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JEE Main 201912 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
12 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Quadratic Equation

JEE Main 2019 · 12 April · Morning Shift · Mathematics · Quadratic Equation

If α and β are the roots of the equation 375 x ^ 2 - 25x - 2 = 0,...

If α\alphaand β\beta are the roots of the equation 375x2−25x−2=0,375{x}^{2} - 25x - 2 = 0, then lim⁡n→∞∑r=1nαr+lim⁡n→∞∑r=1nβr\lim_{n \to \infty}\mathop{∑}_{r = 1}^{n}{\alpha}^{r} + \lim_{n \to \infty}\mathop{∑}_{r = 1}^{n}{\beta}^{r} is equal to:

Options

  1. A112\dfrac{1}{12}
  2. B21346\dfrac{21}{346}
  3. C7116\dfrac{7}{116}
  4. D29358\dfrac{29}{358}

Correct answer

A. 112\dfrac{1}{12}

Step-by-step solution

For given quadratic 375x2−25x−2=0,375{x}^{2} - 25x - 2 = 0, its roots are,

α,β=25±252+2×4×3752×375⇒∣α∣<1,∣β∣<1\alpha, \beta = \dfrac{25 \pm \sqrt{{25}^{2} + 2 \times 4 \times 375}}{2 \times 375} ⇒ \left|\alpha\right| < 1, \left|\beta\right| < 1

Also, α+β=25375,αβ=−2375\alpha + \beta = \dfrac{25}{375}, \alpha\beta = \dfrac{- 2}{375}

Now lim⁡n→∞∑r=1nαr+lim⁡n→∞∑r=1nβr\lim_{n \to \infty}\mathop{∑}_{r = 1}^{n}{\alpha}^{r} + \lim_{n \to \infty}\mathop{∑}_{r = 1}^{n}{\beta}^{r}

= \dfrac{\alpha}{1 - \alpha} + \dfrac{\beta}{1 - \beta} \left[a+ar+ar2+.....infiniteterms=a1−rif∣r∣<1\begin{array}{ll}a + ar + a{r}^{2} + . . . . . infiniteterms \\ = \dfrac{a}{1 - r}if \left|r\right| < 1\end{array}\right]

=(α+β)−2αβ1−(α+β)+αβ=25375+43751−25375−2375=29348=112= \dfrac{\left(\alpha + \beta\right) - 2\alpha\beta}{1 - \left(\alpha + \beta\right) + \alpha\beta} = \dfrac{\dfrac{25}{375} + \dfrac{4}{375}}{1 - \dfrac{25}{375} - \dfrac{2}{375}} = \dfrac{29}{348} = \dfrac{1}{12}

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