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MCQMediumJEE 2024Applications of Integrals (Area)

JEE Mathematics 2024 Question with Solution

The area enclosed by the curves xy+4y=16xy + 4y = 16 and x+y=6x + y = 6 is equal to:

  • A

    28−30log⁡228 - 30 \log 2

  • B

    30−28log⁡230 - 28 \log 2

  • C

    30−32log⁡230 - 32 \log 2

  • D

    32−30log⁡232 - 30 \log 2

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The curves are xy+4y=16xy + 4y = 16 and x+y=6x + y = 6.

Find: The area enclosed between these two curves.

Graph showing the line and hyperbola with shaded enclosed region between intersection points at x = -2 and x = 4, marked axes, asymptote x = -4, and labels (0,6) and (6,0).

Rewrite the curves in terms of yy:

y(x+4)=16y(x+4)=16

so,

y=16x+4y=\frac{16}{x+4}

and from x+y=6x+y=6,

y=6−xy=6-x

Find the points of intersection by solving

16x+4=6−x\frac{16}{x+4}=6-x

Therefore,

16=(6−x)(x+4)16=(6-x)(x+4) 16=6x+24−x2−4x16=6x+24-x^2-4x 16=−x2+2x+2416=-x^2+2x+24 x2−2x−8=0x^2-2x-8=0 (x−4)(x+2)=0(x-4)(x+2)=0

Hence,

x=4,x=−2x=4,\quad x=-2

So the limits are from x=−2x=-2 to x=4x=4.

To determine the upper curve, test a point such as x=0x=0. Then for the hyperbola,

y=160+4=4y=\frac{16}{0+4}=4

and for the line,

y=6−0=6y=6-0=6

Thus, y=6−xy=6-x lies above y=16x+4y=\frac{16}{x+4} on [−2,4][-2,4].

Now the enclosed area is

Area=∫−24[(6−x)−16x+4]dx\text{Area}=\int_{-2}^{4}\left[(6-x)-\frac{16}{x+4}\right]dx

Evaluate the two parts separately:

∫−24(6−x) dx=[6x−x22]−24\int_{-2}^{4}(6-x)\,dx=\left[6x-\frac{x^2}{2}\right]_{-2}^{4} =(24−8)−(−12−2)=30=\left(24-8\right)-\left(-12-2\right)=30

Also,

∫−2416x+4 dx=16[ln⁡∣x+4∣]−24\int_{-2}^{4}\frac{16}{x+4}\,dx=16\left[\ln|x+4|\right]_{-2}^{4} =16(ln⁡8−ln⁡2)=16ln⁡4=32ln⁡2=16(\ln 8-\ln 2)=16\ln 4=32\ln 2

Therefore,

Area=30−32ln⁡2\text{Area}=30-32\ln 2

So the correct option is C.

Why top minus bottom matters

A common sign error is to write the integrand as

16x+4−(6−x)\frac{16}{x+4}-(6-x)

but on the interval [−2,4][-2,4] the line is above the hyperbola. Hence the correct area expression must be

∫−24[(6−x)−16x+4]dx\int_{-2}^{4}\left[(6-x)-\frac{16}{x+4}\right]dx

This ensures the area is positive and matches the required enclosed region.

Common mistakes

  • Using the curves in the wrong order inside the integral. This gives a negative value because y=6−xy=6-x is above y=16x+4y=\frac{16}{x+4} on [−2,4][-2,4]. Always check which curve is upper before forming top minus bottom.

  • Finding the intersection points incorrectly after substitution. From 16x+4=6−x\frac{16}{x+4}=6-x, the quadratic is x2−2x−8=0x^2-2x-8=0, not with altered signs. Expand carefully before factoring.

  • Evaluating the logarithmic integral incorrectly. For ∫16x+4 dx\int \frac{16}{x+4}\,dx, the antiderivative is 16ln⁡∣x+4∣16\ln|x+4|. Do not omit the modulus in the formula, even though x+4>0x+4>0 on the given interval.

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