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

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Limits

Let the slope of the line

45 x + 5 y + 3 = 0

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 ______.

Answer: 12

The line

45 x + 5 y + 3 = 0

can be written

y = - 9 x - \frac{3}{5}

, so its slope is

- 9

.

Hence

27 r_{1} + \frac{9 r _{2}}{2} = - 9

... (*)

Consider

L = x \to 3 lim \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}

. At

x = 3

the numerator is

0

(equal limits), and the denominator is

\frac{9 r _{2}}{2} - 9 r_{2} - 27 r_{1} - 9 = - \frac{9 r _{2}}{2} - 27 r_{1} - 9

.

Using (*),

- (27 r_{1} + \frac{9 r _{2}}{2}) = 9

, so the denominator at

x = 3

9 - 9 = 0

: a

\frac{0}{0}

form, so apply L'Hopital in

x

.

Numerator derivative:

\frac{d}{d x} \int_{3}^{x} 8 t^{2} d t = 8 x^{2}

.

Denominator derivative:

\frac{3 r _{2}}{2} - 2 r_{2} x - 3 r_{1} x^{2} - 3

.

At

x = 3

: numerator

= 8 (9) = 72

; denominator

= \frac{3 r _{2}}{2} - 6 r_{2} - 27 r_{1} - 3 = - \frac{9 r _{2}}{2} - 27 r_{1} - 3 = 9 - 3 = 6

(again using (*)).

L = \frac{72}{6} = 12

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Integer answer