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QJEEAdv24Ma7e4b: JEE Advanced 2024 Mathematics - Class 12 JEEAdv24Ma7e4b Solution: Let the function be defined by Define . Let denote the number…

Let the function f:[1,∞)→Rf:[1, \infty) \rightarrow \mathbb{R} be defined by

f(t)={(−1)n+12,amp; if t=2n−1,n∈N,(2n+1−t)2f(2n−1)+(t−(2n−1))2f(2n+1),amp; if 2n−1lt;tlt;2n+1,n∈N.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)=∫1xf(t)dt,x∈(1,∞)g(x)=\int_1^x f(t) d t, x \in(1, \infty) . Let α\alpha denote the number of solutions of the equation g(x)=0g(x)=0 in the interval (1,8](1,8] and β=lim⁡x→l+g(x)x−1\beta=\lim \limits_{x \rightarrow l+} \frac{g(x)}{x-1} .

Then the value of α+β\alpha+\beta is equal to _______.

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JEE Advanced 2024 Paper 2 Online Mathematics - Application of Integration Question 1 English Explanation

f(t)={2amp;;amp;t=14−2tamp;;amp;1lt;tlt;3−2amp;;amp;t=3−8−2tamp;;amp;3lt;tlt;52amp;;amp;t=512−2tamp;;amp;5lt;tlt;7−2amp;;amp;t=7−16+2tamp;;amp;7lt;tlt;9\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)=∫1xf(t)dt;g′(x)=f(x) for x∈(1,8]g(x)=0⇒x=3,5,7∴α=3β=limx→1+g(x)x−1

Apply L'pital

amp;=g′(1+)1=f(1+)amp;β=2amp;∴α+β=5\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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