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JEE Advanced · Mathematics · 16. Limits

Let the function \(f:[1, \infty) \rightarrow \mathbb{R}\) be defined by
\(f(t)=\left\{\begin{array}{cc}(-1)^{n+1} 2, & \text { if } t=2 n-1, n \in \mathbb{N}, \\ \frac{(2 n+1-t)}{2} f(2 n-1)+\frac{(t-(2 n-1))}{2} f(2 n+1), & \text { if } 2 n-1 < t < 2 n+1, n \in \mathbb{N} .\end{array}\right.\)
Define \(g(x)=\int_1^x f(t) d t, x \in(1, \infty)\). Let \(\alpha\) denote the number of solutions of the equation \(g(x)=0\) in the interval \((1,8]\) and \(\beta=\lim _{x \rightarrow 1^+} \frac{g(x)}{x-1}\). Then the value of \(\alpha+\beta\) is equal to ________.

  1. A 5
  2. B 10
  3. C 12
  4. D 9
Verified Solution

Answer & Solution

Correct Answer

(A) 5

Step-by-step Solution

Detailed explanation

\(\mathrm{f}(\mathrm{t})=\left\{\begin{array}{ccc}2 & ; & \mathrm{t}=1 \\ 4-2 \mathrm{t} & ; & 1 < \mathrm{t} < 3 \\ -2 & ; & \mathrm{t}=3 \\ -8-2 \mathrm{t} & ; & 3 < \mathrm{t} < 5 \\ 2 & ; & \mathrm{t}=5 \\ 12-2 \mathrm{t} & ; & 5 < \mathrm{t} < 7 \\ -2 & ; & \mathrm{t}=7 \\ -16+2 \mathrm{t} & ; & 7 < \mathrm{t} < 9\end{array}\right.\)
\(g(x)=\int_1^x f(t) d t \Rightarrow g^{\prime}(x)=f(x)\)
for \(x \in(1,8]\)
\(g(x)=0 \Rightarrow x=3,5,7 \therefore \alpha=3\)
\(\beta=\lim _{x \rightarrow 1^{+}} \frac{g(x)}{x-1}\)
Apply L'Hospital's
\(\begin{aligned} & =\frac{g^{\prime}\left(1^{+}\right)}{1}=f\left(1^{+}\right) \\ & \beta=2 \\ & \therefore \alpha+\beta=5\end{aligned}\)

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