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

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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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

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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

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4

(3)

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

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

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

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

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

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

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

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

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

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

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

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

[IMAGE 12]---------

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

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

[IMAGE 13]---------

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

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

[IMAGE 14]---------

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

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

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

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

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(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]

0%

(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]

0%

(2)

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

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

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

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

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

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

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

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

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

$It is a quadratic equation.$

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

$Hence, 2 real solutions exist.$

Topic Question Set

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]

0%

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]

0%

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Topic Question Set

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]

0%

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]

0%

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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]