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Limits Continuity and Differentiability question

2019 · 12 Apr · Shift 1 · Q24
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  5. /2019 · 12 Apr · Shift 1 · Q24

Limits Continuity and Differentiability question

2019 · 12 Apr · Shift 1 · Q24

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
If α\alphaα and β\betaβ are the roots of the equation 375x2 – 25x – 2 = 0, then lim⁡n→∞∑r=1nαr+lim⁡n→∞∑r=1nβr\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{\alpha ^r}} + \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{\beta ^r}}n→∞lim​r=1∑n​αr+n→∞lim​r=1∑n​βr is equal to :
  1. A
    7116{7 \over {116}}1167​
  2. B
    29348{{29} \over {348}}34829​
  3. C
    112{1 \over {12}}121​
  4. D
    21346{{21} \over {346}}34621​
View written solutionFree

Correct answer: B

  1. Given quadratic and its roots

The roots α,β\alpha, \betaα,β satisfy 375x2−25x−2=0.375x^2-25x-2=0.375x2−25x−2=0.

We need to find lim⁡n→∞∑r=1nαr+lim⁡n→∞∑r=1nβr.\lim_{n\to\infty}\sum_{r=1}^n \alpha^r+\lim_{n\to\infty}\sum_{r=1}^n \beta^r.limn→∞​∑r=1n​αr+limn→∞​∑r=1n​βr.

  1. Check whether the geometric series converge

First find the roots approximately:

=\frac{25\pm\sqrt{625+3000}}{750} =\frac{25\pm\sqrt{3625}}{750}.$$ Since $\sqrt{3625}=5\sqrt{145}\approx 60.2$, $$\alpha\approx \frac{25+60.2}{750}\approx 0.1136, \qquad \beta\approx \frac{25-60.2}{750}\approx -0.0469.$$ Thus $$|\alpha|<1,\qquad |\beta|<1,$$ so both geometric series converge. Hence, $$\lim_{n\to\infty}\sum_{r=1}^n \alpha^r=\frac{\alpha}{1-\alpha}, \qquad \lim_{n\to\infty}\sum_{r=1}^n \beta^r=\frac{\beta}{1-\beta}.$$ Therefore required value is $$S=\frac{\alpha}{1-\alpha}+\frac{\beta}{1-\beta}.$$ --- 3. **Combine the two terms** $$S=\frac{\alpha(1-\beta)+\beta(1-\alpha)}{(1-\alpha)(1-\beta)} =\frac{\alpha+\beta-2\alpha\beta}{1-(\alpha+\beta)+\alpha\beta}.$$ Now use Vieta's formulas for $$375x^2-25x-2=0.$$ So, $$\alpha+\beta=\frac{-(-25)}{375}=\frac{25}{375}=\frac{1}{15},$$ $$\alpha\beta=\frac{-2}{375}.$$ Substitute into $S$: Numerator: $$\alpha+\beta-2\alpha\beta=\frac{1}{15}-2\left(-\frac{2}{375}\right) =\frac{1}{15}+\frac{4}{375} =\frac{25+4}{375}= rac{29}{375}.$$ Denominator: $$1-(\alpha+\beta)+\alpha\beta =1-\frac{1}{15}-\frac{2}{375} =\frac{375-25-2}{375}= rac{348}{375}.$$ Thus, $$S=\frac{29/375}{348/375}=\frac{29}{348}.$$ --- 4. **Match with the options** $$\boxed{\frac{29}{348}}$$ which corresponds to **Option B**. --- 5. **Comparison with stored answer** Stored correct answer: **B** My derived answer: **B** So they agree.
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