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Calculus, EXT1 EQ-Bank 16

Find the equation of the tangent to the curve  \(y=\tan ^{-1}\left(x^2\right)\)  at the point on the curve where  \(x=1\).   (3 marks)

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\(y=x-1+\dfrac{\pi}{4}\)

Show Worked Solution

\(y=\tan ^{-1}\left(x^2\right) \ \Rightarrow \ \dfrac{dy}{dx}=\dfrac{1}{1+\left(x^2\right)^2} \times 2 x=\dfrac{2 x}{1+x^4}\)

\(\text{At} \ \ x=1:\)

\(y=\tan ^{-1}\left(1^2\right)=\dfrac{\pi}{4}\)

\(\dfrac{dy}{dx}=\dfrac{2}{1+1^4}=1\)
 

\(\text{Find equation of tangent,} \ \ m=1 \ \ \text {through}\ \left(1, \dfrac{\pi}{4}\right):\)

\(y-\dfrac{\pi}{4}\) \(=1(x-1)\)
\(y\) \(=x-1+\dfrac{\pi}{4}\)

Calculus, EXT1 EQ-Bank 19

The function  \(y=f(x)\)  has an inverse function  \(y=f^{-1}(x)\).

The tangent to  \(y=f(x)\) at the point \((2,3)\), \(\ell\), has a gradient of 1.

Show that the tangent to  \(y=f^{-1}(x)\)  at the point \((3,2)\) is parallel to \(\ell\).   (3 marks)

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\(\text{Since tangent to}\  f(x) \ \text {touches at}\ (2,3):\)

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

\(\text{For inverse functions,}\ \ \left(f^{-1}\right)^{\prime}(x)=\dfrac{1}{f^{\prime}\left(f^{-1}(x)\right)}\)

\(\text{Since tangent to} \ \ f(x) \ \ \text{at} \ \ x=2 \ \ \text{has gradient 1}\)

\(m_{\ell}=1 \ \Rightarrow \ f^{\prime}(2)=1\)
 

\(\text{Find gradient of}\ f^{-1}(x) \ \text{at} \ \ x=3:\)

\((f^{-1})^{\prime}(3)=\dfrac{1}{f^{\prime}\left(f^{-1}(3)\right)}=\dfrac{1}{f^{\prime}(2)}=1\)

\(\therefore \text{Gradient of tangent to} \ f^{-1}(x) \ \text {at } x=3\ \ \ \text {is parallel to} \ \ell\).

Show Worked Solution

\(\text{Since tangent to}\  f(x) \ \text {touches at}\ (2,3):\)

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

\(\text{For inverse functions,}\ \ \left(f^{-1}\right)^{\prime}(x)=\dfrac{1}{f^{\prime}\left(f^{-1}(x)\right)}\)

\(\text{Since tangent to} \ \ f(x) \ \ \text{at} \ \ x=2 \ \ \text{has gradient 1}\)

\(m_{\ell}=1 \ \Rightarrow \ f^{\prime}(2)=1\)
 

\(\text{Find gradient of}\ f^{-1}(x) \ \text{at} \ \ x=3:\)

\((f^{-1})^{\prime}(3)=\dfrac{1}{f^{\prime}\left(f^{-1}(3)\right)}=\dfrac{1}{f^{\prime}(2)}=1\)

\(\therefore \text{Gradient of tangent to} \ f^{-1}(x) \ \text {at } x=3\ \ \ \text {is parallel to} \ \ell\).

Calculus, EXT1 EQ-Bank 32

The polynomial  \(h(x)=x^3+2x+1\)  passes through the point \((1,4)\).

Find the gradient of the tangent to  \(f(x)=x h^{-1}(x)\)  at the point where \(x=4\).   (3 marks)

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\(\dfrac{9}{5}\)

Show Worked Solution

\(h(1)=4 \ \ \Rightarrow\ \ h^{-1}(4)=1\)

\(f(x)=x h^{-1}(x)\)

\(\text{Using the product rule:}\)

\(f^{\prime}(x)=h^{-1}(x)+x\cdot \dfrac{d}{dx}\left(h^{-1}(x)\right)\)
 

\(\text{Find}\ \dfrac{d}{dx}\left(h^{-1}(x)\right):\)

\(\text{Let}\ \ y=h^{-1}(x)\ \ \Rightarrow\ \ x=h(y)\)

\(\dfrac{dx}{dy}=h^{\prime}(y)\ \ \Rightarrow\ \ \dfrac{dy}{dx}=\dfrac{1}{h^{\prime}(y)}\)

\(\dfrac{d}{dx}\left(h^{-1}(x)\right)=\dfrac{1}{h^{\prime}\left(h^{-1}(x)\right)}\)
 

\(f^{\prime}(x)=h^{-1}(x)+\dfrac{x}{h^{\prime}\left(h^{-1}(x)\right)}\)

\(f^{\prime}(4)=h^{-1}(4)+\dfrac{4}{h^{\prime}\left(h^{-1}(4)\right)}=1+\dfrac{4}{h^{\prime}(1)}\)
 

\(h(x)=x^3+2x+1\ \ \Rightarrow\ \ h^{\prime}(x)=3x^2+2\)

\(f^{\prime}(4)=1+\dfrac{4}{3(1)^2+2}=\dfrac{9}{5}\)

\(\therefore\ \text{Gradient of tangent}=\dfrac{9}{5}\)

Calculus, EXT1 C2 2025 HSC 10 MC

For the function \(f(x)\), it is known that  \(f(3)=1, f^{\prime}(3)=2\)  and  \(f^{\prime \prime}(3)=4\).

Let  \(g(x)=f^{-1}(x)\).

What is the value of \(g^{\prime \prime}(1)\) ?

  1. \(\dfrac{1}{4}\)
  2. \(-\dfrac{1}{4}\)
  3. \(-\dfrac{1}{2}\)
  4. \(-1\)
Show Answers Only

\(C\)

Show Worked Solution

\(f(3)=1, f^{\prime}(3)=2, f^{\prime \prime}(3)=4\)

\(\text{Given} \ \ g(x)=f^{-1}(x):\)

\(f(g(x))=x \ \ \text{(Definition of an inverse fn)}\)

♦♦♦ Mean mark 22%.

\(\text{Differentiate both sides:}\)

\(g^{\prime}(x) \cdot f^{\prime}(g(x))=1 \ \ \Rightarrow \ \ g^{\prime}(x)=\dfrac{1}{f^{\prime}(g(x))}\)

\(g^{\prime \prime}(x)=\dfrac{d}{d x}\left(\dfrac{1}{f^{\prime}(g(x))}\right)=-\dfrac{f^{\prime \prime}(g(x)) \cdot g^{\prime}(x)}{\left[f^{\prime}(g(x))\right]^2}\)
 

\(\text{When}\ \ x=1:\)

\(g^{\prime}(1)\) \(=\dfrac{1}{f^{\prime}(g(1))}=\dfrac{1}{f^{\prime}(3)}=\dfrac{1}{2}\)
\(g^{\prime \prime}(1)\) \(=-\dfrac{f^{\prime \prime}(g(1)) \cdot g^{\prime}(1)}{\left[f^{\prime}(g(1))\right]^2}=-\dfrac{f^{\prime \prime}(3) \cdot \dfrac{1}{2}}{\left[f^{\prime}(3)\right]^2}=-\dfrac{4 \times \dfrac{1}{2}}{2^2}=-\dfrac{1}{2}\)

 
\(\Rightarrow C\)

Calculus, EXT1 C2 2025 HSC 14d

The function  \(f(x)\)  is defined by  \(f(x)=\cos ^{-1}(\sin x)\)  in the domain \((0, \pi)\).

Find  \(f^{\prime}(x)\)  for those values of \(x\) where it is defined.   (3 marks)

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\(f^{\prime}(x)=-1 \ \text{for} \ \ x \in\left(0, \dfrac{\pi}{2}\right)\)

\(f^{\prime}(x)=1 \ \text{for} \ \ x \in\left(\dfrac{\pi}{2}, \pi\right)\)

Show Worked Solution
\(f(x)\) \(=\cos ^{-1}(\sin x)\)
\(f^{\prime}(x)\) \(=-\dfrac{\cos x}{\sqrt{1-\sin ^2 x}}\)
  \(=-\dfrac{\cos x}{\abs{\cos x}}\)
  \(= \pm 1\)

♦♦ Mean mark 38%.

\(\text{Consider limitations:}\)

\(1-\sin ^2 x \neq 0 \ \Rightarrow \ \sin x \neq \pm 1 \ \Rightarrow \ x \neq \dfrac{\pi}{2}\)

\(\text{In lst quadrant:} \ \ -\dfrac{\cos x}{\abs{\cos x}}=-1\)

\(\text{In 2nd quadrant:}\ \  -\dfrac{\cos x}{\abs{\cos x}}=1\)

\(f^{\prime}(x)=-1 \ \ \text{for} \ \ x \in\left(0, \dfrac{\pi}{2}\right)\)

\(f^{\prime}(x)=1 \ \ \text{for} \ \ x \in\left(\dfrac{\pi}{2}, \pi\right)\)

Calculus, EXT1 C2 EQ-Bank 18

Find the value of \(n\), given

\(\displaystyle \int_0^n-\dfrac{1}{\sqrt{1-x^2}}=-\dfrac{\pi}{6}\).   (2 marks)

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\(n=\dfrac{1}{2}\)

Show Worked Solution
\(\displaystyle\int_0^n-\dfrac{1}{\sqrt{1-x^2}}\) \(=-\dfrac{\pi}{6}\)
\(\left[\cos ^{-1} x\right]_0^n\) \(=-\dfrac{\pi}{6}\)
\(\cos ^{-1}(n)-\cos ^{-1} 0\) \(=-\dfrac{\pi}{6}\)
\(\cos ^{-1}(n)-\dfrac{\pi}{2}\) \(=-\dfrac{\pi}{6}\)
\(\cos ^{-1}(n)\) \(=\dfrac{\pi}{2}-\dfrac{\pi}{6}\)
\(n\) \(=\cos \left(\dfrac{\pi}{3}\right)\)
  \(=\dfrac{1}{2}\)

Calculus, EXT1 C2 EQ-Bank 26

  1. Find  \(\displaystyle \int \frac{1}{\sqrt{4 x-x^2}}\, d x\).   (2 marks)

  2. Determine the values of \(x\) for which the antiderivative  \(\displaystyle \int \dfrac{1}{\sqrt{4 x-x^2}}\, d x\)  is real and finite.   (1 mark)

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a.   \(\sin ^{-1}\left(\dfrac{x-2}{2}\right)+c\)

b.   \(0<x<4\)

Show Worked Solution
a.     \(\displaystyle\int \frac{1}{\sqrt{4 x-x^2}}\, d x\) \(=\displaystyle\int \frac{1}{\sqrt{4-4+4 x-x^2}}\, d x\)
    \(=\displaystyle\int \frac{1}{\sqrt{4-(x-2)^2}}\, d x\)
    \(=\sin ^{-1}\left(\dfrac{x-2}{2}\right)+c\)

 

b.    \(4 x-x^2>0\)

\(x(4-x)>0\)

\(0<x<4\)

Calculus, EXT1 C2 EQ-Bank 17

Find  \(\displaystyle \int \frac{1}{\sqrt{15-2 x-x^2}}\, d x\).   (3 marks)

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\(\displaystyle\sin ^{-1}\left(\frac{x+1}{4}\right)+c\)

Show Worked Solution
\(\displaystyle\int \frac{1}{\sqrt{15-2 x-x^2}}\, d x\) \(=\displaystyle\int \frac{1}{\sqrt{16-1-2 x-x^2}}\, d x\)
  \(=\displaystyle\int \frac{1}{\sqrt{16-(x+1)^2}}\, d x\)
  \(=\displaystyle\sin ^{-1}\left(\frac{x+1}{4}\right)+c\)

Calculus, EXT1 C2 EQ-Bank 6 MC

\(\displaystyle \int \dfrac{1}{\sqrt{5-4 x-x^2}}\, d x=\)
 

  1. \(\sin ^{-1}\left(\dfrac{x+2}{3}\right)+c \)
  2. \( \sin ^{-1}\left(\dfrac{x-2}{3}\right)+c \)
  3. \(\sin ^{-1}\left(\dfrac{x+2}{9}\right)+c \)
  4. \(\sin ^{-1}\left(\dfrac{x-2}{9}\right)+c\)
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\(\Rightarrow A\)

Show Worked Solution
\(\displaystyle \int \frac{1}{\sqrt{5-4 x-x^2}}\, d x\) \(=\displaystyle\int \frac{1}{\sqrt{9-4-4 x-x^2}}\,dx\)
  \(=\displaystyle \int \frac{1}{\sqrt{9-(x+2)^2}}\,dx\)
  \(=\sin ^{-1}\left(\dfrac{x+2}{3}\right)+c\)

\(\Rightarrow A\)

Calculus, EXT1 C2 EQ-Bank 30

Given  \(y=\dfrac{\cos 2 x}{x^2}\),  find the gradient of the tangent of its inverse function at  \(\left(\dfrac{8 \sqrt{2}}{\pi^2}, \dfrac{\pi}{4}\right)\).   (3 marks)

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\(-\dfrac{\pi^2}{32}\)

Show Worked Solution

\(y=\dfrac{\cos2x}{x^2}\)

\(\text{Inverse: Swap} \ \ x \leftrightarrow y\)

\(x=\dfrac{\cos 2 y}{y^2}\)

\(u=\cos 2 y \ \ \quad \quad \quad v=y^2\)

\(v^{\prime}=-2 \sin 2 y \ \ \quad v^{\prime}=2 y\)

\(\dfrac{dx}{dy}=\dfrac{-2 y^2 \sin 2 y-2 y\, \cos 2 y}{y^4}\)

\(\dfrac{dy}{dx}=\dfrac{y^4}{-2 y^2 \sin 2 y-2 y\, \cos 2 y}\)
 

\(\text{At}\ \ y=\dfrac{\pi}{4}:\)

\(\dfrac{d y}{d x}\) \(=\dfrac{\left(\dfrac{\pi}{4}\right)^4}{-2\left(\dfrac{\pi}{4}\right)^2 \sin \left(\dfrac{\pi}{2}\right)-2\left(\dfrac{\pi}{4}\right) \cos \left(\dfrac{\pi}{2}\right)}\)
  \(=-\dfrac{\pi^4}{256} \times \dfrac{8}{\pi^2}\)
  \(=-\dfrac{\pi^2}{32}\)

Calculus, EXT1 C2 2024 HSC 14b

For what values of the constant \(k\) would the function  \(f(x)=\dfrac{k x}{1+x^2}+\arctan x\)  have an inverse?   (3 marks)

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\(f(x)\ \text{has an inverse for}\ \ -1 \leqslant k \leqslant 1\)

Show Worked Solution

  \(f(x)\) \(=\dfrac{k x}{1+x^2}+\arctan x\)
  \(f^{\prime}(x)\) \(=\dfrac{k\left(1+x^2\right)-k x(2 x)}{\left(1+x^2\right)^2}+\dfrac{1}{1+x^2}\)
    \(=\dfrac{k+k x^2-2 k x^2+1+x^2}{\left(1+x^2\right)^2}\)
    \(=\dfrac{x^2(1-k)+k+1}{\left(1+x^2\right)^2}\)
♦♦♦ Mean mark 26%.

\(\text{Inverse function } \Rightarrow f(x) \text { has no SPs}\)

\(x^2(1-k)+k+1 \neq 0\)

\(\text {No solution if } \ \Delta<0:\)

  \(-4(1-k)(k+1)\) \(<0\)
  \((1-k)(k+1)\) \(>0\)

 
\(f(x)\ \text{has an inverse for}\ \ -1 \leq k \leq 1\)

Calculus, EXT1 C2 2024 HSC 11e

Differentiate the function  \(f(x)=\arcsin \left(x^5\right)\).   (1 mark)

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\(f^{\prime}(x)=\dfrac{5 x^4}{\sqrt{1-x^{10}}}\)

Show Worked Solution

\(f(x)=\sin ^{-1}\left(x^5\right)\)

\(f^{\prime}(x)=5 x^4 \times \dfrac{1}{\sqrt{1-\left(x^5\right)^2}}=\dfrac{5 x^4}{\sqrt{1-x^{10}}}\)

Calculus, EXT1 C2 2023 HSC 14a

Let  \(f(x)=2 x+\ln x\), for \(x>0\).

  1. Explain why the inverse of \(f(x)\) is a function.  (1 mark)

  2. Let  \(g(x)=f^{-1}(x)\). By considering the value of \(f(1)\), or otherwise, evaluate \(g^{\prime}(2)\).  (2 mark)

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i.     \(f(x)=2 x+\ln x\)

\(f^{′}(x)=2+\dfrac{1}{x} \)

\(\text{In domain}\ x \gt 0\ \ \Rightarrow f^{-1}(x) \gt 0 \ \ (f(x)\ \text{is monotonically increasing}) \)

\(\text{Since}\ f(x)\ \text{is one-to-one,}\ f^{-1}(x)\ \text{is a function.} \)
 

ii.    \(\dfrac{1}{3}\)

Show Worked Solution

i.     \(f(x)=2 x+\ln x\)

\(f^{′}(x)=2+\dfrac{1}{x} \)

\(\text{In domain}\ x \gt 0\ \ \Rightarrow f^{-1}(x) \gt 0 \ \ (f(x)\ \text{is monotonically increasing}) \)

\(\text{Since}\ f(x)\ \text{is one-to-one,}\ f^{-1}(x)\ \text{is a function.} \)

Mean mark (i) 54%.

 
ii.    \(g(x)=f^{-1}(x) \)

\(f(g(x))=x\)

\(\text{Differentiate both sides:}\)

\(g^{′}(x)\ f^{′}(g(x))\) \(=1\)  
\(g^{′}(x)\) \(=\dfrac{1}{f^{′}(g(x))}\)  
\(g^{′}(2)\) \(=\dfrac{1}{f^{′}(g(2))}\)  

 
\(f(1)=2 \times 1 + \ln1 = 2 \)

\(\Rightarrow g(2)=1 \ \text{(by inverse definition)}\)

\(\therefore g^{′}(2)\) \(= \dfrac{1}{f^{′}(1)} \)  
  \(=\dfrac{1}{2+\frac{1}{1}}\)  
  \(=\dfrac{1}{3} \)  
♦♦ Mean mark (ii) 32%.

Calculus, EXT1 C2 2023 HSC 11d

Find  \( {\displaystyle \int} \dfrac{1}{\sqrt{4-9x^2}}\ dx\)  (2 marks)

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\(\dfrac{1}{3} \sin^{-1} \Big{(}\dfrac{3x}{2} \Big{)} +c \)

Show Worked Solution
\({\displaystyle \int} \dfrac{1}{\sqrt{4-9x^2}}\ dx\) \(=\dfrac{1}{3} {\displaystyle \int} \dfrac{3}{\sqrt{2^2-(3x)^2}}\ dx\)  
  \(=\dfrac{1}{3} \sin^{-1} \Big{(}\dfrac{3x}{2} \Big{)} +c \)  

Calculus, EXT1 C2 2022 HSC 12c

Find the equation of the tangent to the curve  `y=x  text{arctan}(x)`  at the point with coordinates `(1,(pi)/(4))`. Give your answer in the form  `y=mx+c`  (3 marks)

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`y=((2+pi)/4)x-1/2`

Show Worked Solution
`y` `=xtan^(-1)(x)`  
`dy/dx` `=x xx 1/(1+x^2)+tan^(-1)(x)`  

 
`text{When}\ \ x=1:`

`dy/dx=1/2+tan^(-1)(1)=1/2+pi/4=(2+pi)/4`
 

`text{Equation of tangent}\ \ m=(2+pi)/4,\ text{through}\ \ (1,(pi)/(4)):`

`y-pi/4` `=(2+pi)/4 (x-1)`  
`y` `=((2+pi)/4)x-(2+pi)/4+pi/4`  
`y` `=((2+pi)/4)x-1/2`  

Calculus, EXT1 C2 2022 HSC 9 MC

A given function  `f(x)`  has an inverse  `f^{-1}(x)`.

The derivatives of  `f(x)`  and  `f^{-1}(x)`  exist for all real numbers `x`.

The graphs  `y=f(x)`  and  `y=f^{-1}(x)`  have at least one point of intersection.

Which statement is true for all points of intersection of these graphs?

  1. All points of intersection lie on the line  `y=x`.
  2. None of the points of intersection lie on the line  `y=x`.
  3. At no point of intersection are the tangents to the graphs parallel.
  4. At no point of intersection are the tangents to the graphs perpendicular.
Show Answers Only

`D`

Show Worked Solution

`text{By Elimination:}`

`text{Consider}\ \ f(x)=x\ \ =>\ \ f^(-1)(x)=x:`

`text{All POI lie on}\ \ y=x\ \ text{and all tangents are parallel}`

`text{→ Eliminate B and C}`
 

`text{Consider}\ \ f(x)=-x\ \ =>\ \ f^(-1)(x)=-x:`

`text{All POI lie on}\ \ y=-x`

`text{→ Eliminate A}`

`=>D`


♦♦♦ Mean mark 13%.

Calculus, EXT1 C2 2021 HSC 14e

The polynomial  \(g(x)=x^3+4 x-2\)  passes through the point \((1,3)\).

Find the gradient of the tangent to  \(f(x)=x g^{-1}(x)\)  at the point where  \(x=3\).   (3 marks)

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\(\text{Gradient of tangent}=\dfrac{10}{7}\)

Show Worked Solution

\(g(1)=3 \ \ \Rightarrow\ \ g^{-1}(3)=1\)

\(f(x)=x g^{-1}(x)\)

\(\text {Using product rule:}\)

\(f^{\prime}(x)=g^{-1}(x)+x \cdot \dfrac{d}{d x}\left(g^{-1}(x)\right)\)

♦♦♦ Mean mark 10%.

\(\text{Find}\ \ \dfrac{d}{d x}\left(g^{-1}(x)\right):\)

\(\text{Let} \ \ y=g^{-1}(x) \ \ \Rightarrow\ \ x=g(y)\)

\(\dfrac{d x}{d y}=g^{\prime}(y)\ \ \Rightarrow\ \ \dfrac{d y}{d x}=\dfrac{1}{g^{\prime}(y)}\)

\(\dfrac{d}{d x}\left(g^{-1}(x)\right)=\dfrac{1}{g^{\prime}\left(g^{-1}(x)\right)}\)
 

\(f^{\prime}(x)=g^{-1}(x)+\dfrac{x}{g^{\prime}\left(g^{-1}(x)\right)}\)

\(f^{\prime}(3)=g^{-1}(3)+\dfrac{3}{g^{\prime}\left(g^{-1}(3)\right)}=1+\dfrac{3}{g^{\prime}(1)}\)
 

\(g(x)=x^3+4 x-2\ \ \Rightarrow\ \ g^{\prime}(x)=3 x^2+4\)

\(f^{\prime}(3)=1+\dfrac{3}{3(1)^2+4} = \dfrac{10}{7}\)

\(\therefore \ \text{Gradient of tangent}=\dfrac{10}{7}\)

Calculus, EXT1 C2 2020 SPEC1 6

Let  `f(x) = tan^(-1) (3x - 6) + pi`.

  1. Show that  `f^{prime}(x) = 3/(9x^2 - 36x + 37)`.  (1 mark)

  2. Hence, show that the graph of  `f`  has a point of inflection at  `x = 2`.  (2 marks)

  3. Sketch the graph of  `y = f(x)`  on the axes provided below. Label any asymptotes with their equations and the point of inflection with its coordinates.   (2 marks)

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `text(See Worked Solutions)`
Show Worked Solution
a.    `f^{\prime}(x)` `= (d/(dx) (3x – 6))/(1 + (3x – 6)^2)`
    `= 3/(9x^2 – 36x + 37)`

 

b.   `f^{\prime\prime}(x) = (3(18x – 36))/(9x^2 – 36x + 37)^2`

`f^{\prime\prime}(x) = 0\ \ text(when)\ \ 18x – 36 = 0 \ => \ x = 2`

`text(If)\ \ x < 2, 18x – 36 < 0 \ => \ f^{\prime\prime}(x) < 0`

`text(If)\ \ x > 2, 18x – 36 > 0 \ => \ f^{\prime\prime}(x) > 0`

`text(S) text(ince)\ \ f^{\prime\prime}(x)\ \ text(changes sign about)\ \ x = 2,`

`text(a POI exists at)\ \ x = 2`

 

c.   

Calculus, EXT1 C2 2020 HSC 13c

Suppose  `f(x) = tan(cos^(-1)(x))`  and  `g(x) = (sqrt(1-x^2))/x`.

The graph of  `y = g(x)`  is given.
 

  1. Show that  `f^(′)(x) = g^(′)(x)`.   (4 marks)

  2. Using part (i), or otherwise, show that  `f(x) = g(x)`.   (3 marks) 

Show Answers Only

i.    `text(See Worked Solutions)`

ii.   `text(See Worked Solutions)`

Show Worked Solution

i.   `f(x) = tan(cos^(-1)(x))`

♦ Mean mark (i) 50%.
`f^(′)(x)` `= -1/sqrt(1-x^2) · sec^2(cos^(-1)(x))`
  `= -1/sqrt(1-x^2) · 1/(cos^2(cos^(-1)(x)))`
  `= -1/(x^2sqrt(1-x^2))`

 
`g(x) = (1-x^2)^(1/2) · x^(-1)`

`g^(′)(x)` `= 1/2 · -2x(1-x^2)^(-1/2) · x^(-1)-(1-x^2)^(1/2) · x^(-2)`
  `= (-x)/(x sqrt(1-x^2))-sqrt(1-x^2)/(x^2)`
  `= (-x^2-sqrt(1-x^2) sqrt(1-x^2))/(x^2 sqrt(1-x^2))`
  `= (-x^2-(1-x^2))/(x^2sqrt(1-x^2))`
  `= -1/(x^2sqrt(1-x^2))`
  `=f^(′)(x)`

 

ii.   `f^(′)(x) = g^(′)(x)`

♦♦♦ Mean mark (ii) 15%.

`=> f(x) = g(x) + c`
 

`text(Find)\ c:`

`f(1)= tan(cos^(-1) 1)= tan 0=0`

`g(1) = sqrt(1-1)/0 = 0`

`f(1) = g(1) + c\ \ =>\ \ c = 0`

`:. f(x) = g(x)`

Calculus, EXT1 C2 2019 HSC 3 MC

What is the derivative of  `tan^(-1)\ x/2`?

A.     `1/(2(4 + x^2))`

B.     `1/(4 + x^2)`

C.     `2/(4 + x^2)`

D.     `4/(4 + x^2)`

Show Answers Only

`C`

Show Worked Solution
`y` `= tan^(-1)\ x/2`
`(dy)/(dx)` `= (1/2)/(1 + (x/2)^2)`
  `= 1/(2(1 + x^2/4))`
  `= 2/(4 + x^2)`

 
`=>  C`

Calculus, EXT1 C2 2018 HSC 12c

Let  `f(x) = sin^(-1) x + cos^(-1) x`.

  1. Show that  `f^{′}(x) = 0`  (1 mark)

  2. Hence, or otherwise, prove
     
    `qquad sin^(-1) x + cos^(-1) x = pi/2`.  (1 mark)

  3. Hence, sketch
     
    `qquad f(x) = sin^(-1) x + cos^(-1) x`.  (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(See Worked Solutions)`
Show Worked Solution

i.    `f^{′}(x) = 1/sqrt(1-x^2) + (-1/sqrt (1-x^2)) = 0`

 

ii.  `text(S)text(ince)\ \ f^{′}(x) = 0\ \ => f(x)\ \ text(is a constant.)`

♦ Mean mark (ii) 37%.

`text(Substituting)\ \ x=1\ \ text{into the equation  (any value works)}`

  `sin^(-1) 1 + cos^(-1) 1` `= pi/2 + 0`
    `= pi/2\ \ text(… as required)`

 

iii.  `text(Domain restrictions require:)\ \ -1<x<1`

♦ Mean mark (iii) 40%.

 

Calculus, EXT1 C2 2007 HSC 1c

Differentiate  `tan^(–1)(x^4)`  with respect to  `x`.  (2 marks)

Show Answers Only

`(4x^3)/(1 + x^8)`

Show Worked Solution
`y` `= tan^(−1)(x^4)`
`(dy)/(dx)` `= 1/(1 + (x^4)^2) xx d/(dx) (x^4)`
  `= (4x^3)/(1 + x^8)`

Calculus, EXT1 C2 2006 HSC 2a

Let  `f(x) = sin^-1 (x + 5).`

  1. State the domain and range of the function  `f(x).`  (2 marks)

  2. Find the gradient of the graph of  `y = f(x)`  at the point where  `x = -5.`  (2 marks)

  3. Sketch the graph of  `y = f(x).`  (2 marks)

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a.    `text(Domain):\ -6 <= x <= -4, \ \ \ text(Range):\ -pi/2 <= y <= pi/2`

b.    `1`

c.     
   

Show Worked Solution

a.    `f(x) = sin^-1 (x + 5)`

`text(Domain)`

`-1 <= x + 5 <= 1`

`-6 <= x <= -4`

`text(Range)`

`-pi/2 <= y <= pi/2`

 

b.    `y = sin^-1 (x + 5)`

`(dy)/(dx) = 1/sqrt(1-(x + 5)^2)`
 

`text(When)\ \ x = -5`

`(dy)/(dx)` `= 1/sqrt(1-(-5 + 5)^2)`
  `= 1/sqrt(1-0)`
  `= 1`

 
`:.\ text(Gradient of)\ \ y = f(x)\ \ text(at)\ \ x = -5\ \ text(is)\ \ 1.`

 

c.    

 EXT1 2006 2a

Calculus, EXT1 C2 2005 HSC 2a

Find  `d/(dx) (2 sin^-1 5x).`  (2 marks)

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`10/(sqrt (1 – 25x^2))`

Show Worked Solution
`d/(dx) (2 sin^-1 5x)` `= 2 xx d/(dx) (sin^-1 5x)`
  `= 2 xx 1/(sqrt(1 – (5x)^2)) xx d/(dx) (5x)`
  `= 2/(sqrt (1 – 25x^2)) xx 5`
  `= 10/(sqrt(1 – 25 x^2))`

Calculus, EXT1 C2 2015 HSC 13d

Let  `f(x) = cos^(-1)\ (x) + cos^(-1)\ (-x)`, where  `-1 ≤ x ≤ 1`.

  1. By considering the derivative of  `f(x)`, prove that  `f(x)`  is constant.  (2 marks)

  2. Hence deduce that  `cos^(-1)\ (-x) = pi - cos^(-1)\ (x)`.  (1 mark)

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a.    `text(See Worked Solutions.)`

b.     `text(See Worked Solutions.)`

Show Worked Solution

a.    `text(Prove)\ f(x)\ text(is a constant)`

`f(x)` `= cos^(-1)(x) + cos^(-1)(-x), \ \ -1 ≤ x ≤ 1`
`f′(x)` `= (-1)/sqrt(1 – x^2) + (-1)/sqrt(1 -(-x)^2) xx d/(dx) (-x)`
  `= (-1)/sqrt(1 – x^2) + 1/sqrt(1 – x^2)`
  `= 0`

 
`:.\ text(S)text(ince)\ \ f′(x) = 0,  f(x)\ text(must be a constant.)`
 

♦ Mean mark 35%.
b.     `f(0)` `= cos^(−1)(0) + cos^(−1)(0)`
    `= pi/2 + pi/2`
    `= pi`

 
`:.f(x) = pi`

`pi = cos^(−1)(x) + cos^(−1)(−x)`

`:.cos^(−1)(−x) = pi – cos^(−1)(x)\ \ …text(as required)`

Calculus, EXT1 C2 2015 HSC 7 MC

What is the value of `k` such that `int_0^k 1/sqrt(4 − x^2) \ dx= pi/3 ?`

  1. `1`
  2. `sqrt3`
  3. `2`
  4. `2sqrt3`
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`B`

Show Worked Solution
`int_0^1 1/sqrt(4 − x^2)dx` `= pi/3`
`[sin^(−1)\ x/2]_0^k` `= pi/3`
`sin^(−1)\ k/2 − sin^(−1)\ 0` `= pi/3`
`sin^(−1)\ k/2` `= pi/3`
`k/2` `= sin\ pi/3`
  `= sqrt3/2`
`:.k` `= sqrt3`

`⇒ B`

Calculus, EXT1 C2 2014 HSC 6 MC

What is the derivative of  `3 sin^(-1)\ x/2`?

  1. `6/sqrt(4 - x^2)`
  2. `3/sqrt(4 - x^2)`
  3. `3/(2sqrt(4 - x^2))` 
  4. `3/(4sqrt(4 - x^2))`
Show Answers Only

`B`

Show Worked Solution

`y = 3 sin^(-1)\ x/2`

`dy/dx = 3 xx 1/sqrt(4 – x^2)`

`=>  B`

Calculus, EXT1 C2 2013 HSC 11g

Differentiate  `x^2 sin^(–1) 5x`.   (2 marks)

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`(5x^2)/sqrt(1\ – 25x^2) + 2x sin^(-1) 5x`

Show Worked Solution
`y` `= x^2 sin^(-1) 5x`
`text(Using the product rule)`
`(dy)/(dx)` `= x^2 xx 1/sqrt(1\ – (5x)^2) xx d/(dx)(5x) + 2x sin^(-1) 5x`
  `= (5x^2)/sqrt(1\ – 25x^2) + 2x sin^(-1) 5x`

Calculus, EXT1 C2 2013 HSC 11b

Find  `int 1/sqrt (49 - 4x^2)\ dx`.   (2 marks)

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`1/2 sin^(-1) ((2x)/7) + c`

Show Worked Solution

`int 1/sqrt(49 – 4x^2)\ dx`

`= int 1/(2 sqrt(49/4 – x^2))\ dx`

`= 1/2 int 1/sqrt((7/2)^2 – x^2)\ dx`

`= 1/2 sin^(-1) ((2x)/7) + c`

Calculus, EXT1 C2 2010 HSC 5b

Let  `f(x) = tan^(-1)(x) + tan^(-1)(1/x)`  for  `x != 0`. 

  1. By differentiating  `f(x)`, or otherwise, show that  `f(x) = pi/2`  for  `x > 0`.   (3 marks)

  2. Given that  `f(x)`  is an odd function, sketch the graph  `y = f(x)`.     (1 mark)

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a.    `text(Proof)\ \ text{(See Worked Solutions)}`

Inverse Functions, EXT1 2010 HSC 5b Answer

b.    

 

 

 

 

 

Show Worked Solution
♦ Mean mark 48%
MARKER’S COMMENT: The most common errors were not being able to differentiate `tan^(-1) (1/x)` and not recognising the significance of `f′(x)=0`.
a.     `f(x)` `= tan^(-1) (x) + tan^(-1) (1/x)\ text(for)\ x != 0`
  `f prime (x)` `= 1/(1 + x^2) + 1/(1 + (1/x)^2) xx d/(dx) (1/x)`
    `= 1/(1 + x^2) + 1/(1 + 1/(x^2)) xx -1/(x^2)`
    `= 1/(1 + x^2)\ – 1/(x^2 + 1)`
    `= 0`

 

`text(S)text(ince)\ \ f prime (x) = 0`

`=> f(x)\ text(is a constant)`

 

`text(Substitute)\ \ x = 1\ \ text(into)\ \ f(x)` 

`f(1)` `= tan^(-1) 1 + tan^(-1) (1/1)`
  `= pi/4 + pi/4`
  `= pi/2`

 

`:.\ f(x) = pi/2\ \ text(for)\ \ x > 0\ \ \ …\ text(as required)`

 

♦ Mean mark 35%
b.     `text(Given)\ \ f(x)\ \ text(is odd)`
  `f(–x) = -f(x)`

Inverse Functions, EXT1 2010 HSC 5b Answer

Calculus, EXT1 C2 2012 HSC 9 MC

What is the derivative of  `cos^(–1) (3x)`?

  1. `1/(3 sqrt(1 - 9x^2))`  
  2. `(-1)/(3 sqrt(1 - 9x^2))`  
  3. `3/sqrt(1 - 9x^2)`  
  4. `(-3)/sqrt(1 - 9x^2)`  
Show Answers Only

`D`

Show Worked Solution

`y = cos^(-1) (3x)`

`dy/dx` `= (-1)/sqrt(1\ – (3x)^2) xx d/dx (3x)`
  `= (-3)/sqrt(1\ – 9x^2)`

`=>  D`