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

2019 · 11 Jan · Shift 2 · Q31
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. /2019 · 11 Jan · Shift 2 · Q31

Quadratic Equation and Inequalities question

2019 · 11 Jan · Shift 2 · Q31

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
Let α\alphaα and β\betaβ be the roots of the quadratic equation x2 sin θ\thetaθ– x(sin θ\thetaθ cos θ\thetaθ+ 1) + cos θ\thetaθ= 0 (0 <θ\thetaθ< 45o), and α\alphaα<β\betaβ. Then ∑n=0∞(αn+(−1)nβn)\sum\limits_{n = 0}^\infty {\left( {{\alpha ^n} + {{{{\left( { - 1} \right)}^n}} \over {{\beta ^n}}}} \right)}n=0∑∞​(αn+βn(−1)n​) is equal to :
  1. A
    11+cos⁡θ+11−sin⁡θ{1 \over {1 + \cos \theta }} + {1 \over {1 - \sin \theta }}1+cosθ1​+1−sinθ1​
  2. B
    11−cos⁡θ+11+sin⁡θ{1 \over {1 - \cos \theta }} + {1 \over {1 + \sin \theta }}1−cosθ1​+1+sinθ1​
  3. C
    11−cos⁡θ−11+sin⁡θ{1 \over {1 - \cos \theta }} - {1 \over {1 + \sin \theta }}1−cosθ1​−1+sinθ1​
  4. D
    11+cos⁡θ−11−sin⁡θ{1 \over {1 + \cos \theta }} - {1 \over {1 - \sin \theta }}1+cosθ1​−1−sinθ1​
View written solutionFree

Correct answer: B

  1. Given quadratic

The equation is x2sin⁡θ−x(sin⁡θcos⁡θ+1)+cos⁡θ=0,x^2\sin\theta - x(\sin\theta\cos\theta+1)+\cos\theta=0,x2sinθ−x(sinθcosθ+1)+cosθ=0, with roots α,β\alpha,\betaα,β and 0<θ<45∘0<\theta<45^\circ0<θ<45∘.

So, by Vieta: α+β=sin⁡θcos⁡θ+1sin⁡θ,\alpha+\beta=\frac{\sin\theta\cos\theta+1}{\sin\theta},α+β=sinθsinθcosθ+1​, αβ=cos⁡θsin⁡θ=cot⁡θ.\alpha\beta=\frac{\cos\theta}{\sin\theta}=\cot\theta.αβ=sinθcosθ​=cotθ.

We need to find S=∑n=0∞(αn+(−1)nβn).S=\sum_{n=0}^{\infty}\left(\alpha^n+\frac{(-1)^n}{\beta^n}\right).S=∑n=0∞​(αn+βn(−1)n​).

  1. Identify the roots explicitly

Let us check whether simple values satisfy the quadratic.

Substitute x=cos⁡θx=\cos\thetax=cosθ: sin⁡θcos⁡2θ−cos⁡θ(sin⁡θcos⁡θ+1)+cos⁡θ\sin\theta\cos^2\theta-\cos\theta(\sin\theta\cos\theta+1)+\cos\thetasinθcos2θ−cosθ(sinθcosθ+1)+cosθ =sin⁡θcos⁡2θ−sin⁡θcos⁡2θ−cos⁡θ+cos⁡θ=0.=\sin\theta\cos^2\theta-\sin\theta\cos^2\theta-\cos\theta+\cos\theta=0.=sinθcos2θ−sinθcos2θ−cosθ+cosθ=0. So, x=cos⁡θx=\cos\thetax=cosθ is a root.

Substitute x=csc⁡θx=\csc\thetax=cscθ: sin⁡θ csc⁡2θ−csc⁡θ(sin⁡θcos⁡θ+1)+cos⁡θ\sin\theta\,\csc^2\theta-\csc\theta(\sin\theta\cos\theta+1)+\cos\thetasinθcsc2θ−cscθ(sinθcosθ+1)+cosθ =csc⁡θ−cos⁡θ−csc⁡θ+cos⁡θ=0.=\csc\theta-\cos\theta-\csc\theta+\cos\theta=0.=cscθ−cosθ−cscθ+cosθ=0. So, x=csc⁡θx=\csc\thetax=cscθ is the other root.

Hence the roots are α=cos⁡θ,β=csc⁡θ,\alpha=\cos\theta,\qquad \beta=\csc\theta,α=cosθ,β=cscθ, because for 0<θ<45∘0<\theta<45^\circ0<θ<45∘, 0<cos⁡θ<1,csc⁡θ>1,0<\cos\theta<1,\qquad \csc\theta>1,0<cosθ<1,cscθ>1, thus indeed α<β\alpha<\betaα<β.

  1. Rewrite the series

Now, S=∑n=0∞((cos⁡θ)n+(−1)n(csc⁡θ)n).S=\sum_{n=0}^{\infty}\left((\cos\theta)^n+\frac{(-1)^n}{(\csc\theta)^n}\right).S=∑n=0∞​((cosθ)n+(cscθ)n(−1)n​). Since 1csc⁡θ=sin⁡θ\dfrac{1}{\csc\theta}=\sin\thetacscθ1​=sinθ, S=∑n=0∞(cos⁡θ)n+∑n=0∞(−sin⁡θ)n.S=\sum_{n=0}^{\infty}(\cos\theta)^n+\sum_{n=0}^{\infty}(-\sin\theta)^n.S=∑n=0∞​(cosθ)n+∑n=0∞​(−sinθ)n.

Because 0<θ<45∘0<\theta<45^\circ0<θ<45∘, we have ∣cos⁡θ∣<1,∣sin⁡θ∣<1,|\cos\theta|<1, \qquad |\sin\theta|<1,∣cosθ∣<1,∣sinθ∣<1, so both are convergent geometric series.

  1. Sum the geometric series

First, ∑n=0∞(cos⁡θ)n=11−cos⁡θ.\sum_{n=0}^{\infty}(\cos\theta)^n=\frac{1}{1-\cos\theta}.∑n=0∞​(cosθ)n=1−cosθ1​.

Second, ∑n=0∞(−sin⁡θ)n=11+sin⁡θ.\sum_{n=0}^{\infty}(-\sin\theta)^n=\frac{1}{1+\sin\theta}.∑n=0∞​(−sinθ)n=1+sinθ1​.

Therefore, S=11−cos⁡θ+11+sin⁡θ.S=\frac{1}{1-\cos\theta}+\frac{1}{1+\sin\theta}.S=1−cosθ1​+1+sinθ1​.

  1. Match with the options

This is exactly Option B: 11−cos⁡θ+11+sin⁡θ.\boxed{\frac{1}{1-\cos\theta}+\frac{1}{1+\sin\theta}}.1−cosθ1​+1+sinθ1​​.

  1. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.

PreviousNext

More from Quadratic Equation and Inequalities

  • If α, β and γ are three consecutive terms of a non-constant G.P. such that the equations α x 2 + 2 β x + γ= 0 and x2 + x – 1 = 0 have a common root, then α(β+γ) is equal to :2019 · MCQ
  • If λ be the ratio of the roots of the quadratic equation in x, 3m2x2 + m(m – 4)x + 2 = 0, then the least value of m for which λ+λ1​=1, is2019 · MCQ
  • The number of integral values of m for which the quadratic expression, (1 + 2m)x2 – 2(1 + 3m)x + 4(1 + m), x ∈ R, is always positive, is :2019 · MCQ
  • If λ∈ R is such that the sum of the cubes of the roots of the equation, x2 + (2 −λ) x + (10 −λ) = 0 is minimum, then the magnitude of the difference of the roots of this equation is :2018 · MCQ
  • If tanA and tanB are the roots of the quadratic equation, 3x2 − 10x − 25 = 0, then the value of 3 sin2(A + B) − 10 sin(A + B).cos(A + B) − 25 cos2(A + B) is :2018 · MCQ
  • If f(x) is a quadratic expression such that f (1) + f (2) = 0, and − 1 is a root of f (x) = 0, then the other root of f(x) = 0 is :2018 · MCQ
  • Let p, q and r be real numbers (p e q, r e 0), such that the roots of the equation x+p1​+x+q1​=r1​ are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to :2018 · MCQ
  • If an angle A of a Δ ABC satiesfies 5 cosA + 3 = 0, then the roots of the quadratic equation, 9x2 + 27x + 20 = 0 are :2018 · MCQ