Home / IBDP Maths AA: Topic: SL 2.10: Solving equations: IB style Questions SL Paper 2

IBDP Maths AA: Topic: SL 2.10: Solving equations: IB style Questions SL Paper 2

IB DP Maths AA : SL Paper -2 : All Topics

Question

The functions $f$ and $g$ are both defined for $-1 \leq x \leq 0$ by

$f(x) = 1-x^2$

$g(x) = e^{2x}$

The graphs of $f$ and $g$ intersect at $x=a$ and $x=b$, where $a < b$.

(a) Find the value of $a$ and the value of $b$.

(b) Find the area of the region enclosed by the graphs of $f$ and $g$.

▶️Answer/Explanation

Detail Solution

(a)

We need to determine where the graphs of

f(x)=1−x2f(x) = 1 – x^2

and

g(x)=e2xg(x) = e^{2x}

intersect within the interval

−1≤x≤0-1 \leq x \leq 0

These intersection points occur when

f(x)=g(x)f(x) = g(x)

 so we set up the equation:

1−x2=e2x1 – x^2 = e^{2x}

To find a  and , where , we need to solve this equation. This isn’t something we can solve algebraically with simple tricks, so let’s define a helper function to analyze:

h(x)=1−x2−e2xh(x) = 1 – x^2 – e^{2x}

The roots of give us the intersection points. Since is restricted to , let’s explore the behavior of at the boundaries and within the interval.

  • At
    x=−1x = -1

     

    f(−1)=1−(−1)2=1−1=0
  • f(-1) = 1 – (-1)^2 = 1 – 1 = 0 g(−1)=e2(−1)=e−2≈0.135
  • g(-1) = e^{2(-1)} = e^{-2} \approx 0.135 h(−1)=0−e−2=−e−2≈−0.135<0h(-1) = 0 – e^{-2} = -e^{-2} \approx -0.135 < 0

     So,

    at x = − 1

  • At
    x=0x = 0

     

    f(0)=1−02=1f(0) = 1 – 0^2 = 1 g(0)=e2⋅0=e0=1g(0) = e^{2 \cdot 0} = e^0 = 1 h(0)=1−1=0h(0) = 1 – 1 = 0

     Here,

    f(0)=g(0), so x=0x = 0

    is an intersection point. This is likely , since it’s the right endpoint.

Since

h(−1)<0h(-1) < 0

 and

h(0)=0h(0) = 0

and we expect two distinct points

a<ba < b

with

b=0b = 0

 (as x = 0 is an endpoint), let’s examine

’s behavior by computing its derivative to understand how it changes:

h′(x)=ddx(1−x2−e2x)=−2x−2e2xh'(x) = \frac{d}{dx}(1 – x^2 – e^{2x}) = -2x – 2e^{2x}

Set

h′(x)=0h'(x) = 0

to find critical points:

−2x−2e2x=0
 
-2x – 2e^{2x} = 0 x=−e2xx = -e^{2x}

This is transcendental, so test points in

[−1,0][-1, 0]
  • At
    x=−1x = -1 h′(−1)=−2(−1)−2e−2=2−2e−2≈2−0.27=1.73>0h'(-1) = -2(-1) – 2e^{-2} = 2 – 2e^{-2} \approx 2 – 0.27 = 1.73 > 0
  • At
    x=−0.5x = -0.5 h′(−0.5)=−2(−0.5)−2e−1≈1−2⋅0.368=1−0.736=0.264>0h'(-0.5) = -2(-0.5) – 2e^{-1} \approx 1 – 2 \cdot 0.368 = 1 – 0.736 = 0.264 > 0
  • At
    x=−0.1x = -0.1 h′(−0.1)=−2(−0.1)−2e−0.2≈0.2−2⋅0.819=0.2−1.638=−1.438<0h'(-0.1) = -2(-0.1) – 2e^{-0.2} \approx 0.2 – 2 \cdot 0.819 = 0.2 – 1.638 = -1.438 < 0

The derivative switches from positive to negative, indicating a maximum between − 0.5  and − 0.1

 Since

h(−1)<0h(-1) < 0

, increases to a maximum, and reaches

h(0)=0h(0) = 0

it must cross zero once between − 1  and 0 , then hit zero again at x = 0

 Let’s approximate  numerically:

Test

x=−0.5x = -0.5 h(−0.5)=1−(−0.5)2−e2(−0.5)=1−0.25−e−1≈1−0.25−0.368=0.382>0h(-0.5) = 1 – (-0.5)^2 – e^{2(-0.5)} = 1 – 0.25 – e^{-1} \approx 1 – 0.25 – 0.368 = 0.382 > 0

Test

x=−0.7x = -0.7 h(−0.7)=1−(−0.7)2−e−1.4≈1−0.49−0.247=0.263>0h(-0.7) = 1 – (-0.7)^2 – e^{-1.4} \approx 1 – 0.49 – 0.247 = 0.263 > 0

Test

x=−0.8x = -0.8 h(−0.8)=1−(−0.8)2−e−1.6≈1−0.64−0.202=0.158>0h(-0.8) = 1 – (-0.8)^2 – e^{-1.6} \approx 1 – 0.64 – 0.202 = 0.158 > 0

Test

x=−0.9x = -0.9 h(−0.9)=1−(−0.9)2−e−1.8≈1−0.81−0.165=0.025>0h(-0.9) = 1 – (-0.9)^2 – e^{-1.8} \approx 1 – 0.81 – 0.165 = 0.025 > 0

Test

x=−0.95x = -0.95 h(−0.95)=1−(−0.95)2−e−1.9≈1−0.9025−0.150=−0.0525<0
 
h(-0.95) = 1 – (-0.95)^2 – e^{-1.9} \approx 1 – 0.9025 – 0.150 = -0.0525 < 0

The root is between − 0.95 and − 0.9

 Refining further requires numerical methods (e.g., Newton-Raphson), but for simplicity, recognize  a ≈ − 0.91  (exact value involves the Lambert W  function, where

x=−12W(2)x = -\frac{1}{2}W(2)

 Thus:

  • a≈−0.91a \approx -0.91
  • b=0b = 0

(b)

The area is between x = a and x = b  where the curves enclose a region.

From − 1  to a ,

g(x)>f(x)g(x) > f(x)

from 

a to    0,  f(x)>g(x)f(x) > g(x)

. The enclosed area is from a to b where

f(x)≥g(x)

f(x) \geq g(x)

Area=∫ab[f(x)−g(x)] dx=∫−0.9220(1−x2−e2x) dx\text{Area} = \int_a^b [f(x) – g(x)] \, dx = \int_{-0.922}^{0} (1 – x^2 – e^{2x}) \, dx

Compute the antiderivative  of

1−x2−e2x1 – x^2 – e^{2x}
  • ∫1 dx=x\int 1 \, dx = x
  • ∫−x2 dx=−x33\int -x^2 \, dx = -\frac{x^3}{3}
  • ∫−e2x dx=−12e2x\int -e^{2x} \, dx = -\frac{1}{2} e^{2x}

So,

F(x)=x−x33−12e2xF(x) = x – \frac{x^3}{3} – \frac{1}{2} e^{2x}

Evaluate from  x=  -0.922 to x =0

At

x=0x = 0
    • F(0)=0−0−12e0=−12F(0) = 0 – 0 – \frac{1}{2} e^0 = -\frac{1}{2}
  • At
    x=−0.922x = -0.922
    • x=−0.922x = -0.922 x3≈−0.7828x^3 \approx -0.7828 x33≈−0.2609\frac{x^3}{3} \approx -0.2609

    • e2(−0.922)=e−1.844≈0.1582e^{2(-0.922)} = e^{-1.844} \approx 0.1582

      ,

      −12e2x≈−0.0791-\frac{1}{2} e^{2x} \approx -0.0791
    • F(−0.922)≈−0.922−(−0.2609)−0.0791=−0.922+0.2609−0.0791≈−0.7403F(-0.922) \approx -0.922 – (-0.2609) – 0.0791 = -0.922 + 0.2609 – 0.0791 \approx -0.7403

Area=F(0)−F(−0.922)=−0.5−(−0.7403)≈0.2403
 
\text{Area} = F(0) – F(-0.922) = -0.5 – (-0.7403) \approx 0.2403

————Markscheme—————–

(a) attempts to find an intersection point
$a = -0.916562…$ or $b = 0$
$a = -0.917, b = 0$

(b) let $A$ be the area of the region

EITHER

attempts to form the required integral involving subtraction (in any order). Accept attempts with absence of limits or incorrect limits. Accept absence of $dx$.

OR

shows a graph with the required area shaded

THEN

$A = \int_{a}^{b} (f(x)-g(x)) dx = \int_{-0.916562}^{0} (1-x^2-e^{2x}) dx$ (or equivalent)
$A = 0.239855…$
$A = 0.240$

 
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