If the area of the region { ( x , y ) : x 2 a ≤ y ≤ x 1 , 1 ≤ x ≤ 2 , 0 < a < 1 } is ( lo g e 2 ) − 7 1 then the value of 7 a − 3 is equal to:

area ∫ 1 2 ( x 1 − x 2 a ) d x [ ℓ n x + x a ] 1 2 ℓ nn 2 + 2 a − a = lo g e 2 − 7 1 2 − a = − 7 1 a = 7 2 7 a = 2 7 a − 3 = − 1

Mathematics - Area Under Curves Question with Solution | TestHub

MathematicsArea Under CurvesArea bounded by two curvesHard2 minPYQ_2024

MathematicsHardsingle choice

If the area of the region ${(x, y) : \frac{a}{x ^{2}} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2, 0 < a < 1}$ is $(lo g_{e} 2) - \frac{1}{7}$ then the value of $7 a - 3$ is equal to:

Options:

Answer:

Solution:

$area \int_{1}^{2} (\frac{1}{x} - \frac{a}{x ^{2}}) d x [ℓ n x + \frac{a}{x}]_{1}^{2} ℓ nn 2 + \frac{a}{2} - a = lo g_{e} 2 - \frac{1}{7} \frac{- a}{2} = - \frac{1}{7} a = \frac{2}{7} 7 a = 2 7 a - 3 = - 1$

Stream:JEESubject:MathematicsTopic:Area Under CurvesSubtopic:Area bounded by two curves

⏱ 2mℹ️ Source: PYQ_2024

Doubts & Discussion

Loading discussions...