Let the slope of the line 45 x+5 y+3=0 be 27 r_1+\frac9 r_22 for some r_1, r_2 \in \mathbbR . Then \lim_x arrow 3(\int_3^x \frac8 t^2\frac3 r_2 x2-r_2 x^2-r_1 x^3-3 x d t) is equal to \underline\hspace2cm .

Definite Integration PYQ — JEE Main 2024 · 29 Jan · Shift 2 · Q54 with Solution

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Correct answer: 12

Find the slope of the given line

The line is $45x+5y+3=0.$

Rewrite in slope-intercept form: $5y=-45x-3$ $y=-9x-\frac{3}{5}.$

Hence slope is $m=-9.$

Given that $27r_1+\frac{9r_2}{2}=-9.$ Divide by $9$ : $3r_1+\frac{r_2}{2}=-1. \qquad (1)$

Interpret the limit

We need to evaluate $\lim_{x\to 3}\left(\int_3^x \frac{8t^2}{\frac{3r_2x}{2}-r_2x^2-r_1x^3-3x}\,dt\right).$

Inside the integral, the denominator does not depend on $t$ ; it depends only on the upper-limit variable $x$ . So we can treat it as a constant with respect to $t$ :

=\frac{1}{\frac{3r_2x}{2}-r_2x^2-r_1x^3-3x}\int_3^x 8t^2\,dt.$$ Now, $$\int 8t^2\,dt=\frac{8t^3}{3},$$ so $$\int_3^x 8t^2\,dt=\frac{8}{3}(x^3-27).$$ Therefore the expression becomes $$\lim_{x\to 3} \frac{\frac{8}{3}(x^3-27)}{\frac{3r_2x}{2}-r_2x^2-r_1x^3-3x}.$$ --- 3. **Factor and simplify the denominator using relation (1)** Denominator: $$D(x)=\frac{3r_2x}{2}-r_2x^2-r_1x^3-3x.$$ Factor out $x$: $$D(x)=x\left(\frac{3r_2}{2}-r_2x-r_1x^2-3\right).$$ At $x=3$, $$\frac{3r_2}{2}-3r_2-9r_1-3=-\left(9r_1+\frac{3r_2}{2}+3\right).$$ From (1), multiply by $3$: $$9r_1+\frac{3r_2}{2}=-3.$$ Thus $$9r_1+\frac{3r_2}{2}+3=0,$$ so the denominator also vanishes at $x=3$. Hence it is a $0/0$ form. Now rewrite: $$D(x)=x\left[-\left(r_1x^2+r_2x-\frac{3r_2}{2}+3\right)\right].$$ Using $$3r_1+\frac{r_2}{2}=-1 \implies \frac{3r_2}{2}-3=-9r_1-r_2,$$ we get $$\frac{3r_2}{2}-r_2x-r_1x^2-3 = -(r_1x^2+r_2x+9r_1+r_2).$$ So $$D(x)=-x\big(r_1x^2+r_2x+9r_1+r_2\big).$$ But $$r_1x^2+r_2x+9r_1+r_2 = r_1(x^2-9)+r_2(x-3)+18r_1+2r_2.$$ A cleaner way is to factor directly using the condition that $x=3$ is a root: $$r_1x^2+r_2x+9r_1+r_2=(x-3)(r_1x+3r_1+r_2).$$ Hence $$D(x)=-x(x-3)(r_1x+3r_1+r_2).$$ Also, $$x^3-27=(x-3)(x^2+3x+9).$$ Therefore, $$\frac{\frac{8}{3}(x^3-27)}{D(x)} =\frac{\frac{8}{3}(x-3)(x^2+3x+9)}{-x(x-3)(r_1x+3r_1+r_2)} =\frac{\frac{8}{3}(x^2+3x+9)}{-x(r_1x+3r_1+r_2)}.$$ Now let $x\to 3$: $$\lim_{x\to 3} = \frac{\frac{8}{3}(9+9+9)}{-3(3r_1+3r_1+r_2)} =\frac{\frac{8}{3}\cdot 27}{-3(6r_1+r_2)} =\frac{72}{-3(6r_1+r_2)}.

From (1), multiply by $2$ : $6r_1+r_2=-2.$ So $\lim_{x\to 3}=\frac{72}{-3(-2)}=\frac{72}{6}=12.$

Final answer

$\boxed{12}$

More from Definite Integration

View written solutionFree

Correct answer: 12

Find the slope of the given line

The line is $45x+5y+3=0.$

Rewrite in slope-intercept form: $5y=-45x-3$ $y=-9x-\frac{3}{5}.$

Hence slope is $m=-9.$

Given that $27r_1+\frac{9r_2}{2}=-9.$ Divide by $9$ : $3r_1+\frac{r_2}{2}=-1. \qquad (1)$

Interpret the limit

We need to evaluate $\lim_{x\to 3}\left(\int_3^x \frac{8t^2}{\frac{3r_2x}{2}-r_2x^2-r_1x^3-3x}\,dt\right).$

Inside the integral, the denominator does not depend on $t$ ; it depends only on the upper-limit variable $x$ . So we can treat it as a constant with respect to $t$ :

=\frac{1}{\frac{3r_2x}{2}-r_2x^2-r_1x^3-3x}\int_3^x 8t^2\,dt.$$ Now, $$\int 8t^2\,dt=\frac{8t^3}{3},$$ so $$\int_3^x 8t^2\,dt=\frac{8}{3}(x^3-27).$$ Therefore the expression becomes $$\lim_{x\to 3} \frac{\frac{8}{3}(x^3-27)}{\frac{3r_2x}{2}-r_2x^2-r_1x^3-3x}.$$ --- 3. **Factor and simplify the denominator using relation (1)** Denominator: $$D(x)=\frac{3r_2x}{2}-r_2x^2-r_1x^3-3x.$$ Factor out $x$: $$D(x)=x\left(\frac{3r_2}{2}-r_2x-r_1x^2-3\right).$$ At $x=3$, $$\frac{3r_2}{2}-3r_2-9r_1-3=-\left(9r_1+\frac{3r_2}{2}+3\right).$$ From (1), multiply by $3$: $$9r_1+\frac{3r_2}{2}=-3.$$ Thus $$9r_1+\frac{3r_2}{2}+3=0,$$ so the denominator also vanishes at $x=3$. Hence it is a $0/0$ form. Now rewrite: $$D(x)=x\left[-\left(r_1x^2+r_2x-\frac{3r_2}{2}+3\right)\right].$$ Using $$3r_1+\frac{r_2}{2}=-1 \implies \frac{3r_2}{2}-3=-9r_1-r_2,$$ we get $$\frac{3r_2}{2}-r_2x-r_1x^2-3 = -(r_1x^2+r_2x+9r_1+r_2).$$ So $$D(x)=-x\big(r_1x^2+r_2x+9r_1+r_2\big).$$ But $$r_1x^2+r_2x+9r_1+r_2 = r_1(x^2-9)+r_2(x-3)+18r_1+2r_2.$$ A cleaner way is to factor directly using the condition that $x=3$ is a root: $$r_1x^2+r_2x+9r_1+r_2=(x-3)(r_1x+3r_1+r_2).$$ Hence $$D(x)=-x(x-3)(r_1x+3r_1+r_2).$$ Also, $$x^3-27=(x-3)(x^2+3x+9).$$ Therefore, $$\frac{\frac{8}{3}(x^3-27)}{D(x)} =\frac{\frac{8}{3}(x-3)(x^2+3x+9)}{-x(x-3)(r_1x+3r_1+r_2)} =\frac{\frac{8}{3}(x^2+3x+9)}{-x(r_1x+3r_1+r_2)}.$$ Now let $x\to 3$: $$\lim_{x\to 3} = \frac{\frac{8}{3}(9+9+9)}{-3(3r_1+3r_1+r_2)} =\frac{\frac{8}{3}\cdot 27}{-3(6r_1+r_2)} =\frac{72}{-3(6r_1+r_2)}.

From (1), multiply by $2$ : $6r_1+r_2=-2.$ So $\lim_{x\to 3}=\frac{72}{-3(-2)}=\frac{72}{6}=12.$

Final answer

$\boxed{12}$

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