Let A be the area bounded by the curves y2=4k1x,∀ k1∈ [(1/8) , (1/4)] , x2=4k2(.2-y.),∀ k2∈ [(1/4) , 1] and y- axis in the first quadrant. Then the least value of A is

Question

Let A be the area bounded by the curves y2=4k1x,∀ k1∈ [(1/8) , (1/4)] , x2=4k2(.2-y.),∀ k2∈ [(1/4) , 1] and y- axis in the first quadrant. Then the least value of A is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

The area will be least for maximum k1 and minimum k2 . <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-8ueszk7echoiwhom.jpg" /> Required area A= displaystyle ∫ 01((. 2 - x2 .) - √x)dx =2-(1/3)-(2/3)=2-1=1 sq. units

Q. Let $A$ be the area bounded by the curves $y^{2} = 4 k_{1} x, \forall k_{1} \in [\frac{1}{8}, \frac{1}{4}]$ , $x^{2} = 4 k_{2} (2 - y), \forall k_{2} \in [\frac{1}{4}, 1]$ and $y -$ axis in the first quadrant. Then the least value of $A$ is

The area will be least for maximum $k_{1}$ and minimum $k_{2}$ .

Required area $A = \int_{0}^{1} ((2 - x^{2}) - x) d x$
$= 2 - \frac{1}{3} - \frac{2}{3} = 2 - 1 = 1$ sq. units

Q. Let A be the area bounded by the curves y2=4k1​x,∀k1​∈[81​,41​] , x2=4k2​(2−y),∀k2​∈[41​,1] and y− axis in the first quadrant. Then the least value of A is

The area will be least for maximum k1​ and minimum k2​ . Required area A=∫01​((2−x2)−x​)dx =2−31​−32​=2−1=1 sq. units

Q. Let $A$ be the area bounded by the curves $y^{2} = 4 k_{1} x, \forall k_{1} \in [\frac{1}{8}, \frac{1}{4}]$ , $x^{2} = 4 k_{2} (2 - y), \forall k_{2} \in [\frac{1}{4}, 1]$ and $y -$ axis in the first quadrant. Then the least value of $A$ is

The area will be least for maximum $k_{1}$ and minimum $k_{2}$ .

Required area $A = \int_{0}^{1} ((2 - x^{2}) - x) d x$
$= 2 - \frac{1}{3} - \frac{2}{3} = 2 - 1 = 1$ sq. units