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Quadratic Equation and Inequalities question

2019 · Shift 1 · Q26
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  5. /2019 · Shift 1 · Q26

Quadratic Equation and Inequalities question

2019 · Shift 1 · Q26

JEE AdvancedMathematicsQuadratic Equation and InequalitiesMultiple correct+4 / −1
Let α\alphaα and β\betaβ be the roots of x2−x−1=0{x^2} - x - 1 = 0x2−x−1=0, with α\alphaα>β\betaβ. For all positive integers n, define an=αn−βnα−β, n≥1b1=1 and bn=an−1+an+1, n≥2{a_n} = {{{\alpha ^n} - {\beta ^n}} \over {\alpha - \beta }},\,n \ge 1{b_1} = 1\,and\,{b_n} = {a_{n - 1}} + {a_{n + 1}},\,n \ge 2an​=α−βαn−βn​,n≥1b1​=1andbn​=an−1​+an+1​,n≥2 Then which of the following options is/are correct?
  1. A
    ∑n=1∞bn10n=889\sum\limits_{n = 1}^\infty {{{{b_n}} \over {{{10}^n}}}} = {8 \over {89}}n=1∑∞​10nbn​​=898​
  2. B
    bn = α\alphaα n + β\betaβ n for all n ≥\ge≥ 1
  3. C
    a1 + a2 + a3 + ... + an = an+2 −-− 1 for all n ≥\ge≥ 1
  4. D
    ∑n=1∞an10n=1089\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{{10}^n}}}} = {10 \over {89}}n=1∑∞​10nan​​=8910​
View written solutionFree

Correct answer: B, C, D

1. Analyze the Quadratic Equation and its Roots

The given quadratic equation is x2−x−1=0x^2 - x - 1 = 0x2−x−1=0. Let α\alphaα and β\betaβ be its roots. By the quadratic formula, the roots are x=1±1−4(1)(−1)2=1±52x = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 \pm \sqrt{5}}{2}x=21±1−4(1)(−1)​​=21±5​​. Given α>β\alpha > \betaα>β, we have: α=1+52\alpha = \frac{1 + \sqrt{5}}{2}α=21+5​​ and β=1−52\beta = \frac{1 - \sqrt{5}}{2}β=21−5​​.

From Vieta's formulas, we have:

  • Sum of roots: α+β=1\alpha + \beta = 1α+β=1
  • Product of roots: αβ=−1\alpha \beta = -1αβ=−1

Also, since α\alphaα and β\betaβ are roots of the equation, they satisfy it:

  • α2−α−1=0  ⟹  α2=α+1\alpha^2 - \alpha - 1 = 0 \implies \alpha^2 = \alpha + 1α2−α−1=0⟹α2=α+1
  • β2−β−1=0  ⟹  β2=β+1\beta^2 - \beta - 1 = 0 \implies \beta^2 = \beta + 1β2−β−1=0⟹β2=β+1

From these, we can also derive other useful relations:

  • α−β=1+52−1−52=5\alpha - \beta = \frac{1 + \sqrt{5}}{2} - \frac{1 - \sqrt{5}}{2} = \sqrt{5}α−β=21+5​​−21−5​​=5​
  • α−1=1/α\alpha - 1 = 1/\alphaα−1=1/α and β−1=1/β\beta - 1 = 1/\betaβ−1=1/β
  • α−1=−β\alpha^{-1} = -\betaα−1=−β and β−1=−α\beta^{-1} = -\alphaβ−1=−α

2. Analyze the sequences ana_nan​ and bnb_nbn​

Sequence ana_nan​: an=αn−βnα−βa_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}an​=α−βαn−βn​ for n≥1n \ge 1n≥1. This is the Binet's formula for the Fibonacci sequence (FnF_nFn​). Let's check the first few terms:

  • a1=α−βα−β=1a_1 = \frac{\alpha - \beta}{\alpha - \beta} = 1a1​=α−βα−β​=1
  • a2=α2−β2α−β=α+β=1a_2 = \frac{\alpha^2 - \beta^2}{\alpha - \beta} = \alpha + \beta = 1a2​=α−βα2−β2​=α+β=1
  • a3=α3−β3α−β=α2+αβ+β2=(α+β)2−αβ=12−(−1)=2a_3 = \frac{\alpha^3 - \beta^3}{\alpha - \beta} = \alpha^2 + \alpha\beta + \beta^2 = (\alpha+\beta)^2 - \alpha\beta = 1^2 - (-1) = 2a3​=α−βα3−β3​=α2+αβ+β2=(α+β)2−αβ=12−(−1)=2 So, the sequence ana_nan​ is 1,1,2,3,5,…1, 1, 2, 3, 5, \dots1,1,2,3,5,…, which are the Fibonacci numbers, an=Fna_n=F_nan​=Fn​. The sequence satisfies the recurrence relation an=an−1+an−2a_n = a_{n-1} + a_{n-2}an​=an−1​+an−2​ for n≥3n \ge 3n≥3.

Sequence bnb_nbn​: b1=1b_1 = 1b1​=1 and bn=an−1+an+1b_n = a_{n-1} + a_{n+1}bn​=an−1​+an+1​ for n≥2n \ge 2n≥2. Let's analyze bnb_nbn​ for n≥2n \ge 2n≥2: bn=an−1+an+1=αn−1−βn−1α−β+αn+1−βn+1α−βb_n = a_{n-1} + a_{n+1} = \frac{\alpha^{n-1} - \beta^{n-1}}{\alpha - \beta} + \frac{\alpha^{n+1} - \beta^{n+1}}{\alpha - \beta}bn​=an−1​+an+1​=α−βαn−1−βn−1​+α−βαn+1−βn+1​ bn=(αn−1+αn+1)−(βn−1+βn+1)α−βb_n = \frac{(\alpha^{n-1} + \alpha^{n+1}) - (\beta^{n-1} + \beta^{n+1})}{\alpha - \beta}bn​=α−β(αn−1+αn+1)−(βn−1+βn+1)​ bn=αn(α−1+α)−βn(β−1+β)α−βb_n = \frac{\alpha^n(\alpha^{-1} + \alpha) - \beta^n(\beta^{-1} + \beta)}{\alpha - \beta}bn​=α−βαn(α−1+α)−βn(β−1+β)​ Using α−1=−β\alpha^{-1} = -\betaα−1=−β and β−1=−α\beta^{-1} = -\alphaβ−1=−α:

  • α−1+α=−β+α=α−β\alpha^{-1} + \alpha = -\beta + \alpha = \alpha - \betaα−1+α=−β+α=α−β
  • β−1+β=−α+β=−(α−β)\beta^{-1} + \beta = -\alpha + \beta = -(\alpha - \beta)β−1+β=−α+β=−(α−β) Substituting these into the expression for bnb_nbn​: bn=αn(α−β)−βn(−(α−β))α−β=(αn+βn)(α−β)α−β=αn+βnb_n = \frac{\alpha^n(\alpha - \beta) - \beta^n(-(\alpha - \beta))}{\alpha - \beta} = \frac{(\alpha^n + \beta^n)(\alpha - \beta)}{\alpha - \beta} = \alpha^n + \beta^nbn​=α−βαn(α−β)−βn(−(α−β))​=α−β(αn+βn)(α−β)​=αn+βn This holds for n≥2n \ge 2n≥2. For n=1n=1n=1, we are given b1=1b_1=1b1​=1. Let's check our derived formula: α1+β1=1\alpha^1 + \beta^1 = 1α1+β1=1. So the formula bn=αn+βnb_n = \alpha^n + \beta^nbn​=αn+βn is valid for all n≥1n \ge 1n≥1. This sequence represents the Lucas numbers LnL_nLn​.

3. Evaluate the Options

Option B: bn=αn+βnb_n = \alpha^n + \beta^nbn​=αn+βn for all n≥1n \ge 1n≥1 As shown above, this is correct.

Option C: a1+a2+a3+⋯+an=an+2−1a_1 + a_2 + a_3 + \dots + a_n = a_{n+2} - 1a1​+a2​+a3​+⋯+an​=an+2​−1 for all n≥1n \ge 1n≥1 This is a standard identity for Fibonacci numbers. Let's prove it using the explicit formula for aka_kak​. ∑k=1nak=∑k=1nαk−βkα−β=1α−β(∑k=1nαk−∑k=1nβk)\sum_{k=1}^n a_k = \sum_{k=1}^n \frac{\alpha^k - \beta^k}{\alpha - \beta} = \frac{1}{\alpha - \beta} \left( \sum_{k=1}^n \alpha^k - \sum_{k=1}^n \beta^k \right)∑k=1n​ak​=∑k=1n​α−βαk−βk​=α−β1​(∑k=1n​αk−∑k=1n​βk) These are sums of geometric progressions:

  • ∑k=1nαk=ααn−1α−1\sum_{k=1}^n \alpha^k = \alpha \frac{\alpha^n - 1}{\alpha - 1}∑k=1n​αk=αα−1αn−1​
  • ∑k=1nβk=ββn−1β−1\sum_{k=1}^n \beta^k = \beta \frac{\beta^n - 1}{\beta - 1}∑k=1n​βk=ββ−1βn−1​ Since α−1=1/α\alpha-1 = 1/\alphaα−1=1/α and β−1=1/β\beta-1=1/\betaβ−1=1/β (from α2−α−1=0  ⟹  α−1=1/α\alpha^2-\alpha-1=0 \implies \alpha-1=1/\alphaα2−α−1=0⟹α−1=1/α):
  • ∑k=1nαk=ααn−11/α=α2(αn−1)=αn+2−α2\sum_{k=1}^n \alpha^k = \alpha \frac{\alpha^n - 1}{1/\alpha} = \alpha^2 (\alpha^n - 1) = \alpha^{n+2} - \alpha^2∑k=1n​αk=α1/ααn−1​=α2(αn−1)=αn+2−α2
  • ∑k=1nβk=β2(βn−1)=βn+2−β2\sum_{k=1}^n \beta^k = \beta^2(\beta^n-1) = \beta^{n+2} - \beta^2∑k=1n​βk=β2(βn−1)=βn+2−β2 So, the sum is: 1α−β[(αn+2−α2)−(βn+2−β2)]=αn+2−βn+2α−β−α2−β2α−β=an+2−a2\frac{1}{\alpha - \beta} [(\alpha^{n+2} - \alpha^2) - (\beta^{n+2} - \beta^2)] = \frac{\alpha^{n+2} - \beta^{n+2}}{\alpha - \beta} - \frac{\alpha^2 - \beta^2}{\alpha - \beta} = a_{n+2} - a_2α−β1​[(αn+2−α2)−(βn+2−β2)]=α−βαn+2−βn+2​−α−βα2−β2​=an+2​−a2​ Since a2=1a_2 = 1a2​=1, we have ∑k=1nak=an+2−1\sum_{k=1}^n a_k = a_{n+2} - 1∑k=1n​ak​=an+2​−1. This is correct.

Option D: ∑n=1∞an10n=1089\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{{10}^n}}}} = {{10} \over {89}}n=1∑∞​10nan​​=8910​ Let S=∑n=1∞anxnS = \sum_{n=1}^\infty a_n x^nS=∑n=1∞​an​xn with x=1/10x=1/10x=1/10. This is the generating function for ana_nan​. Since ana_nan​ satisfies the recurrence an=an−1+an−2a_n = a_{n-1} + a_{n-2}an​=an−1​+an−2​ for n≥3n \ge 3n≥3 with a1=1,a2=1a_1=1, a_2=1a1​=1,a2​=1, we have: S(1−x−x2)=a1x+(a2−a1)x2=1⋅x+(1−1)x2=xS(1-x-x^2) = a_1 x + (a_2-a_1)x^2 = 1\cdot x + (1-1)x^2 = xS(1−x−x2)=a1​x+(a2​−a1​)x2=1⋅x+(1−1)x2=x. So, S=x1−x−x2S = \frac{x}{1-x-x^2}S=1−x−x2x​. For x=1/10x=1/10x=1/10: S=1/101−1/10−(1/10)2=1/101−1/10−1/100=1/10(100−10−1)/100=1/1089/100=1089S = \frac{1/10}{1 - 1/10 - (1/10)^2} = \frac{1/10}{1 - 1/10 - 1/100} = \frac{1/10}{(100-10-1)/100} = \frac{1/10}{89/100} = \frac{10}{89}S=1−1/10−(1/10)21/10​=1−1/10−1/1001/10​=(100−10−1)/1001/10​=89/1001/10​=8910​ This is correct.

Option A: ∑n=1∞bn10n=889\sum\limits_{n = 1}^\infty {{{{b_n}} \over {{{10}^n}}}} = {8 \over {89}}n=1∑∞​10nbn​​=898​ Let T=∑n=1∞bnxnT = \sum_{n=1}^\infty b_n x^nT=∑n=1∞​bn​xn with x=1/10x=1/10x=1/10. We have bn=αn+βnb_n = \alpha^n + \beta^nbn​=αn+βn. So: T=∑n=1∞(αn+βn)xn=∑n=1∞(αx)n+∑n=1∞(βx)nT = \sum_{n=1}^\infty (\alpha^n + \beta^n)x^n = \sum_{n=1}^\infty (\alpha x)^n + \sum_{n=1}^\infty (\beta x)^nT=∑n=1∞​(αn+βn)xn=∑n=1∞​(αx)n+∑n=1∞​(βx)n These are geometric series. For x=1/10x=1/10x=1/10, ∣αx∣<1|\alpha x| < 1∣αx∣<1 and ∣βx∣<1|\beta x|<1∣βx∣<1, so they converge. T=αx1−αx+βx1−βx=αx(1−βx)+βx(1−αx)(1−αx)(1−βx)T = \frac{\alpha x}{1 - \alpha x} + \frac{\beta x}{1 - \beta x} = \frac{\alpha x(1-\beta x) + \beta x(1-\alpha x)}{(1-\alpha x)(1-\beta x)}T=1−αxαx​+1−βxβx​=(1−αx)(1−βx)αx(1−βx)+βx(1−αx)​ T=αx−αβx2+βx−αβx21−(α+β)x+αβx2=(α+β)x−2αβx21−(α+β)x+αβx2T = \frac{\alpha x - \alpha\beta x^2 + \beta x - \alpha\beta x^2}{1 - (\alpha+\beta)x + \alpha\beta x^2} = \frac{(\alpha+\beta)x - 2\alpha\beta x^2}{1 - (\alpha+\beta)x + \alpha\beta x^2}T=1−(α+β)x+αβx2αx−αβx2+βx−αβx2​=1−(α+β)x+αβx2(α+β)x−2αβx2​ Substituting α+β=1\alpha+\beta=1α+β=1 and αβ=−1\alpha\beta=-1αβ=−1: T=1⋅x−2(−1)x21−1⋅x+(−1)x2=x+2x21−x−x2T = \frac{1 \cdot x - 2(-1)x^2}{1 - 1 \cdot x + (-1)x^2} = \frac{x+2x^2}{1-x-x^2}T=1−1⋅x+(−1)x21⋅x−2(−1)x2​=1−x−x2x+2x2​ For x=1/10x=1/10x=1/10: T=1/10+2(1/100)1−1/10−1/100=12/10089/100=1289T = \frac{1/10 + 2(1/100)}{1 - 1/10 - 1/100} = \frac{12/100}{89/100} = \frac{12}{89}T=1−1/10−1/1001/10+2(1/100)​=89/10012/100​=8912​ The option states the sum is 8/898/898/89, which is incorrect.

Conclusion

Options B, C, and D are correct. Option A is incorrect.

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