One of the points of intersection of the curves y=1+3 x-2 x^2 and y=\frac1x is (\frac12, 2) . Let the area of the region enclosed by these curves be \frac124(l \sqrt5+m)-n \log _e(1+\sqrt5) , where l, ~m, n \in N . Then l+m+n is equal to

Area Under the Curves PYQ — JEE Main 2024 · 4 Apr · Shift 1 · Q32 with Solution

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Correct answer: A

Find the points of intersection

The curves are $y=1+3x-2x^2$ and $y=\frac{1}{x}.$

At intersection, $1+3x-2x^2=\frac{1}{x}, \quad x\neq 0.$

Multiplying by $x$ : $x+3x^2-2x^3=1$ $2x^3-3x^2-x+1=0.$

Since one intersection is given as $\left(\frac12,2\right)$ , $x=\frac12$ is a root. So factorize: $2x^3-3x^2-x+1=(2x-1)(x^2-x-1).$

Thus the other roots are from $x^2-x-1=0,$ so $x=\frac{1\pm\sqrt5}{2}.$

Hence the three intersection $x$ -values are $\frac{1-\sqrt5}{2},\quad \frac12,\quad \frac{1+\sqrt5}{2}.$

The enclosed finite region is between the positive intersections $x=\frac12 \quad \text{and} \quad x=\frac{1+\sqrt5}{2},$ because the left root is negative and does not form the bounded region with the positive branch relevant here.

Determine which curve is above the other

Take a test point, say $x=1$ : $1+3(1)-2(1)^2=2, \qquad \frac11=1.$ So on $\left[\frac12,\frac{1+\sqrt5}{2}\right]$ , the parabola is above the hyperbola.

Therefore area is $A=\int_{1/2}^{(1+\sqrt5)/2}\left(1+3x-2x^2-\frac1x\right)dx.$

Integrate

= x+\frac{3x^2}{2}-\frac{2x^3}{3}-\ln x.$$ So $$A=\left[x+\frac{3x^2}{2}-\frac{2x^3}{3}-\ln x\right]_{1/2}^{(1+\sqrt5)/2}.$$ Let $$\alpha=\frac{1+\sqrt5}{2}.$$ Then $$\alpha^2=\alpha+1, \qquad \alpha^3=\alpha\alpha^2=\alpha(\alpha+1)=2\alpha+1.$$ Hence $$\alpha+\frac{3\alpha^2}{2}-\frac{2\alpha^3}{3} =\alpha+\frac{3(\alpha+1)}{2}-\frac{2(2\alpha+1)}{3}.$$ Taking LCM $6$, $$=\frac{6\alpha+9\alpha+9-8\alpha-4}{6} =\frac{7\alpha+5}{6}.$$ Since $$\alpha=\frac{1+\sqrt5}{2},$$ we get $$\frac{7\alpha+5}{6}= rac{7(1+\sqrt5)+10}{12}= rac{17+7\sqrt5}{12}.$$ Also, $$\ln\alpha=\ln\left(\frac{1+\sqrt5}{2}\right).$$ So at $x=\alpha$, $$F(\alpha)=\frac{17+7\sqrt5}{12}-\ln\left(\frac{1+\sqrt5}{2}\right).$$ Now at $x=\frac12$, $$F\left(\frac12\right)=\frac12+\frac{3}{2}\cdot\frac14-\frac{2}{3}\cdot\frac18-\ln\frac12.$$ That is, $$F\left(\frac12\right)=\frac12+\frac38-\frac1{12}+\ln2 =\frac{6+ ?}{?}$$ Compute carefully: $$\frac12=\frac{6}{12},\quad \frac38=\frac{9}{24}=\frac{18}{48},$$ better use denominator $24$: $$\frac12=\frac{12}{24},\quad \frac38=\frac{9}{24},\quad \frac1{12}=\frac{2}{24}.$$ Therefore $$\frac12+\frac38-\frac1{12}=\frac{12+9-2}{24}=\frac{19}{24}.$$ And since $-\ln\frac12=\ln2$, $$F\left(\frac12\right)=\frac{19}{24}+\ln2.$$ Thus $$A=\left(\frac{17+7\sqrt5}{12}-\ln\frac{1+\sqrt5}{2}\right)-\left(\frac{19}{24}+\ln2\right).$$ Simplify the logarithms: $$-\ln\frac{1+\sqrt5}{2}-\ln2=-\ln(1+\sqrt5).$$ And the constant part: $$\frac{17+7\sqrt5}{12}-\frac{19}{24} =\frac{34+14\sqrt5-19}{24} =\frac{15+14\sqrt5}{24}.$$ So $$A=\frac{1}{24}(14\sqrt5+15)-\ln(1+\sqrt5).$$ Comparing with $$\frac{1}{24}(l\sqrt5+m)-n\log_e(1+\sqrt5),$$ we get $$l=14,\quad m=15,\quad n=1.$$ Therefore, $$l+m+n=14+15+1=30.$$ --- 4. **Check with options** Option A is $30$. So the correct answer is **A**.

More from Area Under the Curves

View written solutionFree

Correct answer: A

Find the points of intersection

The curves are $y=1+3x-2x^2$ and $y=\frac{1}{x}.$

At intersection, $1+3x-2x^2=\frac{1}{x}, \quad x\neq 0.$

Multiplying by $x$ : $x+3x^2-2x^3=1$ $2x^3-3x^2-x+1=0.$

Since one intersection is given as $\left(\frac12,2\right)$ , $x=\frac12$ is a root. So factorize: $2x^3-3x^2-x+1=(2x-1)(x^2-x-1).$

Thus the other roots are from $x^2-x-1=0,$ so $x=\frac{1\pm\sqrt5}{2}.$

Hence the three intersection $x$ -values are $\frac{1-\sqrt5}{2},\quad \frac12,\quad \frac{1+\sqrt5}{2}.$

Determine which curve is above the other

Take a test point, say $x=1$ : $1+3(1)-2(1)^2=2, \qquad \frac11=1.$ So on $\left[\frac12,\frac{1+\sqrt5}{2}\right]$ , the parabola is above the hyperbola.

Therefore area is $A=\int_{1/2}^{(1+\sqrt5)/2}\left(1+3x-2x^2-\frac1x\right)dx.$

Integrate

= x+\frac{3x^2}{2}-\frac{2x^3}{3}-\ln x.$$ So $$A=\left[x+\frac{3x^2}{2}-\frac{2x^3}{3}-\ln x\right]_{1/2}^{(1+\sqrt5)/2}.$$ Let $$\alpha=\frac{1+\sqrt5}{2}.$$ Then $$\alpha^2=\alpha+1, \qquad \alpha^3=\alpha\alpha^2=\alpha(\alpha+1)=2\alpha+1.$$ Hence $$\alpha+\frac{3\alpha^2}{2}-\frac{2\alpha^3}{3} =\alpha+\frac{3(\alpha+1)}{2}-\frac{2(2\alpha+1)}{3}.$$ Taking LCM $6$, $$=\frac{6\alpha+9\alpha+9-8\alpha-4}{6} =\frac{7\alpha+5}{6}.$$ Since $$\alpha=\frac{1+\sqrt5}{2},$$ we get $$\frac{7\alpha+5}{6}= rac{7(1+\sqrt5)+10}{12}= rac{17+7\sqrt5}{12}.$$ Also, $$\ln\alpha=\ln\left(\frac{1+\sqrt5}{2}\right).$$ So at $x=\alpha$, $$F(\alpha)=\frac{17+7\sqrt5}{12}-\ln\left(\frac{1+\sqrt5}{2}\right).$$ Now at $x=\frac12$, $$F\left(\frac12\right)=\frac12+\frac{3}{2}\cdot\frac14-\frac{2}{3}\cdot\frac18-\ln\frac12.$$ That is, $$F\left(\frac12\right)=\frac12+\frac38-\frac1{12}+\ln2 =\frac{6+ ?}{?}$$ Compute carefully: $$\frac12=\frac{6}{12},\quad \frac38=\frac{9}{24}=\frac{18}{48},$$ better use denominator $24$: $$\frac12=\frac{12}{24},\quad \frac38=\frac{9}{24},\quad \frac1{12}=\frac{2}{24}.$$ Therefore $$\frac12+\frac38-\frac1{12}=\frac{12+9-2}{24}=\frac{19}{24}.$$ And since $-\ln\frac12=\ln2$, $$F\left(\frac12\right)=\frac{19}{24}+\ln2.$$ Thus $$A=\left(\frac{17+7\sqrt5}{12}-\ln\frac{1+\sqrt5}{2}\right)-\left(\frac{19}{24}+\ln2\right).$$ Simplify the logarithms: $$-\ln\frac{1+\sqrt5}{2}-\ln2=-\ln(1+\sqrt5).$$ And the constant part: $$\frac{17+7\sqrt5}{12}-\frac{19}{24} =\frac{34+14\sqrt5-19}{24} =\frac{15+14\sqrt5}{24}.$$ So $$A=\frac{1}{24}(14\sqrt5+15)-\ln(1+\sqrt5).$$ Comparing with $$\frac{1}{24}(l\sqrt5+m)-n\log_e(1+\sqrt5),$$ we get $$l=14,\quad m=15,\quad n=1.$$ Therefore, $$l+m+n=14+15+1=30.$$ --- 4. **Check with options** Option A is $30$. So the correct answer is **A**.

Area Under the Curves question

Area Under the Curves question

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