Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Area Under the Curves question

2003 · Shift 0 · Q82
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Area Under the Curves
  5. /2003 · Shift 0 · Q82

Area Under the Curves question

2003 · Shift 0 · Q82

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
The area of the region bounded by the curves y=∣x−1∣y = \left| {x - 1} \right|y=∣x−1∣ and y=3−∣x∣y = 3 - \left| x \right|y=3−∣x∣ is :
  1. A
    666 sq. units
  2. B
    222 sq. units
  3. C
    333 sq. units
  4. D
    444 sq. units
View written solutionFree

Correct answer: D

  1. Given curves

We need the area bounded by y=∣x−1∣y=|x-1|y=∣x−1∣ and y=3−∣x∣.y=3-|x|.y=3−∣x∣.

  1. Write both curves piecewise

For y=∣x−1∣y=|x-1|y=∣x−1∣:

\begin{cases} 1-x, & x<1,\\ x-1, & x\ge 1. \end{cases}$$ For $$y=3-|x|$$: $$3-|x|= \begin{cases} 3+x, & x<0,\\ 3-x, & x\ge 0. \end{cases}$$ 3. **Find points of intersection** We solve $$|x-1|=3-|x|.$$ Consider intervals. - **Case 1:** $$x<0$$ $$|x-1|=1-x, \quad 3-|x|=3+x.$$ So, $$1-x=3+x$$ $$-2=2x$$ $$x=-1.$$ Then $$y=|{-1}-1|=2.$$ Intersection point: $$(-1,2).$$ - **Case 2:** $$0\le x<1$$ $$|x-1|=1-x, \quad 3-|x|=3-x.$$ So, $$1-x=3-x,$$ which gives $$1=3$$, impossible. No intersection here. - **Case 3:** $$x\ge 1$$ $$|x-1|=x-1, \quad 3-|x|=3-x.$$ So, $$x-1=3-x$$ $$2x=4$$ $$x=2.$$ Then $$y=|2-1|=1.$$ Intersection point: $$(2,1).$$ So the bounded region lies between $$x=-1$$ and $$x=2$$. 4. **Determine which curve is above** Take a test point, say $$x=0$$: $$|0-1|=1, \qquad 3-|0|=3.$$ So $$y=3-|x|$$ is above $$y=|x-1|$$ in the region. Hence area is $$A=\int_{-1}^{2}\Big[(3-|x|)-|x-1|\Big]dx.$$ 5. **Split according to critical points** Critical points are $$x=0$$ and $$x=1$$. So, $$A=\int_{-1}^{0}\big[(3+x)-(1-x)\big]dx+\int_{0}^{1}\big[(3-x)-(1-x)\big]dx+\int_{1}^{2}\big[(3-x)-(x-1)\big]dx.$$ Simplify each integrand: - On $$[-1,0]$$: $$ (3+x)-(1-x)=2+2x. $$ - On $$[0,1]$$: $$ (3-x)-(1-x)=2. $$ - On $$[1,2]$$: $$ (3-x)-(x-1)=4-2x. $$ Thus, $$A=\int_{-1}^{0}(2+2x)dx+\int_{0}^{1}2\,dx+\int_{1}^{2}(4-2x)dx.$$ 6. **Evaluate the integrals** First, $$\int_{-1}^{0}(2+2x)dx=\left[2x+x^2\right]_{-1}^{0}=0-(-2+1)=1.$$ Second, $$\int_{0}^{1}2\,dx=2.$$ Third, $$\int_{1}^{2}(4-2x)dx=\left[4x-x^2\right]_{1}^{2}=(8-4)-(4-1)=4-3=1.$$ Therefore, $$A=1+2+1=4.$$ 7. **Final answer** The area bounded by the curves is $$\boxed{4\text{ sq. units}}.$$ So the correct option is **D**.
PreviousNext

More from Area Under the Curves

  • The area bounded by the curves y=lnx,y=ln∣x∣,y=∣lnx∣ and y=∣ln∣x∣∣ is :2002 · MCQ
  • If the area of the region {(x,y):​4−x2​≤y≤x2,y≤4,x≥0} is (α802​​−β),α,β∈N, then α+β is equal to ​…2025 · Numerical
  • The area of the region bounded by the curve y=max{∣x∣,x∣x−2∣}, the x-axis and the lines x=−2 and x=4 is equal to ​2025 · Numerical
  • The area of the region {(x,y):∣x−y∣≤y≤4x​} is2025 · MCQ
  • Let f:[0,∞)→R be a differentiable function such that f(x)=1−2x+∫0x​ex−tf(t)dt for all x∈[0,∞). Then the area of the region bounded by y=f(x) and the coordinate axes is2025 · MCQ
  • If the area of the region {(x,y):∣x−5∣≤y≤4x​} is A, then 3A is equal to ​.2025 · Numerical
  • If the area of the region bounded by the curves y=4−4x2​ and y=2x−4​ is equal to α, then 6α. equals2025 · MCQ
  • If the area of the region {(x,y):1+x2≤y≤min{x+7,11−3x}} is A, then 3A is equal to :2025 · MCQ