Let f denote the area of the region in the first quadrant bounded by x = 0 , x = 1 , y 2 = x and y = | x - 5 | - | 1 - x | + x 2 . Then ( f ( 0 ) + f ( 1 ) ) is equal to [2026]

Question

Let $f (α)$ denote the area of the region in the first quadrant bounded by $x = 0$ , $x = 1$ , $y^{2} = x$ and $y =$ $| α x - 5 |$ $-$ $| 1 - α x |$ $+ α x^{2} .$

Then $(f (0) + f (1))$ is equal to [2026]

Smart Achievers · Accepted Answer

(3)

$at α = 0 \Rightarrow f (0)$

$x = 0, x = 1, y^{2} = x$

$y =$ $| 0 \cdot x - 5 |$ $-$ $| 1 - 0 \cdot x |$ $+ 0 \cdot x^{2}$

$y = 4$

$A_{1} = \int_{0}^{1} (4 - \sqrt{x}) d x$

$\begin{matrix} = \end{matrix}$ $4 x - \frac{x^{3 / 2}}{3 / 2} |_{0}^{1}$

$= 4 - \frac{2}{3} (1) = \frac{10}{3}$

$at α = 1 \Rightarrow f (1)$

$x = 0, x = 1, y^{2} = x$

$y =$ $| x - 5 | - | 1 - x | + x^{2}, x \in (0, 1)$

$y = 5 - x - (1 - x) + x^{2}$

$y = 4 + x^{2}$

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$A_{2} = \int_{0}^{1} [(4 + x^{2}) - \sqrt{x}] d x$

$=$ $4 x + \frac{x^{3}}{3} - \frac{x^{\frac{3}{2}}}{\frac{3}{2}} |_{0}^{1}$

$= 4 + \frac{1}{3} - \frac{2}{3} = \frac{11}{3}$

$| f (0) + f (1) |$ $=$ $| A_{1} + A_{2} |$ $=$ $| \frac{10}{3} + \frac{11}{3} |$ $=$ $| \frac{21}{3} |$ $= 7$

$option (3)$

Solution

Q.
Let $f (α)$ denote the area of the region in the first quadrant bounded by $x = 0$ , $x = 1$ , $y^{2} = x$ and $y =$ $| α x - 5 |$ $-$ $| 1 - α x |$ $+ α x^{2} .$

Then $(f (0) + f (1))$ is equal to [2026]

1 9

2 12

3 7

4 14

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Solution

Q. Let f(α) denote the area of the region in the first quadrant bounded by x=0, x=1, y2=x and y=|αx-5|-|1-αx|+αx2. Then (f(0)+f(1)) is equal to [2026]

1 9

2 12

3 7

4 14

Want Us to Email you About Special Offers & Updates?

Q.
Let $f (α)$ denote the area of the region in the first quadrant bounded by $x = 0$ , $x = 1$ , $y^{2} = x$ and $y =$ $| α x - 5 |$ $-$ $| 1 - α x |$ $+ α x^{2} .$

Then $(f (0) + f (1))$ is equal to [2026]