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Area Under the Curves question

2019 · 10 Apr · Shift 2 · Q38
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  5. /2019 · 10 Apr · Shift 2 · Q38

Area Under the Curves question

2019 · 10 Apr · Shift 2 · Q38

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
The area (in sq.units) of the region bounded by the curves y = 2x and y = |x + 1|, in the first quadrant is :
  1. A
    12{1 \over 2}21​
  2. B
    32{3 \over 2}23​
  3. C
    32−1log⁡e2{3 \over 2} - {1 \over {\log _e^2}}23​−loge2​1​
  4. D
    log⁡e2+32\log _e^2 + {3 \over 2}loge2​+23​
View written solutionFree

Correct answer: \(\FRAC{3}{2}-\FRAC{1}{\LN 2}\), STORED ANSWER C APPEARS INCONSISTENT UNLESS THE OPTION WAS MISPRINTED

  1. Interpret the curves carefully

    The equation is intended as y=2xy=2^xy=2x and y=∣x+1∣.y=|x+1|.y=∣x+1∣.

    We are asked for the area of the region bounded by these curves in the first quadrant.

  2. Simplify ∣x+1∣|x+1|∣x+1∣ in the first quadrant

    In the first quadrant, x≥0x\ge 0x≥0. Hence x+1>0x+1>0x+1>0, so ∣x+1∣=x+1.|x+1|=x+1.∣x+1∣=x+1.

    Therefore, in the first quadrant, the curves are y=2xandy=x+1.y=2^x \quad \text{and} \quad y=x+1.y=2xandy=x+1.

  3. Find their points of intersection

    Solve 2x=x+1.2^x=x+1.2x=x+1.

    We can immediately check:

    • At x=0x=0x=0: 20=12^0=120=1 and 0+1=10+1=10+1=1.
    • At x=1x=1x=1: 21=22^1=221=2 and 1+1=21+1=21+1=2.

    So the curves intersect at (0,1)and(1,2).(0,1) \quad \text{and} \quad (1,2).(0,1)and(1,2).

  4. Determine which curve is above the other on [0,1][0,1][0,1]

    Take x=12x=\tfrac12x=21​:

    \qquad x+1=1.5.$$ So on $0<x<1$, $$x+1 > 2^x.$$ Hence the required area is $$A=\int_0^1 \big[(x+1)-2^x\big]dx.$$

  5. Evaluate the integral

    A=∫01(x+1)dx−∫012xdx.A=\int_0^1 (x+1)dx - \int_0^1 2^x dx.A=∫01​(x+1)dx−∫01​2xdx.

    First part: ∫01(x+1)dx=[x22+x]01=12+1=32.\int_0^1 (x+1)dx=\left[\frac{x^2}{2}+x\right]_0^1=\frac12+1=\frac32.∫01​(x+1)dx=[2x2​+x]01​=21​+1=23​.

    Second part: ∫2xdx=2xln⁡2.\int 2^x dx=\frac{2^x}{\ln 2}.∫2xdx=ln22x​. Therefore,

    \frac{2-1}{\ln 2}= rac{1}{\ln 2}.$$ Thus, $$A=\frac32-\frac{1}{\ln 2}.$$

  6. Match with the options

    The area is 32−1ln⁡2.\boxed{\frac32-\frac{1}{\ln 2}}.23​−ln21​​.

    This corresponds to none of the given options exactly.

    Option C is written as 32−1log⁡e22.\frac32-\frac{1}{\log_e 2^2}.23​−loge​221​. If interpreted as 32−1(ln⁡2)2\frac32-\frac{1}{(\ln 2)^2}23​−(ln2)21​, it is incorrect.

    Most likely, there is a formatting/printing issue in the options, and the intended correct option should have been 32−1ln⁡2.\frac32-\frac{1}{\ln 2}.23​−ln21​.

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