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

### Related Formula textArea = int_x_1^x_2 (y_textupper - y_textlower) dx$$\text{Area} = \int_{x_1}^{x_2} (y_{\text{upper}} - y_{\text{lower}}) dx$$ ### Core Logic Evaluate f(0)$f(0)$ by setting alpha = 0$\alpha = 0$ in the function for y$y$. At alpha = 0 implies y = |0cdot x - 5| - |1 - 0cdot x| + 0cdot x^2$\alpha = 0 \implies y = |0\cdot x - 5| - |1 - 0\cdot x| + 0\cdot x^2$ y = |-5| - |1| = 5 - 1 = 4$$y = |-5| - |1| = 5 - 1 = 4$$ So for f(0)$f(0)$, the bounded area A_1$A_1$ is between y = 4$y = 4$, y = sqrtx$y = \sqrt{x}$, from x = 0$x = 0$ to x = 1$x = 1$.

### Step 1: Calculating f(0) A_1 = int_0^1 (4 - sqrtx) dx$$A_1 = \int_{0}^{1} (4 - \sqrt{x}) dx$$ = left[ 4x - fracx^3/23/2 right]_0^1$$= \left[ 4x - \frac{x^{3/2}}{3/2} \right]_{0}^{1}$$ = 4(1) - frac23(1) = frac103$$= 4(1) - \frac{2}{3}(1) = \frac{10}{3}$$ So, f(0) = frac103$f(0) = \frac{10}{3}$. ### Step 2: Determining curve for f(1) Evaluate f(1)$f(1)$ by setting alpha = 1$\alpha = 1$ in the function for y$y$. At alpha = 1 implies y = |x - 5| - |1 - x| + x^2$\alpha = 1 \implies y = |x - 5| - |1 - x| + x^2$. 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$|1 - x| = 1 - x$. y = (5 - x) - (1 - x) + x^2$$y = (5 - x) - (1 - x) + x^2$$ y = 5 - x - 1 + x + x^2 = 4 + x^2$$y = 5 - x - 1 + x + x^2 = 4 + x^2$$

### Step 3: Calculating f(1) For f(1)$f(1)$, the bounded area A_2$A_2$ is between y = 4 + x^2$y = 4 + x^2$ and y = sqrtx$y = \sqrt{x}$, from x = 0$x = 0$ to x = 1$x = 1$. A_2 = int_0^1 left((4 + x^2) - sqrtxright) dx$$A_2 = \int_{0}^{1} \left((4 + x^2) - \sqrt{x}\right) dx$$ = left[ 4x + fracx^33 - fracx^3/23/2 right]_0^1$$= \left[ 4x + \frac{x^3}{3} - \frac{x^{3/2}}{3/2} \right]_{0}^{1}$$ = left( 4 + frac13 - frac23 right) = 4 - frac13 = frac113$$= \left( 4 + \frac{1}{3} - \frac{2}{3} \right) = 4 - \frac{1}{3} = \frac{11}{3}$$ So, f(1) = frac113$f(1) = \frac{11}{3}$. ### Step 4: Final Sum |f(0) + f(1)| = left| A_1 + A_2 right| = left| frac103 + frac113 right|$$|f(0) + f(1)| = \left| A_1 + A_2 \right| = \left| \frac{10}{3} + \frac{11}{3} \right|$$ = frac213 = 7$$= \frac{21}{3} = 7$$

### 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Area Under the Curve Class 11 Maths: Modulus Function