Question 1: Find the points of intersection of the coordinate axes and the following linear functions graphically: (i) \(y = -5x + 10\) (ii) \(y = 2x - 1\) (iii) \(y = \frac{1}{2}x - 3\) (iv) \(y = 3x + \frac{3}{2}\).
Axes se intersection ka matlab x-intercept aur y-intercept. Y-intercept ke liye \(x = 0\) rakhte hain, x-intercept ke liye \(y = 0\) rakh kar \(x\) solve karte hain. Phir in do points ko plot karke line draw karte hain.
(i) \(y = -5x + 10\)
y-intercept: \(x = 0 \implies y = 10\), point \((0, 10)\). x-intercept: \(y = 0 \implies 0 = -5x + 10 \implies x = 2\), point \((2, 0)\).
(ii) \(y = 2x - 1\)
y-intercept: \(x = 0 \implies y = -1\), point \((0, -1)\). x-intercept: \(0 = 2x - 1 \implies x = \frac{1}{2}\), point \(\left(\frac{1}{2}, 0\right)\).
(iii) \(y = \frac{1}{2}x - 3\)
y-intercept: \(x = 0 \implies y = -3\), point \((0, -3)\). x-intercept: \(0 = \frac{1}{2}x - 3 \implies x = 6\), point \((6, 0)\).
(iv) \(y = 3x + \frac{3}{2}\)
y-intercept: \(x = 0 \implies y = \frac{3}{2}\), point \(\left(0, \frac{3}{2}\right)\). x-intercept: \(0 = 3x + \frac{3}{2} \implies x = -\frac{1}{2}\), point \(\left(-\frac{1}{2}, 0\right)\).
Answer: (i) \((0,10), (2,0)\) (ii) \((0,-1), \left(\frac{1}{2},0\right)\) (iii) \((0,-3), (6,0)\) (iv) \(\left(0,\frac{3}{2}\right), \left(-\frac{1}{2},0\right)\)
Question 2: Find the point(s) of intersection of the following functions graphically: (i) \(f(x) = 2x+5,\; g(x) = -x+5\) (ii) \(f(x) = 3x-2,\; g(x) = 10-x\) (iii) \(f(x) = 2x-4,\; g(x) = 3x-1\) (iv) \(f(x) = -3x-4,\; g(x) = \frac{1}{2}x+3\) (v) \(f(x) = x-1,\; g(x) = x^2-4x+3\) (vi) \(f(x) = 3x+4,\; g(x) = x^2+2x-8\) (vii) \(f(x) = -2x-1,\; g(x) = x^2-4x\) (viii) \(f(x) = -x^2-3x+2,\; g(x) = x+6\).
Intersection par dono functions ki \(y\)-values barabar hoti hain, is liye \(f(x) = g(x)\) rakh kar solve karte hain. Linear-linear se ek point, linear-quadratic se quadratic equation — jiske do, ek (tangent), ya zero real roots ho sakte hain.
(i) \(2x + 5 = -x + 5 \implies 3x = 0 \implies x = 0\). Then \(y = 2(0) + 5 = 5\). Point: \((0, 5)\).
(ii) \(3x - 2 = 10 - x \implies 4x = 12 \implies x = 3\). Then \(y = 3(3) - 2 = 7\). Point: \((3, 7)\).
(iii) \(2x - 4 = 3x - 1 \implies x = -3\). Then \(y = 2(-3) - 4 = -10\). Point: \((-3, -10)\).
(iv) \(-3x - 4 = \frac{1}{2}x + 3 \implies -\frac{7}{2}x = 7 \implies x = -2\). Then \(y = -3(-2) - 4 = 2\). Point: \((-2, 2)\).
(v) \(x - 1 = x^2 - 4x + 3 \implies x^2 - 5x + 4 = 0 \implies (x-1)(x-4) = 0 \implies x = 1, 4\). Then \(y = 0, 3\). Points: \((1, 0)\) and \((4, 3)\).
(vi) \(3x + 4 = x^2 + 2x - 8 \implies x^2 - x - 12 = 0 \implies (x-4)(x+3) = 0 \implies x = 4, -3\). Then \(y = 16, -5\). Points: \((4, 16)\) and \((-3, -5)\).
(vii) \(-2x - 1 = x^2 - 4x \implies x^2 - 2x + 1 = 0 \implies (x-1)^2 = 0 \implies x = 1\) (repeated root). Then \(y = -2(1) - 1 = -3\). The line is tangent to the parabola at \((1, -3)\).
(viii) \(-x^2 - 3x + 2 = x + 6 \implies x^2 + 4x + 4 = 0 \implies (x+2)^2 = 0 \implies x = -2\) (repeated root). Then \(y = -2 + 6 = 4\). Tangent point: \((-2, 4)\).
Answer: (i) \((0,5)\) (ii) \((3,7)\) (iii) \((-3,-10)\) (iv) \((-2,2)\) (v) \((1,0), (4,3)\) (vi) \((4,16), (-3,-5)\) (vii) \((1,-3)\) tangent (viii) \((-2,4)\) tangent
Common mistake: Finding \(x\) but forgetting to compute \(y\) — an intersection point needs both coordinates. Always substitute back into either function.
Question 3: Graph the following functions: (i) \(y = \sqrt{3x}\) (ii) \(y = \sqrt{x}+5\) (iii) \(y = -\frac{1}{2}\sqrt{x}\) (iv) \(y = -\sqrt{x+1}+2\) (v) \(y = \sqrt[3]{2x+1}\) (vi) \(y = 2\sqrt{x}-3\) (vii) \(y = \sqrt[3]{x^2+x-2}\).
Graphing ka tareeqa: pehle domain aur range likhen, phir easy values ki table banayen (square root ke liye perfect squares, cube root ke liye perfect cubes), phir points ko smooth curve se join karen.
(i) \(y = \sqrt{3x}\) — Domain \(x \ge 0\), Range \(y \ge 0\).
| \(x\) | \(0\) | \(3\) | \(12\) |
|---|
| \(y\) | \(0\) | \(3\) | \(6\) |
|---|
Curve starts at \((0,0)\) and rises to the right.
(ii) \(y = \sqrt{x}+5\) — Domain \(x \ge 0\), Range \(y \ge 5\). The graph of \(\sqrt{x}\) shifted 5 units up.
| \(x\) | \(0\) | \(1\) | \(4\) | \(9\) |
|---|
| \(y\) | \(5\) | \(6\) | \(7\) | \(8\) |
|---|
(iii) \(y = -\frac{1}{2}\sqrt{x}\) — Domain \(x \ge 0\), Range \(y \le 0\). Reflected below the x-axis.
| \(x\) | \(0\) | \(1\) | \(4\) | \(9\) |
|---|
| \(y\) | \(0\) | \(-0.5\) | \(-1\) | \(-1.5\) |
|---|
(iv) \(y = -\sqrt{x+1}+2\) — Domain \(x \ge -1\), Range \(y \le 2\). Starts at \((-1, 2)\), falls to the right.
| \(x\) | \(-1\) | \(0\) | \(3\) | \(8\) |
|---|
| \(y\) | \(2\) | \(1\) | \(0\) | \(-1\) |
|---|
(v) \(y = \sqrt[3]{2x+1}\) — Domain \(\mathbb{R}\), Range \(\mathbb{R}\). S-shaped, crosses the x-axis where \(2x+1 = 0\), i.e. \(x = -\frac{1}{2}\).
| \(x\) | \(-4\) | \(-1\) | \(-\frac{1}{2}\) | \(0\) | \(3\) |
|---|
| \(y\) | \(\approx -1.91\) | \(-1\) | \(0\) | \(1\) | \(\approx 1.91\) |
|---|
(vi) \(y = 2\sqrt{x}-3\) — Domain \(x \ge 0\), Range \(y \ge -3\). Starts at \((0, -3)\), rises to the right.
| \(x\) | \(0\) | \(1\) | \(4\) | \(9\) |
|---|
| \(y\) | \(-3\) | \(-1\) | \(1\) | \(3\) |
|---|
(vii) \(y = \sqrt[3]{x^2+x-2}\) — Domain \(\mathbb{R}\), Range \(\mathbb{R}\). Note \(x^2+x-2 = (x+2)(x-1)\), so \(y = 0\) at \(x = -2\) and \(x = 1\).
| \(x\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) |
|---|
| \(y\) | \(0\) | \(\approx -1.26\) | \(\approx -1.26\) | \(0\) | \(\approx 1.59\) |
|---|
Answer: See tables and shape notes above for each graph.
Question 4: A building's height is \(H(t) = 100 + 20t\) metres and a nearby growing tree's height is \(T(t) = 50 + 10t + t^2\) metres, where \(t\) is months. (i) When will both have the same height? (ii) What will that height be? Sketch both graphs and find when the tree overtakes the building.
Same height ke liye \(H(t) = T(t)\) rakh kar quadratic equation solve karenge — quadratic formula se, aur negative time reject kar denge. Height ke liye \(t\) wapas kisi ek function mein rakhenge.
(i) Set \(H(t) = T(t)\):
\[
100 + 20t = 50 + 10t + t^2 \implies t^2 - 10t - 50 = 0
\]
Apply the quadratic formula:
\[
t = \frac{10 \pm \sqrt{100 + 200}}{2} = \frac{10 \pm \sqrt{300}}{2} = \frac{10 \pm 10\sqrt{3}}{2} = 5 \pm 5\sqrt{3}
\]
Time cannot be negative, so \(t = 5 + 5\sqrt{3} \approx 5 + 8.66 = 13.66\) months.
(ii) Substitute into \(H(t)\):
\[
H(13.66) = 100 + 20(13.66) = 373.2 \text{ m}
\]
Sketch: \(H(t)\) is a straight line from \((0, 100)\) with slope \(20\). \(T(t)\) is an upward parabola from \((0, 50)\). Table:
| \(t\) | \(0\) | \(5\) | \(10\) | \(15\) | \(20\) |
|---|
| \(H(t)\) | \(100\) | \(200\) | \(300\) | \(400\) | \(500\) |
|---|
| \(T(t)\) | \(50\) | \(125\) | \(250\) | \(425\) | \(650\) |
|---|
The parabola crosses the line at \(t \approx 13.66\) and stays above it afterwards, so the tree overtakes the building for \(t > 13.66\) months.
Answer: (i) \(t = 5 + 5\sqrt{3} \approx 13.66\) months (ii) ≈ \(373.2\) m; the tree overtakes the building after about \(13.66\) months.
Question 5: A radioactive substance has initial quantity \(Q_0 = 200\) grams and decay model \(Q(t) = Q_0\left(\frac{1}{2}\right)^t\) (with \(t\) in years). Find the remaining quantity after 6 years graphically.
Diye gaye model mein \(t = 6\) substitute karke exact value nikaalte hain. Graphically dikhane ke liye kuch years ki table banayenge — decay curve tezi se girti hai.
Using the given model \(Q(t) = 200\left(\frac{1}{2}\right)^t\):
\[
Q(6) = 200\left(\frac{1}{2}\right)^6 = 200 \cdot \frac{1}{64} = 3.125 \text{ g}
\]
Table for the graph:
| \(t\) (years) | \(0\) | \(2\) | \(4\) | \(6\) | \(8\) |
|---|
| \(Q(t)\) (g) | \(200\) | \(50\) | \(12.5\) | \(3.125\) | \(0.78125\) |
|---|
Plotting these points gives the exponential decay curve; reading the curve at \(t = 6\) gives \(3.125\) g. Note: as written, this model halves the quantity every year (a 2-year half-life would instead be \(Q(t) = 200\left(\frac{1}{2}\right)^{t/2}\)); the value above follows the model stated in the question.
Answer: \(3.125\) grams after 6 years
