# IB DP Maths Topic 9.3 Continuous functions and differentiable functions HL Paper 3

 Marks available 8 Reference code 14M.3ca.hl.TZ0.4

## Question

The function f is defined by $$f(x) = \left\{ \begin{array}{r}{e^{ – x^3}}( – {x^3} + 2{x^2} + x),x \le 1\\ax + b,x > 1\end{array} \right.$$, where $$a$$ and $$b$$ are constants.

Find the exact values of $$a$$ and $$b$$ if $$f$$ is continuous and differentiable at $$x = 1$$.

[8]
a.

(i)     Use Rolle’s theorem, applied to $$f$$, to prove that $$2{x^4} – 4{x^3} – 5{x^2} + 4x + 1 = 0$$ has a root in the interval $$\left] { – 1,1} \right[$$.

(ii)     Hence prove that $$2{x^4} – 4{x^3} – 5{x^2} + 4x + 1 = 0$$ has at least two roots in the interval $$\left] { – 1,1} \right[$$.

[7]
b.

## Markscheme

$$\mathop {{\text{lim}}}\limits_{x \to {1^ – }} {{\text{e}}^{ – {x^2}}}\left( { – {x^3} + 2{x^2} + x} \right) = \mathop {{\text{lim}}}\limits_{x \to {1^ + }} (ax + b)$$   $$( = a + b)$$     M1

$$2{{\text{e}}^{ – 1}} = a + b$$     A1

differentiability: attempt to differentiate both expressions     M1

$$f'(x) = – 2x{{\text{e}}^{ – {x^2}}}\left( { – {x^3} + 2{x^2} + x} \right) + {{\text{e}}^{ – {x^2}}}\left( { – 3{x^2} + 4x + 1} \right)$$   $$(x < 1)$$     A1

(or $$f'(x) = {{\text{e}}^{ – {x^2}}}\left( {2{x^4} – 4{x^3} – 5{x^2} + 4x + 1} \right)$$)

$$f'(x) = a$$   $$(x > 1)$$     A1

substitute $$x = 1$$ in both expressions and equate

$$– 2{{\text{e}}^{ – 1}} = a$$     A1

substitute value of $$a$$ and find $$b = 4{{\text{e}}^{ – 1}}$$     M1A1

[8 marks]

a.

(i)     $$f'(x) = {{\text{e}}^{ – {x^2}}}\left( {2{x^4} – 4{x^3} – 5{x^2} + 4x + 1} \right)$$   (for $$x \leqslant 1$$)     M1

$$f(1) = f( – 1)$$     M1

Rolle’s theorem statement     (A1)

by Rolle’s Theorem, $$f'(x)$$ has a zero in $$\left] { – 1,1} \right[$$     R1

hence quartic equation has a root in $$\left] { – 1,1} \right[$$     AG

(ii)     let $$g(x) = 2{x^4} – 4{x^3} – 5{x^2} + 4x + 1$$.

$$g( – 1) = g(1) < 0$$ and $$g(0) > 0$$     M1

as $$g$$ is a polynomial function it is continuous in $$\left[ { – 1,0} \right]$$ and $$\left[ {0,{\text{ 1}}} \right]$$.     R1

(or $$g$$ is a polynomial function continuous in any interval of real numbers)

then the graph of $$g$$ must cross the x-axis at least once in $$\left] { – 1,0} \right[$$     R1

and at least once in $$\left] {0,1} \right[$$.

[7 marks]

b.

## Examiners report

[N/A]
a.
[N/A]
b.
 Marks available 6 Reference code 18M.3ca.hl.TZ0.2

## Question

The function $$f$$ is defined by

$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\left| {x – 2} \right| + 1}&{x < 2} \\ {a{x^2} + bx}&{x \geqslant 2} \end{array}} \right.$

where $$a$$ and $$b$$ are real constants

Given that both $$f$$ and its derivative are continuous at $$x = 2$$, find the value of $$a$$ and the value of $$b$$.

## Markscheme

considering continuity at $$x = 2$$

$$\mathop {{\text{lim}}}\limits_{x \to {2^ – }} f\left( x \right) = 1$$ and $$\mathop {{\text{lim}}}\limits_{x \to {2^ + }} f\left( x \right) = 4a + 2b$$    (M1)

$$4a + 2b = 1$$     A1

considering differentiability at $$x = 2$$

$$f’\left( x \right) = \left\{ {\begin{array}{*{20}{c}} { – 1}&{x < 2} \\ {2ax + b}&{x \geqslant 2} \end{array}} \right.$$    (M1)

$$\mathop {{\text{lim}}}\limits_{x \to {2^ – }} f’\left( x \right) = – 1$$ and $$\mathop {{\text{lim}}}\limits_{x \to {2^ + }} f’\left( x \right) = 4a + b$$     (M1)

Note: The above M1 is for attempting to find the left and right limit of their derived piecewise function at $$x = 2$$.

$$4a + b = – 1$$     A1

$$a = – \frac{3}{4}$$ and $$b = 2$$     A1

[6 marks]

[N/A]
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