Questions on exponential functions and e

Question 1

For the curve y = 3^x, write down the y-intercept, the value of y at x = 2, and the equation of the asymptote.

Accepted Answer

The intercept is (0, 1), since 3⁰ = 1. At x = 2, y = 9. The asymptote is y = 0, the x-axis, which the curve approaches on the left without ever touching. B1 B1 B1 for the three answers. Every curve y = a^x passes through (0, 1) whatever a is, so that intercept is worth remembering rather than working out.

Question 2

Explain why the equation 5^x = 0 has no solution.

Accepted Answer

Every exponential is positive everywhere: 5^x takes every value above zero and no others. The curve hugs the x-axis as x heads left but never reaches it, so zero is approached, not achieved. B1 for the range being strictly positive, B1 for the asymptote argument.

Question 3

Explain why the curves y = 2^x and y = (½)^x are reflections of each other in the y-axis.

Accepted Answer

(½)^x = 2^−x, since raising a reciprocal to the power x is the same as raising the original to the power −x. B1 for that identity. Replacing x by −x in a graph reflects it in the y-axis, so y = 2^−x is the mirror image of y = 2^x across that axis, and (½)^x is exactly this curve under another name. B1 for the reflection rule. Both curves still pass through (0, 1), since 2⁰ = (½)⁰ = 1: that point sits on the mirror line and maps to itself, while growth on one side becomes decay on the other. B1 for identifying (0, 1) as the fixed point of the reflection.

Question 4

Explain why the base a = 1 is excluded from the family y = a^x.

Accepted Answer

1^x = 1 for every value of x, so the graph is the flat line y = 1, not a curve that grows or decays. B1 for the constant value. A flat line has no horizontal asymptote to approach, since every point on it already sits at height 1, so the two features that define the rest of the family, a one-sided approach to y = 0 and a strictly increasing or strictly decreasing curve, are both absent. B1 for stating that the curve has no asymptote and is neither increasing nor decreasing.

Question 5

For the curve y = e^4x, write down the gradient function, the gradient at x = 0, and the gradient at x = 0.5 to 3 significant figures.

Accepted Answer

Differentiating e^kx multiplies by k, so the gradient function is 4e^4x. At x = 0 the gradient is 4e⁰ = 4, and at x = 0.5 it is 4e² = 29.6 to 3 significant figures, four times the height there. B1 for the gradient function, B1 for 4, B1 for 29.6.

Question 6

For the curve y = e^−2x, write down the gradient function and the value of y at x = 1, and describe the curve's behaviour as x grows.

Accepted Answer

The gradient function is −2e^−2x, negative everywhere: the curve falls at every point. At x = 1, y = e⁻² = 0.135 to 3 significant figures. As x grows the curve decays towards the asymptote y = 0, steeply at first and ever more gently. B1 for the gradient function, B1 for 0.135, B1 for the decay to the asymptote.

Question 7

A population of bacteria triples every hour, starting from 500. Write down a model for the population P after t hours, and find the population after 4 hours.

Accepted Answer

P = 500 × 3^t, since each unit step in t multiplies the population by the same ratio 3. After 4 hours, P = 500 × 81 = 40 500. B1 for the model, M1 for substituting t = 4, A1 for the value after four hours. The starting value belongs in front and the growth factor in the base, so P = 500t³ or P = 3 × 500^t both fail the first check, which is that P = 500 at t = 0.

Question 8

For the curve y = e^x+1 − 3, state the equation of the asymptote, find the y-intercept to 3 significant figures, and describe the transformations that turn y = e^x into this curve.

Accepted Answer

The −3 translates every point down 3, so the asymptote moves from y = 0 to y = −3. B1 for the asymptote. At x = 0, y = e¹ − 3 = 2.71828… − 3 = −0.282 to 3 significant figures. M1 for substituting x = 0, A1 for −0.282. The curve is y = e^x translated 1 unit in the negative x-direction and 3 units in the negative y-direction, that is, 1 left and 3 down; either order gives the same curve, since a horizontal and a vertical translation do not interact. B1 for both translations named with their directions.

Question 9

A radioactive sample's mass in grams is modelled by m = 90e^−0.06t, with t in days. State the initial mass, find the mass after 15 days to 3 significant figures, and describe what the model predicts happens to m as t grows very large.

Accepted Answer

The initial mass is the front constant, since e⁰ = 1: 90 g. B1. After 15 days, m = 90e^−0.9 = 36.6 g to 3 significant figures. M1 for substituting t = 15, A1 for 36.6 g. As t grows without bound, e^−0.06t approaches 0, so m decays towards the asymptote m = 0 without ever reaching it: the model predicts a mass that keeps halving over and over rather than one that reaches zero at some finite time. B1 for the long-run behaviour with the asymptote.

Question 10

The temperature, in °C, of a cooling cup of tea in a 20°C room is modelled by T = 20 + 60e^−0.08t, with t in minutes. State the temperature at t = 0, find the temperature after 10 minutes to 3 significant figures, and state the temperature the model predicts in the long run, with a reason.

Accepted Answer

At t = 0, T = 20 + 60e⁰ = 20 + 60 = 80°C. B1. After 10 minutes, T = 20 + 60e^−0.8 = 47.0°C to 3 significant figures. M1 for substituting t = 10, A1 for 47.0°C. As t grows without bound, e^−0.08t approaches 0 and T approaches 20°C, the room temperature: the tea cools towards its surroundings rather than towards zero, because the model has a constant added on rather than a bare exponential. B1 for 20°C with the reason.

Question 11

Two population models, both starting at 200, are P₁ = 200e^0.4t and P₂ = 200e^0.25t, t in hours. Find the gradient of each model at t = 0, and explain what the difference reveals about the two populations at that instant.

Accepted Answer

By the gradient property, the gradient of P₁ at t = 0 is 0.4 × 200 = 80 per hour, and the gradient of P₂ at t = 0 is 0.25 × 200 = 50 per hour. M1 for using gradient = k × value at t = 0 for each model, A1 for both values. Although the two populations start equal, population 1 is growing faster at that instant, since a larger proportionality constant k means a steeper climb for the same starting size. B1 for the comparison.

Question 12

For y = 5e^0.2x, show that the gradient at every point equals 0.2y, and find the height and gradient at x = 5, each to 3 significant figures.

Accepted Answer

The gradient function is 5 × 0.2e^0.2x = 0.2 × 5e^0.2x = 0.2y: the rate of change is proportional to the amount, with constant 0.2. At x = 5 the height is 5e = 13.6 and the gradient is 0.2 × 5e = e = 2.72. M1 for differentiating, A1 for the link to 0.2y, A1 for 13.6, A1 for 2.72.

Question 13

The curves y = 2^x and y = 3^x have gradients of about 0.69 and 1.10 at their common point (0, 1). Explain what this suggests about the number e, and state the defining property of y = e^x.

Accepted Answer

A gradient of exactly 1 at (0, 1) must belong to some base between 2 and 3, and that base is e = 2.718…. The defining property is that y = e^x is its own gradient function: at every point the gradient equals the height, and (0, 1) is only the most convenient place to notice it. B1 for the base lying between 2 and 3, B1 for e = 2.718, B1 for the self-derivative property.

Question 14

A colony of bacteria is modelled by P = Ae^kt, t in hours, for constants A and k. At t = 0 the population is 300, and at that instant the population is increasing at a rate of 45 per hour. Using the gradient property of e^kt, find A and k, and hence find the rate at which the population is increasing when it reaches 900, without finding the value of t at which that happens.

Accepted Answer

At t = 0, P = Ae⁰ = A, so A = 300. B1. Differentiating, dP/dt = kAe^kt = kP at every instant, not only at t = 0, since Ae^kt is P itself. M1 for that general statement. At t = 0 this gives kA = 45, so k = 45/300 = 0.15. A1. Because dP/dt = kP holds throughout, the rate when P = 900 is k × 900 = 0.15 × 900 = 135 per hour. M1 for applying dP/dt = kP at the new population, A1 for 135.

Question 15

Two chemical concentrations decay according to M₁ = 50e^−0.1t and M₂ = 50e^−0.2t, for t ≥ 0, in the same units. Show algebraically that M₁ > M₂ for every t > 0. Then use the gradient property of e^kt to find the rate of change of each at t = 0, and state which concentration is falling faster at that instant.

Accepted Answer

M₁ − M₂ = 50e^−0.1t − 50e^−0.2t = 50e^−0.1t(1 − e^−0.1t). M1 for that factorisation. For t > 0, 0 < e^−0.1t < 1, since e^−0.1t is strictly decreasing in t and equals 1 only at t = 0, so 1 − e^−0.1t > 0 while 50e^−0.1t > 0 too. A1 for both factors being positive. A positive times a positive is positive, so M₁ > M₂ for every t > 0. A1 for the conclusion. By the gradient property, dM₁/dt = −0.1 × 50e^−0.1t and dM₂/dt = −0.2 × 50e^−0.2t; at t = 0 these are −0.1 × 50 = −5 and −0.2 × 50 = −10 per unit time. M1 for both gradient functions, A1 for both values at t = 0. Since |−10| > |−5|, M₂ is falling faster at that instant, even though M₁ stays the larger of the two concentrations throughout. B1 for the comparison stated the right way round.