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

 

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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 \(b\).

Answer/Explanation

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]

Examiners report

[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.
Answer/Explanation

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.

Examiners report

[N/A]

a.

[N/A]

b.

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