Let f(α)$f(\alpha)$ denote the area of the region in the first quadrant bounded by x = 0$x = 0$, x = 1$x = 1$, y² = x$y^2 = x$ and y = |α x - 5| - |1 - α x| + α x²$y = |\alpha x - 5| - |1 - \alpha x| + \alpha x^2$. Then (f(0) + f(1))$(f(0) + f(1))$ is equal to

Related Formula

Core Logic

Evaluate f(0)$f(0)$ by setting α = 0$\alpha = 0$ in the function for y$y$.

At α = 0 y = |0· x - 5| - |1 - 0· x| + 0· x²$\alpha = 0 \implies y = |0\cdot x - 5| - |1 - 0\cdot x| + 0\cdot x^2$

So for f(0)$f(0)$, the bounded area A₁$A_1$ is between y = 4$y = 4$, y = √(x)$y = \sqrt{x}$, from x = 0$x = 0$ to x = 1$x = 1$.

Step 1: Calculating f(0)

So, f(0) = (10)/(3)$f(0) = \frac{10}{3}$.

Step 2: Determining curve for f(1)

Evaluate f(1)$f(1)$ by setting α = 1$\alpha = 1$ in the function for y$y$.

At α = 1 y = |x - 5| - |1 - x| + x²$\alpha = 1 \implies y = |x - 5| - |1 - x| + x^2$. Since the area is restricted to x in [0, 1]$x \in [0, 1]$: x - 5$x - 5$ is negative, so |x - 5| = 5 - x$|x - 5| = 5 - x$. 1 - x$1 - x$ is positive, so |1 - x| = 1 - x$|1 - x| = 1 - x$.

Step 3: Calculating f(1)

For f(1)$f(1)$, the bounded area A₂$A_2$ is between y = 4 + x²$y = 4 + x^2$ and y = √(x)$y = \sqrt{x}$, from x = 0$x = 0$ to x = 1$x = 1$.

So, f(1) = (11)/(3)$f(1) = \frac{11}{3}$.

Step 4: Final Sum

Pattern Recognition

Modulus functions combined with parametric limits should be instantly resolved by plugging in the bounds of x$x$ (here [0,1]$[0,1]$) to drop the absolute value signs before attempting any integral.

Chapter Mix

Class 12 Maths: Area Under the Curve Class 11 Maths: Modulus Function