Calculus Question Bank : : ( Qn 01 – 05)

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Question No: 1 [ Maximum Mark: 6] (Do not use GDC)

Consider the function \(f(x)=\sqrt{x}-\frac{4}{x^2} \)
(a)	Find \( f^{\prime}(x)\)	[3 marks]
(b)	Find the gradient of the function f(x) at the point where \(x=4 \)	[3 marks]

Answer : (a) \( f^{\prime}(x)=\frac{1}{2 \sqrt{x}}+\frac{4}{x^3}\) (b) \( f^{\prime}(4)=\frac{5}{16} \)

(a) Find \( f^{\prime}(x)\)

\( \begin{aligned}
& f(x)=\sqrt{x}-\frac{4}{x^2} \\
& f(x)=\sqrt{x}-4 x^{-2} \\
& f^{\prime}(x)=\frac{1}{2 \sqrt{x}}-4 x-2 x^{-3} \\
& f^{\prime}(x)=\frac{1}{2 \sqrt{x}}+\frac{4}{x^3}
\end{aligned}\)

(b) gradient of the function f(x) at the point where \(x=4 \)

\( \begin{aligned}
f^{\prime}(4) & =\frac{1}{2 \sqrt{4}}+\frac{4}{4^3} \\
& =\frac{1}{2 \times 2}+\frac{1}{16} \\
& =\frac{4+1}{16} \\
& =\frac{5}{16}
\end{aligned}\)

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Question No: 2 [ Maximum Mark: 7 ] (Do not use GDC)

Consider the function \(f(x)=2 x^2(x-5) \). The graph of f(x) crosses the x-axis at A and B.
(a)	Find the coordinates of points A and B	[3 marks]
(b)	Find the gradient of f(x) at the points A and B	[4 marks]

Formula Sheet

Answer : (a) \(A(0, 0) \) and \(B(5,0) \) (b) \( f^{\prime}(0)=0 \) and \( f^{\prime}(5)=50\)

GDC Support

(a) solve \( f(x)=0\) to get the x-intercepts

\(f(x)=2 x^2(x-5) \)

\( f(x)=0 \Rightarrow 2 x^2(x-5)=0\)

\(2 x^2=0 \Rightarrow x=0 \) and

\(x-5=0 \Rightarrow x=5 \)

Therefore the coordinates of A and B are \(A(0, 0) \) and \(B(5,0) \)

(b) Gradient funciton of f(x)

\( \begin{gathered}
f(x)=2 x^2(x-5) \\
f(x)=2 x^3-10 x^2 \\
f^{\prime}(x)=6 x^2-20 x
\end{gathered}\)

Gradient of f(x) at \(A(0, 0) \) is \( f^{\prime}(0)=0\). and

Gradient of f(x) at \(A(0, 0) \) is

\( \begin{aligned}
f^{\prime}(5) & =6\left(5^2\right)-20(5) \\
& =6 \times 25-100 \\
& =150-100 \\
f^{\prime}(5) & =50
\end{aligned}\)

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Question No: 3 [ Maximum Mark: 6 ] (Do not use GDC)

Consider the function \( f(x)=x^3+6 x^2+5\).

Find the coordinates of the point at which f(x) has the gradient -9

Formula Sheet

Answer : (a) \( (-1,10) \text { and }(-3,32)\)

GDC Support

Given \( f(x)=x^3+6 x^2+5\)

Differentiate with respect to x we get\( f(x)=x^3+6 x^2+5\)

\( f^{\prime}(x)=-9 \Rightarrow 3 x^2+12 x=-9\)

\( \begin{gathered}
3 x^2+12 x+9=0 \\
x^2+4 x+3=0 \\
(x+1) \quad(x+3)=0 \\
x=-1 \quad x=-3
\end{gathered}\)

when \( x=-1 ; \quad y=(-1)^3+6(-1)^2+5=10\)

when \( x=-3 ; y=(-3)^3+6(-3)^2+5=32\)

Therefore f(x) has a gradient of -1 at \( (-1,10) \text { and }(-3,32)\)

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Question No: 4 [ Maximum Mark: 7 ] (Do not use GDC)

Consider the funciton \( f(x)=3 x^2-2 x+1\)
(a)	Find the gradient of f(x) at \(x=1\)	[2 marks]
(b)	Find the equation of the tangent to \( f(x)\) at \(x=1\)	[2 marks]
(c)	Find the equation of the normal to \( f(x)\) at \(x=1\)	[2 marks]

Formula Sheet

Answer : (a) \( 4 \) (b) \(y=4 x-2 \) (c) \( x+4 y=9\)

GDC Support

(a) Differentiate \(f(x)=3 x^2-2 x+1 \) with respect to x ,

we get \( f^{\prime}(x)=6 x-2\)

the gradient of f(x) at \(x=1\)

is \( \begin{aligned}
& f^{\prime}(1)=6(1)-2 \\
& f^{\prime}(1)=4
\end{aligned} \)

Therefore the gradient of \( f(x) \) at \(x=1\) is 4.

(b) When \(x=1\) , \( f(1)=3 (1)^2-2 (1)+1=2\)

Equation of tangent to \( f(x) \) at \( x=1, y=2\)

\(\begin{aligned}
& y=m x+c \\
& y=4 x+c \\
& 2=4+c \\
& c=-2 \\
& y=4 x-2
\end{aligned} \)

Equation of Normal to \( f(x) \) at \( x=1, y=2\)

\(\begin{gathered}
y=m x+c \\
y=-\frac{1}{4} x+c \\
2=-\frac{1}{4}(1)+c \\
y=2+\frac{1}{4} \\
y=\frac{9}{4} \\
y=\frac{-1}{4} x+\frac{9}{4} \\
y=\frac{-1 x+9}{4} \\
4 y=-1 x+9 \\
x+4 y=9
\end{gathered} \)

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Question No: 5 [ Maximum Mark: 7 ] (Do not use GDC)

Find the equation of normal to the curve \( y=x+\frac{1}{x} \) at \( x=2\) in the form of \( a x+b y=d\), where \(a , b , d \in Z \)

Formula Sheet

Answer : \(4 x+6 y=23 \)

GDC Support

(a) given \(f(x)=x+\frac{1}{x} \) to find y coordinate when \( x=2 \)

\( \begin{aligned}
& f(x)=x+\frac{1}{x} \\
& f(2)=2+\frac{1}{2} \\
& f(2)=\frac{5}{2}
\end{aligned}\)

Differentiate \( f(x)=x+\frac{1}{x}\) with respect to x

we get \( f^{\prime}(x)=1-\frac{1}{x^2}\)

the gradient of the tangent to \( f(x)=x+\frac{1}{x}\) at \( x=2 \) is

\( \begin{aligned}
& f^{\prime}(2)=1-\frac{1}{2^2} \\
& f^{\prime}(2)=\frac{3}{2}
\end{aligned}\)

The gradient of normal \( -\frac{2}{3}\)

The equation of normal is,

\( \begin{aligned}
& y=m x+c \\
& y=-\frac{2}{3} x+c \\
& \frac{5}{2}=\frac{-2}{3} \times 2+c \\
& \frac{5}{2}=-\frac{4}{3}+c \\
& c=\frac{5}{2}+\frac{4}{3} \\
& c=\frac{23}{6}
\end{aligned}\)

\( \begin{aligned}
& y=\frac{-2}{3} x+\frac{23}{6} \\
& y=\frac{-4 x+23}{6} \\
& 6 y=-4 x+23 \\
& 4 x+6 y=23
\end{aligned}\)

Therefore the equation of normal to \( f(x)=x+\frac{1}{x}\) at \( x=2 \) is \( 4 x+6 y=23\)

Video

Question No: 1 [ Maximum Mark: 6] (Do not use GDC)

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Question No: 2 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 3 [ Maximum Mark: 6 ] (Do not use GDC)

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Question No: 4 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 5 [ Maximum Mark: 7 ] (Do not use GDC)

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Curriculum

Calculus Question Bank : : ( Qn 01 – 05)

Question No: 1 [ Maximum Mark: 6] (Do not use GDC)

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Question No: 2 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 3 [ Maximum Mark: 6 ] (Do not use GDC)

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Question No: 4 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 5 [ Maximum Mark: 7 ] (Do not use GDC)

Courses

General

Contact

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