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

2019 · 8 Apr · Shift 2 · Q36
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. /2019 · 8 Apr · Shift 2 · Q36

Area Under the Curves question

2019 · 8 Apr · Shift 2 · Q36

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
Let S(α\alphaα) = {(x, y) : y2 ≤\le≤ x, 0 ≤\le≤ x ≤α\le \alpha≤α} and A(α\alphaα) is area of the region S(α\alphaα). If for a λ\lambdaλ, 0 < λ\lambdaλ < 4, A(λ\lambdaλ) : A(4) = 2 : 5, then λ\lambdaλ equals
  1. A
    2(425)132{\left( {{4 \over {25}}} \right)^{{1 \over 3}}}2(254​)31​
  2. B
    2(25)132{\left( {{2 \over {5}}} \right)^{{1 \over 3}}}2(52​)31​
  3. C
    4(425)134{\left( {{4 \over {25}}} \right)^{{1 \over 3}}}4(254​)31​
  4. D
    4(25)134{\left( {{2 \over {5}}} \right)^{{1 \over 3}}}4(52​)31​
View written solutionFree

Correct answer: C

  1. Interpret the region

Given S(α)={(x,y):y2≤x, 0≤x≤α}.S(\alpha)=\{(x,y): y^2\le x,\ 0\le x\le \alpha\}.S(α)={(x,y):y2≤x, 0≤x≤α}.

The inequality y2≤xy^2\le xy2≤x means −x≤y≤x.-\sqrt{x}\le y\le \sqrt{x}.−x​≤y≤x​. So for each fixed x∈[0,α]x\in[0,\alpha]x∈[0,α], the vertical length of the region is x−(−x)=2x.\sqrt{x}-(-\sqrt{x})=2\sqrt{x}.x​−(−x​)=2x​.

Hence the area is A(α)=∫0α2x dx.A(\alpha)=\int_0^{\alpha}2\sqrt{x}\,dx.A(α)=∫0α​2x​dx.

  1. Compute A(α)A(\alpha)A(α)
=2\left[\frac{2}{3}x^{3/2}\right]_0^{\alpha} =\frac{4}{3}\alpha^{3/2}.$$ So, $$A(\alpha)=\frac{4}{3}\alpha^{3/2}.$$ 3. **Use the given ratio** We are given $$A(\lambda):A(4)=2:5.$$ Now, $$A(\lambda)=\frac{4}{3}\lambda^{3/2},\qquad A(4)=\frac{4}{3}\cdot 4^{3/2}.$$ Since $$4^{3/2}=(\sqrt{4})^3=2^3=8,$$ we get $$A(4)=\frac{4}{3}\cdot 8=\frac{32}{3}.$$ Thus, $$\frac{A(\lambda)}{A(4)}=\frac{\frac{4}{3}\lambda^{3/2}}{\frac{32}{3}}=\frac{\lambda^{3/2}}{8}.$$ Given this equals $\frac{2}{5}$, $$\frac{\lambda^{3/2}}{8}=\frac{2}{5}.$$ So, $$\lambda^{3/2}=\frac{16}{5}.$$ 4. **Solve for $\lambda$** Raise both sides to the power $\frac{2}{3}$: $$\lambda=\left(\frac{16}{5}\right)^{2/3}.$$ Now rewrite: $$\left(\frac{16}{5}\right)^{2/3}=\left(\frac{2^4}{5}\right)^{2/3}=\frac{2^{8/3}}{5^{2/3}}=4\cdot \frac{2^{2/3}}{5^{2/3}}=4\left(\frac{4}{25}\right)^{1/3}.$$ Therefore, $$\boxed{\lambda=4\left(\frac{4}{25}\right)^{1/3}}.$$ 5. **Match with options** This is **Option C**. 6. **Comparison with stored answer** Stored correct answer: **C** My derived answer: **C** So they agree.
PreviousNext

More from Area Under the Curves

  • The area (in sq. units) of the region A = {(x, y) : x2 ≤ y ≤ x + 2} is2019 · MCQ
  • The area (in sq. units) of the region A = {(x, y) : 2y2​≤ x ≤ y + 4} is :-2019 · MCQ
  • The area (in sq. units) bounded by the parabolae y = x2 – 1, the tangent at the point (2, 3) to it and the y-axis is :2019 · MCQ
  • The area of the region A = {(x, y) : 0 ≤ y ≤ x |x| + 1 and − 1 ≤ x ≤ 1} in sq. units, is :2019 · MCQ
  • The area (in sq.units) of the region bounded by the curves y = 2x and y = |x + 1|, in the first quadrant is :2019 · MCQ
  • If the area enclosed between the curves y = kx2 and x = ky2, (k > 0), is 1 square unit. Then k is -2019 · MCQ
  • The area (in sq. units) of the region bounded by the curve x2 = 4y and the straight line x = 4y – 2 is :2019 · MCQ
  • The area (in sq. units) in the first quadrant bounded by the parabola, y = x2 + 1, the tangent to it at the point (2, 5) and the coordinate axes is :2019 · MCQ