Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Quadratic Equation and Inequalities question

2021 · 31 Aug · Shift 1 · Q30
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Quadratic Equation and Inequalities
  5. /2021 · 31 Aug · Shift 1 · Q30

Quadratic Equation and Inequalities question

2021 · 31 Aug · Shift 1 · Q30

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
cosec18 ∘^\circ∘ is a root of the equation :
  1. A
    x2 + 2x −-− 4 = 0
  2. B
    4x2 + 2x −-− 1 = 0
  3. C
    x2 −-− 2x + 4 = 0
  4. D
    x2 −-− 2x −-− 4 = 0
View written solutionFree

Correct answer: D

  1. We need to check which equation has root x=csc⁡18∘x = \csc 18^\circx=csc18∘.

  2. First find the exact value of sin⁡18∘\sin 18^\circsin18∘.

A standard value is:

sin⁡18∘=5−14\sin 18^\circ = \frac{\sqrt{5}-1}{4}sin18∘=45​−1​

So,

csc⁡18∘=1sin⁡18∘=45−1\csc 18^\circ = \frac{1}{\sin 18^\circ} = \frac{4}{\sqrt{5}-1}csc18∘=sin18∘1​=5​−14​

Rationalizing,

csc⁡18∘=4(5+1)5−1=5+1\csc 18^\circ = \frac{4(\sqrt{5}+1)}{5-1} = \sqrt{5}+1csc18∘=5−14(5​+1)​=5​+1

Thus,

x=5+1x = \sqrt{5}+1x=5​+1
  1. Now find the quadratic equation satisfied by x=1+5x = 1+\sqrt{5}x=1+5​.

Subtract 1 from both sides:

x−1=5x-1 = \sqrt{5}x−1=5​

Squaring,

(x−1)2=5(x-1)^2 = 5(x−1)2=5 x2−2x+1=5x^2 - 2x + 1 = 5x2−2x+1=5 x2−2x−4=0x^2 - 2x - 4 = 0x2−2x−4=0
  1. Compare with the options:
  • A: x2+2x−4=0x^2 + 2x - 4 = 0x2+2x−4=0
  • B: 4x2+2x−1=04x^2 + 2x - 1 = 04x2+2x−1=0
  • C: x2−2x+4=0x^2 - 2x + 4 = 0x2−2x+4=0
  • D: x2−2x−4=0x^2 - 2x - 4 = 0x2−2x−4=0

So the correct option is:

D\boxed{D}D​
  1. Verification by substitution: For x=1+5x=1+\sqrt{5}x=1+5​,
x2−2x−4=(1+5)2−2(1+5)−4x^2 - 2x - 4 = (1+\sqrt{5})^2 - 2(1+\sqrt{5}) - 4x2−2x−4=(1+5​)2−2(1+5​)−4 =(1+25+5)−2−25−4=0= (1+2\sqrt{5}+5) - 2 - 2\sqrt{5} - 4 = 0=(1+25​+5)−2−25​−4=0

Hence verified.

  1. Comparison with stored correct answer: Stored correct answer is DDD, which matches our result.
PreviousNext

More from Quadratic Equation and Inequalities

  • The sum of the roots of the equation x+1−2log2​(3+2x)+2log4​(10−2−x)=0, is :2021 · MCQ
  • Let α and β be the roots of the equation 5x2 + 6x – 2 = 0. If Sn =α n + β n, n = 1, 2, 3...., then :2020 · MCQ
  • Let f(x) be a quadratic polynomial such that f(–1) + f(2) = 0. If one of the roots of f(x) = 0 is 3, then its other root lies in :2020 · MCQ
  • If α and β are the roots of the equation x2 + px + 2 = 0 and α1​ and β1​ are the roots of the equation 2x2 + 2qx + 1 = 0, then (α−α1​)(β−β1​)(α+β1​)(β+α1​)…2020 · MCQ
  • The set of all real values of λ for which the quadratic equations, (λ 2 + 1)x2 – 4 λ x + 2 = 0 always have exactly one root in the interval (0, 1) is :2020 · MCQ
  • Let α and β be the roots of x2 - 3x + p=0 and γ and δ be the roots of x2 - 6x + q = 0. If α,β,γ,δ form a geometric progression.Then ratio (2q + p) : (2q - p) is:2020 · MCQ
  • Let [t] denote the greatest integer ≤ t. Then the equation in x, [x]2 + 2[x+2] - 7 = 0 has :2020 · MCQ
  • Let λe0 be in R. If α and β are the roots of the equation, x2 - x + 2 λ= 0 and α and γ are the roots of the equation, 3x2−10x+27λ=0, then λβγ​ is…2020 · MCQ