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Application of Integration question

2025 · Shift 2 · Q18
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Application of Integration question

2025 · Shift 2 · Q18

JEE AdvancedMathematicsApplication of IntegrationMCQ+3 / −1
Let ℝ denote the set of all real numbers. Then the area of the region $ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x}, 5x - 4y - 1 > 0, 4x + 4y - 17 is
  1. A
    1716−log⁡e4\frac{17}{16} - \log_e{4}1617​−loge​4
  2. B
    338−log⁡e4\frac{33}{8} - \log_e{4}833​−loge​4
  3. C
    578−log⁡e4\frac{57}{8} - \log_e{4}857​−loge​4
  4. D
    172−log⁡e4\frac{17}{2} - \log_e{4}217​−loge​4
View written solutionFree

Correct answer: B

We need the area of the region

{(x,y)∈R2:x>0, y>1x, 5x−4y−1>0, 4x+4y−17<0}.\{(x,y)\in \mathbb R^2: x>0,\ y>\tfrac1x,\ 5x-4y-1>0,\ 4x+4y-17<0\}.{(x,y)∈R2:x>0, y>x1​, 5x−4y−1>0, 4x+4y−17<0}.

The last inequality is clearly intended as 4x+4y−17<0  ⟺  x+y<174.4x+4y-17<0 \iff x+y<\frac{17}{4}.4x+4y−17<0⟺x+y<417​. So the region is bounded by:

  1. x>0x>0x>0
  2. y>1xy>\dfrac1xy>x1​
  3. y<5x−14y<\dfrac{5x-1}{4}y<45x−1​
  4. y<174−xy<\dfrac{17}{4}-xy<417​−x

Thus for a given x>0x>0x>0, admissible yyy satisfy

1x<y<min⁡(5x−14,174−x).\frac1x<y<\min\left(\frac{5x-1}{4},\frac{17}{4}-x\right).x1​<y<min(45x−1​,417​−x).

So we first find where these boundary curves intersect.

1. Intersection of the two lines

Solve

5x−14=174−x.\frac{5x-1}{4}=\frac{17}{4}-x.45x−1​=417​−x.

Multiplying by 444:

5x−1=17−4x  ⟹  9x=18  ⟹  x=2.5x-1=17-4x \implies 9x=18 \implies x=2.5x−1=17−4x⟹9x=18⟹x=2.

Then

So the two lines meet at (2,94)(2,\tfrac94)(2,49​).

2. Intersection of hyperbola with line y=5x−14y=\dfrac{5x-1}{4}y=45x−1​

Solve

1x=5x−14.\frac1x=\frac{5x-1}{4}.x1​=45x−1​.

Multiplying by 4x4x4x:

4=5x2−x  ⟹  5x2−x−4=0.4=5x^2-x \implies 5x^2-x-4=0.4=5x2−x⟹5x2−x−4=0.

Factor:

5x2−x−4=(x−1)(5x+4)=0.5x^2-x-4=(x-1)(5x+4)=0.5x2−x−4=(x−1)(5x+4)=0.

Since x>0x>0x>0, we get

and then

So they meet at (1,1)(1,1)(1,1).

3. Intersection of hyperbola with line y=174−xy=\dfrac{17}{4}-xy=417​−x

Solve

1x=174−x.\frac1x=\frac{17}{4}-x.x1​=417​−x.

Multiplying by 4x4x4x:

4=17x−4x2  ⟹  4x2−17x+4=0.4=17x-4x^2 \implies 4x^2-17x+4=0.4=17x−4x2⟹4x2−17x+4=0.

Solve:

4x2−17x+4=(4x−1)(x−4)=0.4x^2-17x+4=(4x-1)(x-4)=0.4x2−17x+4=(4x−1)(x−4)=0.

Thus

Corresponding points are (14,4),(4,14).\left(\frac14,4\right),\quad (4,\tfrac14).(41​,4),(4,41​).

Relevant for the enclosed region will be $x=4$ on the right side. --- ### 4. Determine the $x$-intervals For the region to exist, we need

\frac1x<\min\left(\frac{5x-1}{4},\frac{17}{4}-x\right).

Since the two lines cross at $x=2$: - for $1\le x\le 2$, the smaller upper bound is $\dfrac{5x-1}{4}$, - for $2\le x\le 4$, the smaller upper bound is $\dfrac{17}{4}-x$. Also, from the hyperbola-line intersections, the valid interval is from $x=1$ to $x=4$. Hence area is

A=\int_1^2\left(\frac{5x-1}{4}-\frac1x\right)dx +\int_2^4\left(\frac{17}{4}-x-\frac1x\right)dx.

--- ### 5. Evaluate the first integral

I_1=\int_1^2\left(\frac{5x-1}{4}-\frac1x\right)dx =\int_1^2\left(\frac54x-\frac14-\frac1x\right)dx.

Antiderivative:Antiderivative:Antiderivative:

\int\left(\frac54x-\frac14-\frac1x\right)dx =\frac58x^2-\frac14x-\ln x.

SoSoSo

I_1=\left[\frac58x^2-\frac14x-\ln x\right]_1^2.

At $x=2$:

\frac58(4)-\frac14(2)-\ln 2=\frac52-\frac12-\ln2=2-\ln2.

At $x=1$:

\frac58-\frac14-0=\frac38.

ThereforeThereforeTherefore

I_1=2-\ln2-\frac38=\frac{13}{8}-\ln2.

--- ### 6. Evaluate the second integral

I_2=\int_2^4\left(\frac{17}{4}-x-\frac1x\right)dx.

Antiderivative:Antiderivative:Antiderivative:

\int\left(\frac{17}{4}-x-\frac1x\right)dx =\frac{17}{4}x-\frac12x^2-\ln x.

SoSoSo

I_2=\left[\frac{17}{4}x-\frac12x^2-\ln x\right]_2^4.

At $x=4$:

\frac{17}{4}(4)-\frac12(16)-\ln4=17-8-\ln4=9-\ln4.

At $x=2$:

\frac{17}{4}(2)-\frac12(4)-\ln2=\frac{17}{2}-2-\ln2=\frac{13}{2}-\ln2.

ThusThusThus

I_2=(9-\ln4)-\left(\frac{13}{2}-\ln2\right) =\frac52-\ln2.

--- ### 7. Total area

A=I_1+I_2=\left(\frac{13}{8}-\ln2\right)+\left(\frac52-\ln2\right).

NowNowNow

\frac52=\frac{20}{8},

sososo

A=\frac{33}{8}-2\ln2=\frac{33}{8}-\ln4.

Thustheareais Thus the area isThustheareais

\boxed{\frac{33}{8}-\log_e 4}.

Sothecorrectoptionis∗∗B∗∗. So the correct option is **B**.Sothecorrectoptionis∗∗B∗∗.
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