IB DP Maths Topic 9.3 Continuity and differentiability of a function at a point. HL Paper 3

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]

Question

Let $$f(x) = 2x + \left| x \right|$$ , $$x \in \mathbb{R}$$ .

Prove that f is continuous but not differentiable at the point (0, 0) .

[7]
a.

Determine the value of $$\int_{ – a}^a {f(x){\text{d}}x}$$ where $$a > 0$$ .

[3]
b.

Markscheme

we note that $$f(0) = 0,{\text{ }}f(x) = 3x$$ for $$x > 0$$ and $$f(x) = x{\text{ for }}x < 0$$

$$\mathop {\lim }\limits_{x \to {0^ – }} f(x) = \mathop {\lim }\limits_{x \to {0^ – }} x = 0$$     M1A1

$$\mathop {\lim }\limits_{x \to {0^ – }} f(x) = \mathop {\lim }\limits_{x \to {0^ – }} 3x = 0$$     A1

since $$f(0) = 0$$ , the function is continuous when x = 0     AG

$$\mathop {\lim }\limits_{x \to {0^ – }} \frac{{f(0 + h) – f(0)}}{h} = \mathop {\lim }\limits_{x \to {0^ – }} \frac{h}{h} = 1$$     M1A1

$$\mathop {\lim }\limits_{x \to {0^ + }} \frac{{f(0 + h) – f(0)}}{h} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{3h}}{h} = 3$$     A1

these limits are unequal     R1

so f is not differentiable when x = 0     AG

[7 marks]

a.

$$\int_{ – a}^a {f(x){\text{d}}x = \int_{ – a}^0 {x{\text{d}}x + \int_0^a {3x{\text{d}}x} } }$$     M1

$$= \left[ {\frac{{{x^2}}}{2}} \right]_{ – a}^0 + \left[ {\frac{{3{x^2}}}{2}} \right]_0^a$$     A1

$$= {a^2}$$     A1

[3 marks]

b.

[N/A]

a.

[N/A]

b.