Let S- x-y- y2 - x- 0 - x - and A- is area of the region S- If for a - 0 - 4- A- A4-2-5- then - equals -

A(α)=2 ∫_0^α√(x) dx=43α^3/2 A(λ)A(4)=25 ⇒λ^3/24^3/2=25 ⇒ nbsp; λ^3/2= 165 ⇒λ= (165)^2/3 ⇒λ= 4(425)^1/3 ...

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Byju's Answer

Standard XIII

Mathematics

Area between Two Curves

Let Sα= x,y: ...

Question

Let $S (α) = {(x, y) : y^{2} \leq x, 0 \leq x \leq α}$ and $A (α)$ is area of the region $S (α)$ . If for a $λ, 0 < λ < 4, A (λ) : A (4) = 2 : 5$ , then $λ$ equals :

4 {(\frac{4}{25})}^{1 / 3}

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4 {(\frac{2}{5})}^{1 / 3}

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2 {(\frac{2}{5})}^{1 / 3}

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2 {(\frac{4}{25})}^{1 / 3}

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Solution

The correct option is A $4 {(\frac{4}{25})}^{1 / 3}$
$A (α) = 2 α \int 0 \sqrt{x} d x = \frac{4}{3} α^{3 / 2}$
$\frac{A (λ)}{A (4)} = \frac{2}{5}$
$\Rightarrow \frac{λ^{3 / 2}}{4^{3 / 2}} = \frac{2}{5}$
$\Rightarrow λ^{3 / 2} = \frac{16}{5}$
$\Rightarrow λ = {(\frac{16}{5})}^{2 / 3}$
$\Rightarrow λ = 4 {(\frac{4}{25})}^{1 / 3}$

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Let S(α)={(x,y):y2≤x,0≤x≤α} and A(α) is area of the region S(α). If for a λ,0<λ<4,A(λ):A(4)=2:5, then λ equals :

The correct option is A 4(425)1/3A(α)=2α∫0√x dx=43α3/2 A(λ)A(4)=25 ⇒λ3/243/2=25 ⇒λ3/2=165 ⇒λ=(165)2/3 ⇒λ=4(425)1/3

Let $S (α) = {(x, y) : y^{2} \leq x, 0 \leq x \leq α}$ and $A (α)$ is area of the region $S (α)$ . If for a $λ, 0 < λ < 4, A (λ) : A (4) = 2 : 5$ , then $λ$ equals :