IBDP Maths SL 1.8 The sum of infinite geometric sequences AA HL Paper 2- Exam Style Questions- New Syllabus
Question
(ii) State a corresponding expansion for \( f_4(x) \) using increasing powers of \( x \).
(ii) Represent \( f(x) \) as a rational function in the form \( \frac{1}{a + bx^2} \), determining the values of constants \( a \) and \( b \).
(ii) Derive the expression for \( g^{-1}(x) \) and specify its domain.
Syllabus Topic Codes (IB Mathematics AA HL):
• SL 1.8: Sum of infinite convergent geometric sequences— parts (c)(i), (c)(ii)
• AHL 2.14: Odd and even functions; finding the inverse function including domain restriction— parts (a), (d)(i), (d)(ii)
• SL 5.11: Areas of a region enclosed by a curve and the axes; areas between curves— part (e)
▶️ Answer/Explanation
(a)
\( f_n(-x) = \sum_{r=0}^n (-2(-x)^2)^r = \sum_{r=0}^n (-2x^2)^r = f_n(x) \).
Since \( f_n(-x) = f_n(x) \) for all \( x \), \( f_n \) is even for all \( n \).
Shown.
(b)(i)
\( f_3(x) = \sum_{r=0}^3 (-2x^2)^r = 1 + (-2x^2) + (-2x^2)^2 + (-2x^2)^3 \)
\( = 1 – 2x^2 + 4x^4 – 8x^6 \).
Shown.
(b)(ii)
\( f_4(x) = f_3(x) + (-2x^2)^4 = 1 – 2x^2 + 4x^4 – 8x^6 + 16x^8 \).
\( \boxed{1 – 2x^2 + 4x^4 – 8x^6 + 16x^8} \)
(c)(i)
\( f_n(x) \) is a finite geometric series with first term 1, common ratio \( r = -2x^2 \).
For the infinite series to converge: \( |r| < 1 \) ⇒ \( |-2x^2| < 1 \) ⇒ \( 2x^2 < 1 \) ⇒ \( x^2 < \frac{1}{2} \).
Thus \( -\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} \).
Largest possible \( k \) is \( K = \frac{1}{\sqrt{2}} \) (or \( \frac{\sqrt{2}}{2} \)).
\( \boxed{K = \frac{1}{\sqrt{2}}} \)
(c)(ii)
For \( |x| < K \), \( f(x) = \sum_{r=0}^\infty (-2x^2)^r \) is an infinite geometric series.
Sum = \( \frac{1}{1 – (-2x^2)} = \frac{1}{1 + 2x^2} \).
Thus \( a = 1, b = 2 \).
\( \boxed{f(x) = \frac{1}{1+2x^2}} \)
(d)(i)
For \( 0 \leq x < K \), \( g(x) = \frac{1}{1+2x^2} \) is strictly decreasing because its derivative \( g'(x) = -\frac{4x}{(1+2x^2)^2} < 0 \) for \( x > 0 \).
Hence \( g \) is one-to-one on this interval, so \( g^{-1} \) exists.
(d)(ii)
Let \( y = g(x) = \frac{1}{1+2x^2} \).
Swap \( x \) and \( y \): \( x = \frac{1}{1+2y^2} \).
Solve: \( 1 + 2y^2 = \frac{1}{x} \) ⇒ \( 2y^2 = \frac{1}{x} – 1 = \frac{1-x}{x} \).
\( y^2 = \frac{1-x}{2x} \) ⇒ \( y = \sqrt{\frac{1-x}{2x}} \) (since \( y \geq 0 \) for original domain).
Thus \( g^{-1}(x) = \sqrt{\frac{1-x}{2x}} \).
Domain of \( g^{-1} \): range of \( g \). For \( 0 \leq x < K \), \( g(x) \) decreases from 1 to \( \frac{1}{1+2K^2} = \frac{1}{1+1} = \frac{1}{2} \).
So domain of \( g^{-1} \) is \( \frac{1}{2} < x \leq 1 \).
\( \boxed{g^{-1}(x) = \sqrt{\frac{1-x}{2x}}, \quad \text{domain: } \frac{1}{2} < x \leq 1} \)
(e)
The curves \( y = g(x) \) and \( y = g^{-1}(x) \) intersect at \( x = y \) (since they are inverses).
Solve \( g(x) = x \): \( \frac{1}{1+2x^2} = x \) ⇒ \( 1 = x + 2x^3 \).
Numerically: intersection at \( x \approx 0.5897545 \).
METHOD 1: Add areas to left and right of intersection.
Area = \( \int_0^{0.58975} g(x) dx + \int_{0.58975}^1 g^{-1}(x) dx \).
First integral: \( \int_0^{0.58975} \frac{1}{1+2x^2} dx \approx 0.491548 \).
Second: \( \int_{0.58975}^1 \sqrt{\frac{1-x}{2x}} dx \approx 0.143738 \).
Total \( \approx 0.635286 \region R \).
METHOD 2: Use symmetry.
Area between \( g(x) \) and \( y = x \) from 0 to intersection ≈ 0.317643.
Double it: ≈ 0.635286.
Rounded to three significant figures:
\( \boxed{0.635} \)