Let the slope of the line 45 x + 5 y + 3 = 0 be 27 r 1 + 9 r 2 2 for some r 1 , r 2 R . Then lim x 3 ( 3 x 8 t 2 3 r 2 x 2 - r 2 x 2 - r 1 x 3 - 3 x ? d t ) is equal to [2024]

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Solution

Q.
$Let the slope of the line 45 x + 5 y + 3 = 0 be 27 r_{1} + \frac{9 r_{2}}{2} for some r_{1}, r_{2} \in R .$ $Then \lim_{x \to 3}$ $(\int_{3}^{x} \frac{8 t^{2}}{\frac{3 r_{2} x}{2} - r_{2} x^{2} - r_{1} x^{3} - 3 x} d t)$ $is equal to______.$ [2024]

Ans.
(12)

Slope of line $45 x + 5 y + 3 = 0$ is $\frac{- 45}{5} i . e, - 9$

Now, $- 9 = 27 (- 2) + \frac{9 \times 10}{2} .$ So, $r_{1} = - 2$ and $r_{2} = 10,$ we have $\lim_{x \to 3} \int_{3}^{x} \frac{8 t^{2}}{15 x - 10 x^{2} + 2 x^{3} - 3 x} d t$

$= \lim_{x \to 3} \frac{1}{15 x - 10 x^{2} + 2 x^{3} - 3 x} {[\frac{8 t^{3}}{3}]}_{3}^{x}$

$= \lim_{x \to 3} \frac{1}{15 x - 10 x^{2} + 2 x^{3} - 3 x} [\frac{8 x^{3}}{3} - 72]$

$= \lim_{x \to 3} \frac{8 x^{2}}{15 - 20 x + 6 x^{2} - 3} = \frac{72}{15 - 60 + 54 - 3} = 12$

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