Math Worksheet A4919F

1. **Problem:** Define odd and even functions with examples. **Explanation:** - An **even function** satisfies $f(-x) = f(x)$ for all $x$ in its domain. - An **odd function** satisfies $f(-x) = -f(x)$ for all $x$ in its domain. **Examples:** - Even: $f(x) = x^2$, since $(-x)^2 = x^2$ - Odd: $f(x) = x^3$, since $(-x)^3 = -x^3$ 2. **Problem:** Given $f(x) = x^2 + 6x + 10$, find $f(2) + f(-2)$. **Step 1:** Calculate $f(2) = 2^2 + 6(2) + 10 = 4 + 12 + 10 = 26$ **Step 2:** Calculate $f(-2) = (-2)^2 + 6(-2) + 10 = 4 - 12 + 10 = 2$ **Step 3:** Sum: $f(2) + f(-2) = 26 + 2 = 28$ 3. **Problem:** Differentiate w.r.t $x$: $a^x + e^x + \log_{10} x + \log_a a$ **Step 1:** Recall derivatives: - $\frac{d}{dx} a^x = a^x \ln a$ - $\frac{d}{dx} e^x = e^x$ - $\frac{d}{dx} \log_{10} x = \frac{1}{x \ln 10}$ - $\log_a a = 1$ (constant), derivative $= 0$ **Step 2:** Combine: $$\frac{d}{dx} = a^x \ln a + e^x + \frac{1}{x \ln 10} + 0$$ 4. **Problem:** Find range and coefficient of range for data: $40,52,47,28,45,36,47,50$ **Step 1:** Find max and min: - Max = $52$ - Min = $28$ **Step 2:** Range = Max - Min = $52 - 28 = 24$ **Step 3:** Coefficient of range = $\frac{\text{Range}}{\text{Max} + \text{Min}} = \frac{24}{52 + 28} = \frac{24}{80} = 0.3$ 5. **Problem:** Given two sets: - Set I: $\bar{x} = 82.5$, $\sigma = 7.3$ - Set II: $\bar{x} = 48.75$, $\sigma = 8.35$ Which set is more consistent? **Step 1:** Calculate coefficient of variation (CV): $$CV = \frac{\sigma}{\bar{x}}$$ **Step 2:** For Set I: $$CV_1 = \frac{7.3}{82.5} \approx 0.0885$$ **Step 3:** For Set II: $$CV_2 = \frac{8.35}{48.75} \approx 0.1713$$ **Step 4:** Lower CV means more consistency, so Set I is more consistent. 6. **Problem:** Mean = 82, standard deviation = 7, find coefficient of variance. $$CV = \frac{7}{82} \approx 0.0854$$ 7. **Problem:** Marks of ten students: $40,72,83,57,94,49,65,79,87,64$. Find mean. **Step 1:** Sum marks: $$40 + 72 + 83 + 57 + 94 + 49 + 65 + 79 + 87 + 64 = 690$$ **Step 2:** Mean = $\frac{690}{10} = 69$ 8. **Problem:** Find $\frac{dy}{dx}$ if $13x^3 + 2x^2 y + y^3 = 1$ **Step 1:** Differentiate implicitly w.r.t $x$: $$39x^2 + 2(2x y + x^2 \frac{dy}{dx}) + 3y^2 \frac{dy}{dx} = 0$$ **Step 2:** Expand: $$39x^2 + 4x y + 2x^2 \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = 0$$ **Step 3:** Group $\frac{dy}{dx}$ terms: $$2x^2 \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = -39x^2 - 4x y$$ **Step 4:** Factor: $$\frac{dy}{dx} (2x^2 + 3y^2) = -39x^2 - 4x y$$ **Step 5:** Solve: $$\frac{dy}{dx} = \frac{-39x^2 - 4x y}{2x^2 + 3y^2}$$ 9. **Problem:** A metal wire 36 cm long is bent to form a rectangle. Find dimensions for maximum area. **Step 1:** Let length = $l$, breadth = $b$ **Step 2:** Perimeter constraint: $$2(l + b) = 36 \Rightarrow l + b = 18$$ **Step 3:** Area: $$A = l b = l(18 - l) = 18l - l^2$$ **Step 4:** Maximize $A$ by differentiating w.r.t $l$: $$\frac{dA}{dl} = 18 - 2l = 0 \Rightarrow l = 9$$ **Step 5:** Then $b = 18 - 9 = 9$ **Step 6:** Dimensions are $9$ cm by $9$ cm (square). 10. **Problem:** Find equation of tangent to curve $4x^2 + 9y^2 = 40$ at $(3,2)$. **Step 1:** Differentiate implicitly: $$8x + 18y \frac{dy}{dx} = 0$$ **Step 2:** Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx} = -\frac{8x}{18y} = -\frac{4x}{9y}$$ **Step 3:** At $(3,2)$: $$m = -\frac{4(3)}{9(2)} = -\frac{12}{18} = -\frac{2}{3}$$ **Step 4:** Equation of tangent: $$y - 2 = -\frac{2}{3}(x - 3)$$ 11. **Problem:** Find radius of curve $ay = c$ at point $(c,c)$. **Step 1:** Rewrite as $y = \frac{c}{a}$ (constant), so curve is a horizontal line. **Step 2:** Radius of curvature for a line is infinite. 12. **Problem:** Find $\frac{dy}{dx}$ if $x = a(\theta + \sin \theta)$ and $y = a(1 - \cos \theta)$ at $\theta = \frac{\pi}{2}$. **Step 1:** Compute derivatives w.r.t $\theta$: $$\frac{dx}{d\theta} = a(1 + \cos \theta)$$ $$\frac{dy}{d\theta} = a \sin \theta$$ **Step 2:** Compute $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a \sin \theta}{a(1 + \cos \theta)} = \frac{\sin \theta}{1 + \cos \theta}$$ **Step 3:** At $\theta = \frac{\pi}{2}$: $$\sin \frac{\pi}{2} = 1, \quad \cos \frac{\pi}{2} = 0$$ $$\frac{dy}{dx} = \frac{1}{1 + 0} = 1$$ 13. **Problem:** Find mean deviation about mean for marks intervals and frequencies: | Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | |-------|-------|--------|--------|--------|--------| | No. of students | 5 | 8 | 15 | 16 | 6 | **Step 1:** Find midpoints: $$5, 15, 25, 35, 45$$ **Step 2:** Calculate mean: $$\bar{x} = \frac{5\times5 + 8\times15 + 15\times25 + 16\times35 + 6\times45}{5+8+15+16+6} = \frac{25 + 120 + 375 + 560 + 270}{50} = \frac{1350}{50} = 27$$ **Step 3:** Calculate absolute deviations: $$|5-27|=22, |15-27|=12, |25-27|=2, |35-27|=8, |45-27|=18$$ **Step 4:** Mean deviation: $$= \frac{5\times22 + 8\times12 + 15\times2 + 16\times8 + 6\times18}{50} = \frac{110 + 96 + 30 + 128 + 108}{50} = \frac{472}{50} = 9.44$$ 14. **Problem:** Find standard deviation for data: | Class Interval | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 | |---------------|-----|------|-------|-------|-------|-------|-------|-------| | Frequency | 3 | 5 | 9 | 15 | 20 | 16 | 10 | 2 | **Step 1:** Find midpoints: $$2.5, 7.5, 12.5, 17.5, 22.5, 27.5, 32.5, 37.5$$ **Step 2:** Calculate mean: $$\bar{x} = \frac{3\times2.5 + 5\times7.5 + 9\times12.5 + 15\times17.5 + 20\times22.5 + 16\times27.5 + 10\times32.5 + 2\times37.5}{3+5+9+15+20+16+10+2}$$ $$= \frac{7.5 + 37.5 + 112.5 + 262.5 + 450 + 440 + 325 + 75}{80} = \frac{1710}{80} = 21.375$$ **Step 3:** Calculate $\sum f x^2$: $$3\times(2.5)^2 + 5\times(7.5)^2 + 9\times(12.5)^2 + 15\times(17.5)^2 + 20\times(22.5)^2 + 16\times(27.5)^2 + 10\times(32.5)^2 + 2\times(37.5)^2$$ $$= 3\times6.25 + 5\times56.25 + 9\times156.25 + 15\times306.25 + 20\times506.25 + 16\times756.25 + 10\times1056.25 + 2\times1406.25$$ $$= 18.75 + 281.25 + 1406.25 + 4593.75 + 10125 + 12100 + 10562.5 + 2812.5 = 41900$$ **Step 4:** Use variance formula: $$\sigma^2 = \frac{\sum f x^2}{\sum f} - \bar{x}^2 = \frac{41900}{80} - (21.375)^2 = 523.75 - 456.39 = 67.36$$ **Step 5:** Standard deviation: $$\sigma = \sqrt{67.36} \approx 8.21$$