IBDP Maths AI: Topic SL 5.2: Increasing and decreasing function: IB style Questions SL Paper 1

Question

The function \(f(x)\) is such that \(f'(x) < 0\) for \(1 < x < 4\). At the point \({\text{P}}(4{\text{, }}2)\) on the graph of \(f(x)\) the gradient is zero.

Write down the equation of the tangent to the graph of \(f(x)\) at \({\text{P}}\).[2]

a.

State whether \(f(4)\) is greater than, equal to or less than \(f(2)\).[2]

b.

Given that \(f(x)\) is increasing for \(4 \leqslant x < 7\), what can you say about the point \({\text{P}}\)?[2]

c.
Answer/Explanation

Markscheme

\(y = 2\).     (A1)(A1)     (C2)

Note: Award (A1) for \(y = \ldots \), (A1) for \(2\).
Accept \(f(x) = 2\) and \(y = 0x + 2\)

a.

Less (than).     (A2)     (C2)[2 marks]

b.

Local minimum (accept minimum, smallest or equivalent)     (A2)     (C2)

Note: Award (A1) for stationary or turning point mentioned.
No mark is awarded for \({\text{gradient}} = 0\) as this is given in the question.

c.

Question

The table given below describes the behaviour of f ′(x), the derivative function of f (x), in the domain −4 < x < 2.

State whether f (0) is greater than, less than or equal to f (−2). Give a reason for your answer.[2]

a.

The point P(−2, 3) lies on the graph of f (x).

Write down the equation of the tangent to the graph of f (x) at the point P.[2]

b.

The point P(−2, 3) lies on the graph of f (x).

From the information given about f ′(x), state whether the point (−2, 3) is a maximum, a minimum or neither. Give a reason for your answer.[2]

c.
Answer/Explanation

Markscheme

greater than     (A1)

Gradient between x = −2 and x = 0 is positive.     (R1)

OR

The function is increased between these points or equivalent.     (R1)     (C2)

Note: Accept a sketch. Do not award (A1)(R0).[2 marks]

a.

y = 3     (A1)(A1)     (C2)

Note: Award (A1) for y = a constant, (A1) for 3.[2 marks]

b.

minimum     (A1)

Gradient is negative to the left and positive to the right or equivalent.     (R1)     (C2) 

Note: Accept a sketch. Do not award (A1)(R0).[2 marks]

c.

Question

\[f(x) = \frac{1}{3}{x^3} + 2{x^2} – 12x + 3\]

Find \(f'(x)\) .[3]

a.

Find the interval of \(x\) for which \(f(x)\) is decreasing.[3]

b.
Answer/Explanation

Markscheme

\(f'(x) = {x^2} + 4x – 12\)     (A1)(A1)(A1)     (C3)

Notes: Award (A1) for each term. Award at most (A1)(A1)(A0) if other terms are seen.[3 marks]

a.

\( – 6 \leqslant x \leqslant 2\)     OR     \( – 6 < x < 2\)     (A1)(ft)(A1)(ft)(A1)     (C3)

Notes: Award (A1)(ft) for \( – 6\), (A1)(ft) for \(2\), (A1) for consistent use of strict (\(<\)) or weak (\(\leqslant\)) inequalities. Final (A1) for correct interval notation (accept alternative forms). This can only be awarded when the left hand side of the inequality is less than the right hand side of the inequality. Follow through from their solutions to their \(f'(x) = 0\) only if working seen.[3 marks]

b.

Question

Consider the graph of the function \(f(x) = {x^3} + 2{x^2} – 5\).

Label the local maximum as \({\text{A}}\) on the graph.[1]

a.

Label the local minimum as B on the graph.[1]

b.

Write down the interval where \(f'(x) < 0\).[1]

c.

Draw the tangent to the curve at \(x = 1\) on the graph.[1]

d.

Write down the equation of the tangent at \(x = 1\).[2]

e.
Answer/Explanation

Markscheme

 

correct label on graph     (A1)     (C1)[1 mark]

a.

correct label on graph     (A1)     (C1)[1 mark]

b.

\( – 1.33 < x < 0\)   \(\left( { – \frac{4}{3} < x < 0} \right)\)     (A1)     (C1)[1 mark]

c.

 

tangent drawn at \(x = 1\) on graph     (A1)     (C1)[1 mark]

d.

\(y = 7x – 9\)     (A1)(A1)     (C2)

Notes: Award (A1) for \(7\), (A1) for \(-9\).

If answer not given as an equation award at most (A1)(A0).[2 marks]

e.

Question

Consider the graph of the function \(y = f(x)\) defined below.

Write down all the labelled points on the curve

that are local maximum points;[1]

a.

where the function attains its least value;[1]

b.

where the function attains its greatest value;[1]

c.

where the gradient of the tangent to the curve is positive;[1]

d.

where \(f(x) > 0\) and \(f'(x) < 0\) .[2]

e.
Answer/Explanation

Markscheme

B, F     (C1)

a.

H     (C1)

b.

F     (C1)

c.

A, E     (C1)

d.

C     (C2)

e.

[MAI 5.2-5.3] BASIC DERIVATIVES – TANGENT AND NORMAL-lala

Question

[Maximum mark: 20]
Differentiate the following functions:

Answer/Explanation


Ans.

 

Question

[Maximum mark: 6]
Let \(f(x)=5x^{2}+3\)
(a) Find \({f}'(x)\).
(b) Find the gradient of the curve \(y = f (x)\) at \(x = 1\).
(c) Find the coordinates of the point where the gradient is 20.

Answer/Explanation

Ans.

(a) \({f}'(x)=10x\)
(b) \({f}'(1)=10\)
(c) \({f}'(x)=20\Leftrightarrow 10x=20\Leftrightarrow x=2\), \(y=23\), thus (5,23)

Question

[Maximum mark: 6]
Let \(f(x)=4\sqrt{x}\)
(a) Find \({f}'(x)\).
(b) Find the gradient of the curve \(y = f (x)\) at \(x = 1\).
(c) Find the coordinates of the point where the gradient is 1.

Answer/Explanation

Ans.

(a) \({f}'(x)=\frac{2}{\sqrt{x}}\)
(b) \({f}'(1)=2\)
(c) \({f}'(x)=1\Leftrightarrow \frac{2}{\sqrt{x}}=1\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\), \(y=8\), thus (4,8)

Question

[Maximum mark: 4]
Let \(f(x)=x^{3}-2x^{2}-1\).
(a) Find \({f}'(x)\).
(b) Find the gradient of the curve of \(f (x)\) at the point (2, –1).

Answer/Explanation

Ans.

(a) \({f}'(x)=3x^{2}-4x-0=3x^{2}-4x\)
(b) Gradient=\({f}'(2)=3×4-4×2=4\)

Question

[Maximum mark: 6]
Given the function \(f(x)=x^{2}-3bx+(c+2)\), determine the values of b and c such that
\(f (1) = 0\) and \({f}'(3) = 0\) .

Answer/Explanation

Ans.

\({f}'(x)=2x-3b\)
\(f(1)=0\Leftrightarrow 1^{2}-3b+c+2=0\Leftrightarrow -3b+c+3=0\)
\({f}'(3)=0\Leftrightarrow 6-3b=0\Rightarrow 3b=6\Rightarrow b=2\)
\(-6+c+3=0\Rightarrow c=3\)

 

Question

[Maximum mark: 7]
Given the following values at \(x = 1\)

(a) Find the value of each function below at \(x = 1\).
(i) \(y = f (x) + g(x)\)
(ii) \(y = 2 f (x) + 3g(x)\)
(iii) \(y=f(x)+5x^{2}\)
(b) Calculate the derivatives of the following functions at \(x = 1\)
(i) \(y = f (x) + g(x)\)
(ii) \(y = 2 f (x) + 3g(x)\)
(iii) \(y=f(x)+5x^{2}\)

Answer/Explanation

Ans.

(a) At \(x = 1\)
(i) \(y = f (1) + g(1) = 2 + 3 = 5\)
(ii) \(y = 2 f (1) + 3g(1) = 4 + 9 = 13\)
(iii) \(y = f (1) + 5 = 2 + 5 = 7\)

(b)   (i) \(\frac{dy}{dx} = {f}'(x)+{g}'(x)\).At \(x=1,\frac{dy}{dx}={f}'(1)+{g}'(1)=4+5=9\)

(ii) \(\frac{dy}{dx} = 2{f}'(x)+3{g}'(x)\).At \(x=1,\frac{dy}{dx}=2{f}'(1)+3{g}'(1)=8+15=23\)

(iii) \(\frac{dy}{dx}={f}'(x)+10x\). At \(x=1, \frac{dy}{dx}={f}'(1)+10=4+10=14\)

 

Question

[Maximum mark: 4]
The graph of the function \(y = f (x)\) , 0 ≤ x ≤ 4 , is shown below.

(a) Write down the value of  (i) \(f (1)\)              (ii) \(f (3)\)
(b) Write down the value of  (i) \({f}'(1)\)          (ii) \({f}'(3)\)

Answer/Explanation

Ans.

(a)  (i) 1              (ii) 0.5

(b)  (i) 0             (ii) \(-\frac{1}{2}\)

 

Question

[Maximum mark: 6]
Part of the graph of the periodic function \(f\) is shown below. The domain of \(f\) is
0 ≤ x ≤ 15 and the period is 3.

(a) Find   (i) \(f (2)\)                    (ii) \({f}'(6.5)\)                      (iii) \({f}'(14)\)

(b) How many solutions are there to the equation \(f (x) = 1\) over the given domain?

Answer/Explanation

Ans.

(a)  (i)  1
(ii)  2
(iii) \({f}'(14)={f}'(2)(or \;{f}'(5)\;or\;{f}'(8))=-1\)

(b) There are five repeated periods of the graph, each with two solutions,
ie number of solutions is 5 × 2 = 10

Question

[Maximum mark: 8]
Let \(f(x)=2x^{2}-12x+10\).

(a) Find \({f}'(x)\).
(b) Find the equations of the tangent line and the normal line at \(x = 2\) .
(c) Find the equations of the tangent line and the normal line at \(x = 3\).

Answer/Explanation

Ans.

\(f(x)=2x^{2}-12x+10\)

(a) \({f}'(x)=4x-12\)
(b)  At \(x=2, y=6\), Point (2,6)

\(m_{T}=-4\), \(m_{N}=\frac{1}{4}\)
Tangent line: \(y+6=-4(x-2)\)             \((i.e.\; y=-4x+2)\)
Normal line: \(y+6=\frac{1}{4}(x-2)\)            \((i.e.\; y=\frac{1}{4}x-\frac{13}{2})\)

(c) At \(x = 3\),  \(y =- 8\) , Point (3, –8)
\(m_{T}=0\)
Tangent line: \(y = – 8\) , Normal line: \(x = 3\)

Question

[Maximum mark: 9]
Let \(f(x)=2x^{2}-12x+10\).

(a) Find\({f}'(x)\) .
(b) The line L1 with equation \(y = 4x – 22\) is tangent to the curve.
(i) Write down the gradient of the line L1.
(ii) Find the coordinates of the point where the line L1 touches the curve.
(c) The line L2 is tangent to the curve and parallel to the line \(y = 8x + 3\).
(i) Write down the gradient of the line L2.
(ii) Find the coordinates of the point where the line L2 touches the curve.
(iii) Find the equation of L2 in the form \(y = mx + c\) .

Answer/Explanation

Ans.

\(f(x)=2x^{2}-12x+10\)

(a) \({f}'(x)=4x-12\)
(b) (i) \(m_{1}=4\)
(ii) \(4x-12=4\Leftrightarrow x=4\), thus \(y=-6\). Point (4,-6)
(c) (i) \(m_{2}=8\)
(ii) \(4x-12=8\Leftrightarrow x=5\), thus \(y=0\). Point (5,0)
(iii) \(y-0=8(x-5)\Leftrightarrow y=8x-40\)

Question

[Maximum mark: 6]
Find the equation of the tangent line and the equation of the normal to the curve with
equation\(y=x^{3}+1\) at the point (1,2).

Answer/Explanation

Ans.

\(y=x^{3}+1\)               \(\frac{dy}{dx}=3x^{2}\)

At   \(x = 1\),   \( m_{T}=3\), \(m_{N}=-\frac{1}{3}\)

Equation of tangent: \(y-2=3(x-1)\Rightarrow y=3x-1\)

Equation of normal: \(y-2=-\frac{1}{3}(x-1)\) (OR \(x+3y-7=0\) OR \(y=-\frac{1}{3}x+2\frac{1}{3})\)

Question

[Maximum mark: 6]
Consider the function \(f(x)=4x^{3}+2x\) . Find the equation of the normal to the curve of
\(f\) at the point where \(x = 1\).

Answer/Explanation

Ans.

\({f}'(x)=12x^{2}+2\)

When \(x = 1\), \(f (1)\) = 6, Point (1,6)
When \(x = 1\),\({f}'(1)\)=14, \( m_{T}=14\)

Equation is \(y-6=-\frac{1}{14}(x-1)\left ( y=-\frac{1}{14}x+\frac{85}{14},y=-0.0714x+6.07\right )\)

Question

[Maximum mark: 4]
Find the coordinates of the point on the graph of \(y=x^{2}-x\) at which the tangent is
parallel to the line \(y = 5x\) .

Answer/Explanation

Ans.

\(y=x^{2}-x\)           \(\frac{dy}{dx}=2x-1\)

Line parallel to  \(y=5x\Rightarrow 2x-1=5\Rightarrow x=3\) so \(y=6, Point (3,6)\)

Question

[Maximum mark: 6]
Let \(f(x)=kx^{4}\) . The point P(1, k) lies on the curve of \(f\) . At P, the normal to the curve is
parallel to \(y=-\frac{1}{8}x\) . Find the value of \(k\).

Answer/Explanation

Ans.

\({f}'(x)=4kx^{3}\)
\(m_{N}=-\frac{1}{8}\), thus \(m_{T}=8\)
\(4kx^{3}=8\Rightarrow kx^{3}=2\)
substituting \(x\)=1, \(k\)=2

Question

[Maximum mark: 6]
Consider the function \(f:x \mapsto 3x^2-5x+k\).
The equation of the tangent to the graph of \(f\) at \(x = p\) is \(y = 7x – 9\) .
(a) Write down\({f}'(x)\) .
(b) Find the value of    (i) \(p\) ;            (ii) \(k\) .

Answer/Explanation

Ans.

(a)\({f}'(x)=6x-5\)
(b)\({f}'(p)=7\Rightarrow 6p-5=7\Rightarrow p=2\)
(c) Setting (2)=\(f(2)\)
\(k+2=5\Rightarrow k=3\)

Question

[Maximum mark: 6]
Consider the curve with equation \(f(x)=px^{2}+qx\) , where \(p\) and \(q\) are constants.
The point A(1, 3) lies on the curve. The tangent to the curve at A has gradient 8.
Find the value of \(p\) and of \(q\) .

Answer/Explanation

Ans.

(a) \(f(1)=3\Rightarrow p+q=3\)
\({f}'(x)=2px+q\)
\({f}'(1)=8\Rightarrow 2p+q=8\)
\(p=5,q=-2\)

 

Question

[Maximum mark: 6]
Consider the tangent to the curve \(y=x^{3}+4x^{2}+x-6\) .
(a) Find the equation of this tangent at the point where \(x = – 1\).
(b) Find the coordinates of the point where this tangent meets the curve again.

Answer/Explanation

Ans.

(a) METHOD 1 (directly by GDC)
The equation of the tangent is  \(y = –4x – 8\).
METHOD 2
For \(x=-1, y=-4\) and \(\frac{dy}{dx}=3x^{2}+8x+1, m_{T}=-4\)

Therefore, the tangent equation is \(y+4=-4(x+1)\Rightarrow y=-4x-8\).

(b) This tangent meets the curve when \(-4x-8=x^{3}+4x^{2}+x-6\)
The required point of intersection is (–2, 0).

Question

[Maximum mark: 6]
The line \(y = 16x – 9\) is a tangent to the curve \(y=2x^{3}+ax^{2}+bx-9\) at the point (1,7).
Find the values of \(a\) and \(b\) .

Answer/Explanation

Ans.

For the curve, \(y = 7\) when \(x = 1 ⇒ a + b = 14\), and

\(\frac{dy}{dx}=6x^{2}+2ax+b=16\) when \(x=1\Rightarrow 2a+b=10\).

Solving gives \(a\) = –4 and \(b\) = 18.

Question

[Maximum mark: 11]
The following diagram shows part of the graph of a quadratic function, with equation in
the form \(y = (x – p)(x – q)\) , where \(p,q\in \mathbb{Z}\).

(a) (i) Write down the value of \(p\) and of \(q\) .
(ii) Write down the equation of the axis of symmetry of the curve.
(b) Find the equation of the function in the form \(y=(x-h)^{2}+k\) , where \(h,k\in \mathbb{Z}\).
(c) Find \(\frac{dy}{dx}\)
(d) Let T be the tangent to the curve at the point (0, 5). Find the equation of T.

Answer/Explanation

Ans.

(a) (i) \(p = 1, q = 5\) (or \(p = 5, q = 1\))
(ii) \(x = 3\) (must be an equation)
(b) \(y=(x-1)(x-5)=x^{2}-6x+5=(x-3)^{2}-4\)           (\(h\)=3,\(k\)=-4)
(c) \(\frac{dy}{dx}=2(x-3)(=2x-6)\)
(d) When x = 0, \(\frac{dy}{dx}=-6\)
\(y-5=-6(x-0)\)           (\(y\)=-6x+5 or equivalent)

Question

[Maximum mark: 13]
The function \(f (x)\) is defined as \(f(x)=-(x-h)^{2}+k\). The diagram below shows part of
the graph of \(f (x)\) . The maximum point on the curve is P (3, 2).

(a) Write down the value of  (i) \(h\)         (ii) \(k\)
(b) Show that \(f (x)\) can be written as \(f(x)=-x^{2}+6x-7\).
(c) Find \({f}'(x)\).
The point Q lies on the curve and has coordinates (4, 1). A straight line L, through Q,
is perpendicular to the tangent at Q.
(d) (i) Find the equation of L.
(ii) The line L intersects the curve again at R. Find the x-coordinate of R.

Answer/Explanation

Ans.

(a) \(h = 3\)       \(k = 2\)

(b) \(f(x)=-(x-3)^{2}+2=-x^{2}+6x-9+2=-x^{2}+6x-7\)

(c) \({f}'(x)=-2x+6\)

(d)   (i) tangent gradient = – 2 gradient of  \(L=\frac{1}{2}\)

\(y-1=\frac{1}{2}(x-4)\Rightarrow y=\frac{1}{2}x-1\)

(ii) \(-x^{2}+6x-7=\frac{1}{2}x-1\Leftrightarrow 2x^{2}-11x+12=0\Leftrightarrow\) \(x=1.5\) or \(x=4\) so \(x=1.5\)

(OR by GDC \(-x^{2}+6x-7=\frac{1}{2}x-1\Rightarrow x=1.5\)

Question

[Maximum mark: 11]
The function \(f\) is defined by \(f:x \mapsto -0.5x^{2}+2x+2.5\).
(a) Write down     (i) \({f}'(x)\) ;              (ii) \({f}'(0)\) .
(b) Let N be the normal to the curve at the point where the graph intercepts the
y-axis. Show that the equation of N may be written as \(y=-0.5x+2.5\).
Let \(g:x \mapsto -0.5x+2.5\)
(c) (i) Find the solutions of \(f (x) = g(x)\)
(ii) Hence find the coordinates of the other point of intersection of the normal
and the curve.

Answer/Explanation

Ans.

(a) (i) \({f}'(x)=-x+2\)
(ii) \({f}'(0)=2\)

(b) Gradient of tangent at y-intercept = \({f}'(0)=2\)
⇒ gradient of normal = \(\frac{1}{2}(=-0.5)\)
Therefore, equation of the normal is \(y\) – 2.5 = –0.5 (\(x\) – 0) ⇒ \(y\) = –0.5x + 2.5

(c)  (i) \(-0.5x^{2}+2x+2.5=-0.5x+2.5\Rightarrow x=0\) or \(x=5\)
(ii) Curve and normal intersect when \(x\) = 0 or \(x\) = 5
Other point is when \(x=5\Rightarrow y=-0.5(5)+2.5=0\) (so other point (5, 0)

Question

[Maximum mark: 15]
The equation of a curve may be written in the form \(y = a(x – p)(x – q)\) . The curve
intersects the x-axis at A(–2, 0) and B(4, 0). The curve of \(y = f (x)\) is shown in the
diagram below.

(a) (i) Write down the value of \(p\) and of \(q\).
(ii) Given that the point (6, 8) is on the curve, find the value of \(a\).
(iii) Write the equation of the curve in the form \(y=ax^{2}+bx+c\).
(b) A tangent is drawn to the curve at a point P. The gradient of this tangent is 7.
Find the coordinates of P.
(c) The line L passes through B(4, 0), and is normal to the curve at B.
(i) Find the equation of L.
(ii) Find the x-coordinate of the point where L intersects the curve again.

Answer/Explanation

Ans.

(a) (i) \(p\) =-2   \(q\) = 4  (or \(p\) = 4,  \(q\) =-2 )

(ii)  \(y=a(x+2)(x-4)\Leftrightarrow 8=a(6+2)(6-4)\Leftrightarrow 8=16a\Leftrightarrow a=\frac{1}{2}\)

(iii) \(y=\frac{1}{2}(x+2)(x-4)\Rightarrow y=\frac{1}{2}(x^{2}-2x-8)\Rightarrow y=\frac{1}{2}x^{2}-x-4\)

(b) \(\frac{dy}{dx}=x-1\)

\(x – 1 = 7 ⇔ x = 8, y = 20\) (P is (8, 20))

(c) (i) when x = 4, \(m_{T}=4-1=3\Rightarrow m_{N}=-\frac{1}{3}\)
\(y-0=-\frac{1}{3}(x-4) \left ( y=-\frac{1}{3}x+\frac{4}{3} \right )\)

(ii) \(\frac{1}{2}x^{2}-x-4=-\frac{1}{3}x+\frac{4}{3}\Leftrightarrow x=-\frac{8}{3}\) or \(x=4\)
\(x=-\frac{8}{3}(-2.67)\)

[Maximum mark: 9]
Let \(f(x)=x^{3}-3x^{2}-24x+1\).
(a) Find \({f}'(x)\)
The tangents to the curve of \(f\) at the points P and Q are parallel to the x -axis, where
P is to the left of Q.
(b) Calculate the coordinates of P and of Q.
Let N1 and N2 be the normals to the curve at P and Q respectively.
(c) Write down the coordinates of the points where
(i) the tangent at P intersects N2;
(ii) the tangent at Q intersects N1.

Answer/Explanation

Ans.

(a) \({f}'(x)=3x^{2}-6x-24\)
(b) Tangents parallel to the x-axis mean maximum and minimum (see graph)
EITHER by GDC P(-2, 29) and Q(4, -79)
OR \({f}'(x)=0\Leftrightarrow 3x^{2}-6x-24=0\Leftrightarrow x=-2\) or \(x=4\)
Coordinates are P(-2, 29) and Q(4, -79)
(c)

Question

[Maximum mark: 10]
Consider the curve with equation \(f(x)=3x^{2}\) . The point P(a,3a2 ) lies on the curve.
(a) Find the gradient to the curve at \(x = a\).
(b) Show that the equation of the tangent line to the curve at point P(a,3a2 ) has
equation \(y=6ax-3a^{2}\).
(c) Given that the tangent line passes through the point A(0,-3) find the possible
values of \(a\) .
(d) Hence, find the equations of the tangent lines passing through A(0,-3).

Answer/Explanation

Ans.

\(f(x)=3x^{2}\)

(a) \({f}'(x)=6x\),  \({f}'(a)=6a\)
(b) At \(x=a\), \(y=3a^{2}\), \(m_{T}=6a\)
Tangent line: \(y-3a^{2}=6a(x-a)\Rightarrow y-3a^{2}=6ax-6a^{2}\)
\(\Rightarrow y=6ax-3a^{2}\)

(c) \(-3a^{2}=-3\Leftrightarrow a^{2}-1\Leftrightarrow a=\pm 1\)
(d) For \(a\) = 1, the line is \(y = 6x – 3\)
For \(a\) = -1, the line is \(y =- 6x – 3\)

 

Question

[Maximum mark: 10]
Consider the curve with equation \(f(x)=2x^{3}\) .
(a) Find the equation of the tangent line to the curve at \(x = 1\) .
(b) Find in terms of \(a\) the equation of the tangent line to the curve at \(x = a\).
(c) Hence, find the equation of the tangent line passing through the point A(0, 4) .

Answer/Explanation

Ans.

\(f(x)=2x^{3}\)

(a) \({f}'(x)=6x^{2}\)
At \(x\) = 1, \(y\) = 2 , Point (1,2)
\(m_{T}=6\)
Tangent line: \(y – 2 = 6(x – 1)\)      (i.e. \(y = 6x – 4\) )

(b) At  \(x=a\),  \(y=2a^{3}\) , Point (a,2a3 )
\(m_{T}=6a^{2}\)
Tangent line: \(y-2a^{3}=6a^{2}(x-a)\Rightarrow y-2a^{3}=6a^{2}x-6a^{3}\)
\(\Rightarrow y=6a^{2}x-4a^{3}\)

(c) \(-4a^{3}=4\Leftrightarrow a^{3}=-1\Leftrightarrow a=-1\)

Hence, the tangent line is \(y = 6x + 4\)

 

[MAI 5.2-5.3] BASIC DERIVATIVES – TANGENT AND NORMAL-loyola

Question

[Maximum mark: 20]
Differentiate the following functions

FunctionDerivative
\(y=7x^{3}+5x^{2}+2x+3\) 
\(y=\frac{7}{3}x^{3}-\frac{5}{2}x^{2}\frac{1}{3}x+\frac{4}{5}\) 
\(y=\frac{7x^{3}}{3}-\frac{5x^{2}}{2}+\frac{x}{3}x+\frac{4}{5}\) 
\(y=1+\frac{2}{x}+\frac{3}{x^{2}}\) 
\(y=\frac{1}{3}+\frac{2}{5x}+\frac{3}{7x^{2}}\) 
\(y=x^{2}(1+\frac{2}{x}+\frac{3}{x^{2}})\) 
\(y=\sqrt{x}+\sqrt[3]{x}\) 
\(y=\sqrt{x^{3}}+\sqrt[3]{x^{2}}\) 
\(y=\frac{1+x+x^{2}}{x^{2}}\) 
\(y=\frac{3+5x+7x^{2}}{2x^{2}}\) 
Answer/Explanation

Answer:

Question

[Maximum mark: 6]
Let f (x) = 5x2 + 3
(a) Find f ‘(x) . [2]
(b) Find the gradient of the curve y = f (x) at x = 1. [1]
(c) Find the coordinates of the point where the gradient is 20. [3]

Answer/Explanation

Answer:

(a) f ‘(x) = 10x
(b) f ‘(1) = 10
(c) f ‘(x) = 20⇔10x = 20 ⇔ x = 2 , y = 23, thus (5, 23)

Question

[Maximum mark: 6]
Let \(f(x)=4\sqrt{x}\)

(a) Find f ‘(x) . [2]
(b) Find the gradient of the curve y = f (x) at x = 1 . [1]
(c) Find the coordinates of the point where the gradient is 1. [3]

Answer/Explanation

Answer:

(a) \(f'(x)=\frac{2}{\sqrt{x}}\)

(b) f ‘(1) = 2

(c) \(f'(x)=1\Leftrightarrow \frac{2}{\sqrt{x}}=1\Leftrightarrow \sqrt{x}=2\Leftrightarrow x=4, y=8, thus (4,8)\)

Question

[Maximum mark: 4]

Let \(f(x)=x^{3}-2x^{2}-1\)

(a) Find f ‘(x) [2]
(b) Find the gradient of the curve of f (x) at the point (2, –1). [2]

Answer/Explanation

Answer:

(a) f ‘(x) = 3x2 – 4x – 0= 3x2 – 4x
(b) Gradient = f ‘(2) = 3 × 4 – 4 × 2= 4

Question

[Maximum mark: 6]
Given the function 2 f (x) = x2 = 3bx + (c + 2) , determine the values of b and c such that f (1) = 0 and f ‘(3) = 0 .

Answer/Explanation

Answer:

f ‘(x) = 2x – 3b
f (1) = 0 ⇔ 12 – 3b + c + 2 = 0 ⇔ – 3b + c + 3 = 0
f ‘(3) = 0 ⇔ 6 – 3b = 0 ⇒ 3b = 6 ⇒ b = 2
– 6 + c + 3 = 0 ⇒ c = 3

Question

[Maximum mark: 7]
Given the following values at x =1

(a) Find the value of each function below at x  1.
(i) y = f (x) + g(x)
(ii) y = 2 f (x) + 3g(x)
(iii)  y = f (x)  5x2

(b) Calculate the derivatives of the following functions at x =1
(i) y = f (x) + g(x)
(ii) y + 2 f (x) + 3g(x)
(iii) y = f (x) + 5x2

Answer/Explanation

Answer:

(a) At x =1
(i) y = f (1) + g(1) = 2 + 3 = 5
(ii) y = 2 f (1) + 3g(1) = 4 + 9 =13
(iii) y = f (1) + 5 = 2 + 5 = 7

(b)   (i)  \(\frac{dy}{dx}=f'(x)+g'(x)\) At x=1, \(\frac{dy}{dx}=f'(1)+g'(1)=4+5=9\)

(ii) \(\frac{dy}{dx}=2f'(x)+3g'(x)\)  At x=1, \(\frac{dy}{dx}=2f'(1)+3g'(1)=8+15=23\)

(iii) \(\frac{dy}{dx}=2f'(x)+10x\)   At x=1, \(\frac{dy}{dx}=f'(1)+10=4+10=14\)

Question

[Maximum mark: 4]
The graph of the function y = f (x) , 0 ≤ x ≤ 4 , is shown below.

(a) Write down the value of       (i) f (1)                 (ii) f (3)
(b) Write down the value of       (i) f ‘(1)                (ii) f ‘(3)

Answer/Explanation

Answer:

(a)   (i) 1        (ii) 0.5
(b)   (i) 0       (ii) \(-\frac{1}{2}\)

Question

[Maximum mark: 6]
Part of the graph of the periodic function  is shown below. The domain of  is
0 ≤ x ≤15 and the period is 3.

(a) Find        (i) f (2)         (ii) f ‘(6.5)             (iii) f ‘(14)
(b) How many solutions are there to the equation f (x) =1 over the given domain?

Answer/Explanation

Answer:

(a)    (i) 1
(ii) 2
(iii) f ‘(14) = f ‘(2) (or f ‘(5) or f ‘(8)= –1
(b) There are five repeated periods of the graph, each with two solutions,
ie number of solutions is 5 × 2 =10

Question

[Maximum mark: 8]
Let f (x) = 2x2  – 12 x + 10
(a) Find f ‘(x) . [1]
(b) Find the equations of the tangent line and the normal line at x = 2 . [4]
(c) Find the equations of the tangent line and the normal line at x = 3. [3]

Answer/Explanation

Answer:

f (x) = 2x2 – 12x + 10
(a) f ‘(x) = 4x -12
(b) At x = 2 , y = 6 , Point (2,6)
m= -4, mN = \(\frac{1}{4}\)
Tangent line: y + 6 = -4(x – 2)      (i.e.    y = -4x + 2 )
Normal line: y + 6 =  \(\frac{1}{4}\) (x-2)     (i.e.  \( y=\frac{1}{4}\times -\frac{13}{2}\)

(a) At x = 3, y = -8 , Point (3, –8)
mT = 0
Tangent line: y = -8 ,     Normal line: x = 3

Question

[Maximum mark: 9]
Let f (x ) = 2x2  – 12x +10
(a) Find f ‘(x) . [1]
(b) The line L1 with equation y = 4x – 22 is tangent to the curve.
(i) Write down the gradient of the line L1.
(ii) Find the coordinates of the point where the line L1 touches the curve. [3]
(c) The line L2 is tangent to the curve and parallel to the line y = 8x + 3.
(i) Write down the gradient of the line L2.
(ii) Find the coordinates of the point where the line L2 touches the curve.
(iii) Find the equation of L2 in the form y = mx + c . [5]

Answer/Explanation

Answer:

f (x) = 2x2 – 12 x +10
(a) f ‘(x) = 4x -12
(b) (i)  m1 = 4
(ii) 4x – 12 = 4 ⇔ x = 4 , thus y = -6 . Point (4,-6)
(c) (i) m2 = 8
(ii) 4x -12 = 8⇔ x = 5 , thus y = 0. Point (5,0)
(iii) y – 0 = 8(x – 5) ⇔ y = 8x – 40

Question

[Maximum mark: 6]
Find the equation of the tangent line and the equation of the normal to the curve with equation y = x3 +1 at the point (1,2).

Answer/Explanation

Answer:

y = x3 + 1    \(\frac{dy}{dx}=3x^{2}\)

At x = 1,   mT = 3 ,   mN = \(-\frac{1}{3}\)
Equation of tangent: y – 2 = 3(x – 1) ⇒ y = 3x – 1

Equation of normal: y – 2 = \(-\frac{1}{3}(x-1)\)    (OR x + 3y – 7 = 0   OR y = \(-\frac{1}{3}x+2\frac{1}{3})\)

Question

[Maximum mark: 6]
Consider the function  f (x) = 4x3 + 2x . Find the equation of the normal to the curve of  at the point where x =1.

Answer/Explanation

Answer:

f ‘(x) = 12x2 + 2
When x = 1, f (1) = 6, Point (1,6)
When x = 1, f ‘ (1) = 14,  mT =  14

Equation is y – 6 = \(-\frac{1}{14}(x-1)\left ( y=-\frac{1}{14}x+\frac{85}{14,}y=-0.0714x+6.07 \right )\)

Question

[Maximum mark: 4]
Find the coordinates of the point on the graph of y = x2 – x at which the tangent is parallel to the line y = 5x .

Answer/Explanation

Answer:

\(y=x^{2}-x\)    \(\frac{dy}{dx}=2x-1\)

Line parallel to y = 5x ⇒ 2x – 1 = 5 ⇒  x = 3    so    y = 6, Point (3, 6)

Question

[Maximum mark: 6]
Let  f (x) = kx4 . The point P(1, k) lies on the curve of  f  . At P, the normal to the curve is parallel to  \(y=-\frac{1}{8}x\)  . Find the value of k.

Answer/Explanation

Answer:

f ′(x) = 4kx3
mN = \(-\frac{1}{8}\) , thus  mT = 8
4kx3 = 8 ⇒ kx3 = 2
substituting x = 1 , k = 2

Question

[Maximum mark: 6]
Consider the function \(f:x \mapsto 3x^{2}-5x+k\).
The equation of the tangent to the graph of  at  x = p   is   y = 7x – 9 .
(a) Write down f ‘(x) .
(b) Find the value of          (i)   p ;           (ii)   k .

Answer/Explanation

Answer:

(a) f ‘(x) = 6x – 5
(b) f ‘(p) = 7 ⇒ 6p -5 = 7 ⇒ p = 2
(c) Setting y (2) = f (2)
k + 2 = 5  ⇒ k = 3

Question

[Maximum mark: 6]
Consider the curve with equation f (x) = px2 + qx , where  and  q  are constants.
The point A(1, 3) lies on the curve. The tangent to the curve at A has gradient 8.
Find the value of  p  and of  q

Answer/Explanation

.
Answer:

a) f (1) = 3 ⇒ p + q = 3
f ′(x) = 2px + q
f ′(1) = 8 ⇒ 2p + q = 8
p = 5, q = –2

Question

[Maximum mark: 6]
Consider the tangent to the curve y = x3 + 4x2 + x – 6 .
(a) Find the equation of this tangent at the point where x = -1.
(b) Find the coordinates of the point where this tangent meets the curve again.

Answer/Explanation


Answer:

(a) METHOD 1 (directly by GDC)
The equation of the tangent is y = –4x – 8.
METHOD 2
For x = -1, y = –4 and  \(\frac{dy}{dx}=3x^{2}+8x+1, m_{T}=-4\)
Therefore, the tangent equation is y + 4 = -4(x +1) ⇒ y = –4x – 8.
(b) This tangent meets the curve when –4x – 8 = x3 + 4x2 + x – 6
The required point of intersection is (–2, 0).

Question

[Maximum mark: 6]
The line y =16x – 9 is a tangent to the curve  y = 2x3 + ax + bx – 9 at the point (1,7). Find the values of a and b .

Answer/Explanation

Answer:

For the curve, y = 7 when x = 1 ⇒ a + b = 14, and
\(\frac{dy}{dx}\) = 6x2 + 2ax + b = 16 when x = 1 ⇒ 2a + b = 10.
Solving gives a = –4 and b = 18.

Question

[Maximum mark: 11]
The following diagram shows part of the graph of a quadratic function, with equation in the form  \(y=(x-p)(x-q)\), where  \(p,q\epsilon \mathbb{Z}\)

(a) (i) Write down the value of p and of q .
(ii) Write down the equation of the axis of symmetry of the curve. [3]
(b) Find the equation of the function in the form  y = (x – h)2 + k , where h, k ∈ \(\mathbb{Z}\) . [2]
(c) Find \(\frac{dy}{dx}\) [3]
(d) Let T be the tangent to the curve at the point (0, 5). Find the equation of T. [3]

Answer/Explanation


Answer:

(a) (i) p = 1, q = 5 (or p = 5, q = 1)
(ii) x = 3 (must be an equation)
(b) y = (x – 1)(x – 5) = x2 – 6x + 5 = (x – 3)2 – 4 (h = 3, k = -4)
(c) \(\frac{dy}{dx}\) = 2 (x –  3) (= 2x – 6)
(d) When x = 0, \(\frac{dy}{dx}\) = -6
y – 5 = -6(x – 0)      (y = -6x + 5 or equivalent)

Question

[Maximum mark: 13]
The function f (x) is defined as  f (x) = -(x – h)2 + k . The diagram below shows part of the graph of f (x) . The maximum point on the curve is P (3, 2).

(a) Write down the value of      (i) h         (ii) k [2]
(b) Show that f (x) can be written as  f (x) = -x2 + 6x – 7 . [1]
(c) Find f ‘(x) . [2]
The point Q lies on the curve and has coordinates (4, 1). A straight line L, through Q, is perpendicular to the tangent at Q.
(d) (i) Find the equation of L.
(ii) The line L intersects the curve again at R. Find the x-coordinate of R. [8]

Answer/Explanation

Answer:

a) h = 3 k = 2
(b) f (x) = – (x – 3)2 + 2 = -x2 + 6x – 9 + 2 = -x2 + 6x – 7
(c) f ‘(x) = -2x + 6
(d) (i) tangent gradient = -2 gradient of L = \(\frac{1}{2}\)
\(y-1=\frac{1}{2}(x-4)\Rightarrow y=\frac{1}{2}x-1\)

(ii) \(-x+6x-7=\frac{1}{2}x-1\Leftrightarrow 2x^{2}-11x+12-0\Leftrightarrow x=1.5\)  or x=4   so  x=1.5

Question

[Maximum mark: 11]
The function  is defined by \(f:x \mapsto – 0.5x^{2}+2x+2.5\)

(a) Write down          (i) f ‘(x) ;              (ii) f ‘(0) . [2]
(b) Let N be the normal to the curve at the point where the graph intercepts the
y-axis. Show that the equation of N may be written as y = -0.5x + 2.5 .

Let  \(g:x \mapsto – 0.5x+2.5\)

(c) (i) Find the solutions of f (x) = g(x)
(ii) Hence find the coordinates of the other point of intersection of the normal
and the curve. [6]

Answer/Explanation

 

Answer:

(a) (i) f ‘(x) = –x + 2
(ii) f ‘(0) = 2
(b) Gradient of tangent at y-intercept = f ‘(0) = 2
⇒ gradient of normal = \(\frac{1}{2}(=0.5) \)

Therefore, equation of the normal is y – 2.5 = –0.5 (x – 0) ⇒ y = –0.5x + 2.5

c) (i) –0.5x2 + 2x + 2.5 = –0.5x + 2.5 ⇒ x = 0   or   x = 5
(ii) Curve and normal intersect when x = 0   or   x = 5
Other point is when x = 5 ⇒ y = –0.5(5) + 2.5 = 0 (so other point (5, 0)

Question

[Maximum mark: 15]
The equation of a curve may be written in the form y = a(x – p)(x – q) . The curve
intersects the x-axis at A(–2, 0) and B(4, 0). The curve of y = f (x) is shown in the
diagram below.

(i) Write down the value of p and of q.
(ii) Given that the point (6, 8) is on the curve, find the value of a.
(iii) Write the equation of the curve in the form  y = ax2 + bx + c . [5]
(b) A tangent is drawn to the curve at a point P. The gradient of this tangent is 7.
Find the coordinates of P. [4]
(c) The line L passes through B(4, 0), and is normal to the curve at B.
(i) Find the equation of L.
(ii) Find the x-coordinate of the point where L intersects the curve again. [6]

Answer/Explanation

Answer:

(a) (i)     p = -2   q = 4 (or p = 4, q = -2 )

(ii)  \(y=a(x+2)(x-4)\Leftrightarrow 8=a(6+2)(6-4)\Leftrightarrow 8=16a\Leftrightarrow a=\frac{1}{2} \)

(iii) \(y=\frac{1}{2}(x+2)(x-4)\Leftrightarrow y=\frac{1}{2}(x^{2}-2x-8)\frac{1}{2}x^{2}-x-4\)

(b) \(\frac{dy}{dx}=x-1\)

\(x-1=7\Leftrightarrow x=8, y=20 (P is (8, 20))\)

(c) (i) when x = 4, mT = 4 – 1 = 3 ⇒ mN = \(-\frac{1}{3}\)

\(y-0=-\frac{1}{3}(x-4)\left ( y=-\frac{1}{3}x+\frac{4}{3} \right )\)

(ii) \(\frac{1}{2}x^{2}-x-4=-\frac{1}{3}x+\frac{4}{3}\Leftrightarrow x=-\frac{8}{3}\)  or  x=4

\(x=-\frac{8}{3}(-2.67)\)

Question

[Maximum mark: 9]
Let f (x) = x3 – 24x +1.
(a) Find f ‘(x) [2]
The tangents to the curve of f at the points P and Q are parallel to the x -axis, where P is to the left of Q.
(b) Calculate the coordinates of P and of Q. [3]
Let N1 and N2 be the normals to the curve at P and Q respectively.
(c) Write down the coordinates of the points where
(i) the tangent at P intersects N2;
(ii) the tangent at Q intersects N1. [4]

Answer/Explanation

Answer:

(a) f’ (x) = 3x2 – 6x – 24
(b) Tangents parallel to the x-axis mean maximum and minimum (see graph)
EITHER by GDC P(-2, 29) and Q(4, -79)
OR f’ (x) = 0 ⇔ 3x2 – 6x – 24 = 0 ⇔ x = -2 or x = 4
Coordinates are P(-2, 29) and Q(4, -79)

(c)

(i)     (4, 29)                  (ii) (-2, -79)

Question

[Maximum mark: 10]
Consider the curve with equation f (x) = 3x2 . The point P 2 (a,3a ) lies on the curve.
(a) Find the gradient to the curve at x = a . [2]
(b) Show that the equation of the tangent line to the curve at point P (a, 3a2 ) has
equation y = 6ax – 3a2 . [3]
(c) Given that the tangent line passes through the point A(0,-3) find the possible
values of a . [3]
(d) Hence, find the equations of the tangent lines passing through A(0,- 3) . [2]

Answer/Explanation

Answer:

f (x) = 3x2
(a) f ‘(x) = 6x , f ‘(a) = 6a
(b) At x = a,  y = 3a2 ,  mT = 6a
Tangent line: y – 3a2 = 6a(x – a) ⇒  y -3a2 = 6ax – 6a2

⇒ y = 6ax – 3a2
(c) -3a2 = -3⇔ a2 =1 ⇔ a = ±1
(d) For a =1, the line is y = 6x – 3
For a = -1, the line is y = -6x – 3

Question

[Maximum mark: 10]
Consider the curve with equation f (x) = 2x3 .
(a) Find the equation of the tangent line to the curve at x =1 . [3]
(b) Find in terms of a the equation of the tangent line to the curve at x = a . [3]
(c) Hence, find the equation of the tangent line passing through the point A(0, 4) . [4]

Answer/Explanation


Answer:

f (x) = 2x3
(a)  f ‘(x) = 6x2
At x =1, y = 2 , Point (1,2)
mT = 6
Tangent line: y – 2 = 6(x -1)       (i.e. y = 6x – 4 )
(b) At x  a , 3 y  2a , Point 3 (a,2a )
mT = 6a2
Tangent line: y = 2a3 = 6a2 (x – a) ⇒  y = 2a3 = 6a2 x – 6a3
⇒ y = 6a2 x – 4a3
(c) 4a3 = 4⇔ a3 =-1⇔ a = -1
Hence, the tangent line is y = 6x + 4

 

 

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