Limits Question 6
Question 6 - 2024 (29 Jan Shift 2)
Let the slope of the line $45 x+5 y+3=0$ be $27 r _1+\frac{9 r _2}{2}$ for some $r _1, \quad r _2 \in R$. Then
$\operatorname{Lim} _{x \rightarrow 3}\left(\int _3^{x} \frac{8 t^{2}}{\frac{3 r _2 x}{2}-r _2 x^{2}-r _1 x^{3}-3 x} d t\right)$ is equal to
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Answer (12)
Solution
According to the question,
$27 r _1+\frac{9 r _2}{2}=-9$
$\lim _{x \rightarrow 3} \frac{\int _3^{x} 8 t^{2} d t}{\frac{3 r _2 x}{2}-r _2 x^{2}-r _1 x^{3}-3 x}$
$=\lim _{x \rightarrow 3} \frac{8 x^{2} \text { }}{\frac{3 r _2^{2}}{2}-2 r _2 x-3 r _1 x^{2}-3}$ (using LH’ Rule)
$=\frac{\text { } 72}{\frac{3 r _2}{2}-6 r _2-27 r _1-3}$
$=\frac{\text { math } 72}{-\frac{9 r _2}{2}-27 r _1-3}$
$=\frac{72}{9-3}=12$
