [2026]

Q 30 : The sum of all the roots of the equation (x-1)2-5|x-1|+6=0 is [2026]

Want Us to Email you About Special Offers & Updates?

Q 1 :

The sum of all the solutions of the equation ${(8)}^{2 x} - 16 \cdot {(8)}^{x} + 48 = 0$ is : [2024]

Q 2 :

Let $α, β$ be the roots of the equation $x^{2} + 2 \sqrt{2} x - 1 = 0 .$ The quadratic equation, whose roots are $α^{4} + β^{4}$ and $\frac{1}{10} (α^{6} + β^{6}),$ is: [2024]

Q 3 :

Let $α, β : α > β,$ be the roots of the equation $x^{2} - \sqrt{2} x - \sqrt{3} = 0 .$ Let $P_{n} = α^{n} - β^{n}, n \in N .$ Then $(11 \sqrt{3} - 10 \sqrt{2}) P_{10} + (11 \sqrt{2} + 10) P_{11} - 11 P_{12}$ is equal to [2024]

Q 4 :

Let $α, β$ are the roots of the equation, $x^{2} - x - 1 = 0$ and $S_{n} = 2023 α^{n} + 2024 β^{n},$ then : [2024]

Q 5 :

Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the solutions of the equation $4 x^{4} + 8 x^{3} - 17 x^{2} - 12 x + 9 = 0$ and $(4 + x_{1}^{2}) (4 + x_{2}^{2}) (4 + x_{3}^{2}) (4 + x_{4}^{2}) = \frac{125}{16} m .$ Then the value of $m$ is _______ . [2024]

Q 6 :

If $α$ satisfies the equation $x^{2} + x + 1 = 0$ and ${(1 + α)}^{7} = A + B α + C α^{2}, A, B, C \geq 0,$ then $5 (3 A - 2 B - C)$ is equal to _______ . [2024]

Q 7 :

Let $α, β$ be the roots of the equation $x^{2} - x + 2 = 0$ with $Im (α) > Im (β) .$ Then $α^{6} + α^{4} + β^{4} - 5 α^{2}$ is equal to ____ . [2024]

Q 9 :

Let $a, b, c$ be the lengths of three sides of a triangle satisfying the condition $(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0 .$ If the set of all possible values of $x$ is the interval, $(α, β)$ then $12 (α^{2} + β^{2})$ is equal to _____ . [2024]

Q 10 :

The number of real solutions of the equation $x (x^{2} + 3 | x | + 5 | x - 1 | + 6 | x - 2 |) = 0$ is ______________. [2024]

Q 11 :

Consider the equation $x^{2} + 4 x - n = 0$ , where n $\in$ [20, 100] is a natural number. Then the number of all distinct values of n, for which the given equation has integral roots, is equal to [2025]

Q 12 :

The product of all solutions of the equation $e^{5 {(\log_{e} x)}^{2} + 3} = x^{8}, x > 0$ , is: [2025]

Q 13 :

The product of all the rational roots of the equation ${(x^{2} - 9 x + 11)}^{2} - (x - 4) (x - 5) = 3$ , is equal to [2025]

Q 14 :

The number of real solution(s) of the equation $x^{2} + 3 x + 2 = \min {| x - 3 |, | x + 2 |}$ is : [2025]

Q 15 :

The sum, of the squares of all the roots of the equation $x^{2} + | 2 x - 3 | - 4 = 0$ , is [2025]

Q 16 :

If $α + i β$ and $γ + i δ$ are the roots of $x^{2} - (3 - 2 i) x - (2 i - 2) = 0, i = \sqrt{- 1}$ , then $α γ + β δ$ is equal to : [2025]

Q 17 :

The number of solutions of the equation $(\frac{9}{x} - \frac{9}{\sqrt{x}} + 2) (\frac{2}{x} - \frac{7}{\sqrt{x}} + 3) = 0$ is : [2025]

Q 18 :

If the set of all $a \in R$ , for which the equation $2 x^{2} + (a - 5) x + 15 = 3 a$ has no real root, is the interval $(α, β)$ , and $X = {x \in Z : α < x < β}$ , then $\sum_{x \in X} x^{2}$ is equal to : [2025]

Q 19 :

Let $α, β$ be the roots of the equation $x^{2} - a x - b = 0$ with $I m (α) < I m (β)$ . Let $P_{n} = α^{n} - β^{n}$ . If $P_{3} = - 5 \sqrt{7} i, P_{4} = - 3 \sqrt{7} i, P_{5} = 11 \sqrt{7} i and P_{6} = 45 \sqrt{7} i$ , then $| α^{4} + β^{4} |$ is equal to __________. [2025]

Q 20 :

Let $α, β, γ$ be the three roots of the equation $x^{3} + b x + c = 0 .$ If $β γ = 1 = - α$ , then the value of $b^{3} + 2 c^{3} - 3 α^{3} - 6 β^{3} - 8 γ^{3}$ is equal to [2023]

Q 21 :

Let $α, β$ be the roots of the quadratic equation $x^{2} + \sqrt{6} x + 3 = 0 .$ Then $\frac{α^{23} + β^{23} + α^{14} + β^{14}}{α^{15} + β^{15} + α^{10} + β^{10}}$ is equal to [2023]

Q 22 :

Let $α, β$ be the roots of the equation $x^{2} - \sqrt{2} x + 2 = 0 .$ Then $α^{14} + β^{14}$ is equal to [2023]

Q 23 :

Let $S = {x : x \in R, {(\sqrt{3} + \sqrt{2})}^{x^{2} - 4} + {(\sqrt{3} - \sqrt{2})}^{x^{2} - 4} = 10} .$ Then $n (S)$ is equal to [2023]

Q 24 :

The number of real solutions of the equation $3 (x^{2} + \frac{1}{x^{2}}) - 2 (x + \frac{1}{x}) + 5 = 0$ , is [2023]

Q 25 :

The number of real roots of the equation $\sqrt{x^{2} - 4 x + 3} + \sqrt{x^{2} - 9} = \sqrt{4 x^{2} - 14 x + 6},$ is [2023]

Q 26 :

The equation $e^{4 x} + 8 e^{3 x} + 13 e^{2 x} - 8 e^{x} + 1 = 0, x \in R$ has [2023]

Q 27 :

If $a$ and $b$ are the roots of the equation $x^{2} - 7 x - 1 = 0$ , then the value of $\frac{a^{21} + b^{21} + a^{17} + b^{17}}{a^{19} + b^{19}}$ is equal to ____ . [2023]

Q 28 :

The number of points, where the curve $f (x) = e^{8 x} - e^{6 x} - 3 e^{4 x} - e^{2 x} + 1, x \in ℝ$ cuts the $x$ -axis, is equal to ______ . [2023]

Q 29 :

If $α$ and $β (α < β)$ are the roots of the equation $(- 2 + \sqrt{3}) (| \sqrt{x} - 3 |) + (x - 6 \sqrt{x}) + (9 - 2 \sqrt{3}) = 0, x \geq 0,$ then $\sqrt{\frac{β}{α}} + \sqrt{α β}$ is equal to: [2026]

Q 30 :

The sum of all the roots of the equation ${(x - 1)}^{2} - 5$ $| x - 1 |$ $+ 6 = 0$ is [2026]