Inverse Functions Calculus

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 \(g^{\prime \prime}(1)\) ?

\(\dfrac{1}{4}\)
\(-\dfrac{1}{4}\)
\(-\dfrac{1}{2}\)
\(-1\)

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\(C\)

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

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

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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

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

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

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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 \(x>0\).

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

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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?

All points of intersection lie on the line `y=x`.
None of the points of intersection lie on the line `y=x`.
At no point of intersection are the tangents to the graphs parallel.
At no point of intersection are the tangents to the graphs perpendicular.

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`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 + 4x - 2` passes through the point (1, 3).

Find the gradient of the tangent to `f(x) = xg^(-1)(x)` at the point where `x = 3`. (2 marks)

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`10/7`

Show Worked Solution

`g(x) = x^3 + 4x – 2`

♦♦♦ Mean mark 10%.

`g′(x) = 3x^2 + 4`

`f(x) = x g^(-1)(x)`

`f′(x) = x · d/(dx) g^(-1)(x) + g^(-1)(x)`

COMMENT: The reciprocal relationship of gradients between `g(x)` and `g^(-1)(x)` is critical here.

`g(x)\ text(passes through)\ (1, 3)`

`=> g^(-1)(x)\ text(passes through)\ (3, 1)`

`g′(1) = 3 + 4 = 7`

`=> d/(dx) g^(-1)(3) = 1/(d/(dy) g(y)) = 1/(g′(1)) = 1/7`

`:. f′(x)\|_(x = 3)`	`= 3 · 1/7 + 1`
	`= 10/7`

Calculus, EXT1 C2 2021 HSC 11f

Evaluate `int_0^sqrt3 1/(sqrt(4 - x^2))\ dx`. (2 marks)

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`pi/3`

Show Worked Solution

	`int_0^sqrt3 1/(sqrt(4 – x^2))\ dx`	`= [sin^(-1)\ x/2]_0^sqrt3`
		`= sin^(-1)\ sqrt3/2 – sin^(-1) 0`
		`= pi/3`

Calculus, EXT1 C2 2020 SPEC1 6

Let `f(x) = tan^(-1) (3x - 6) + pi`.

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

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

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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)

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

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.

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

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

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`text(See Worked Solutions)`
`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`.

Show that `f^{′}(x) = 0` (1 mark)

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Hence, or otherwise, prove

`qquad sin^(-1) x + cos^(-1) x = pi/2`. (1 mark)

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Hence, sketch

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

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`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`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 2016 HSC 11c

Differentiate `3tan^(−1)(2x)`. (2 marks)

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`6/(1 + 4x^2)`

Show Worked Solution

`y`	`= 3 tan^-1 (2x)`
`(dy)/(dx)`	`= 3/(1 + (2x)^2) xx 2`
	`= 6/(1 + 4x^2)`

Calculus, EXT1 C2 2007 HSC 1c

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

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`(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 2004 HSC 2b

Find `d/(dx)\ cos^(−1)\ (3x^2).` (2 marks)

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`(−6x)/(sqrt(1 − 9x^4))`

Show Worked Solution

`d/(dx)\ cos^(−1)\ (3x^2)`

`= (−1)/sqrt(1 − (3x^2)^2) xx d/(dx) (3x^2)`

`= (−6x)/(sqrt(1 − 9x^4))`

Calculus, EXT1 C2 2004 HSC 1d

Find `int_0^1(dx)/(sqrt(4 − x^2))`. (2 marks)

Show Answers Only

`pi/6`

Show Worked Solution

`int_0^1(dx)/(sqrt(4 − x^2))`	`= [sin^(−1)\ x/2]_0^1`
	`= sin^(−1)(1/2) − sin^(−1)(0)`
	`= pi/6`

Calculus, EXT1 C2 2006 HSC 2a

Let `f(x) = sin^-1 (x + 5).`

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

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

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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`

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.`

Calculus, EXT1 C2 2006 HSC 1a

Find `int (dx)/(49 + x^2).` (2 marks)

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`1/7 tan^-1\ x/7 + c`

Show Worked Solution

`int (dx)/(49 + x^2)`

`= 1/7 tan^-1\ x/7 + c`

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 2005 HSC 1a

Find `int 1/(x^2 + 49)\ dx.` (1 mark)

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`1/7\ tan^-1\ x/7 + c`

Show Worked Solution

`int 1/(x^2 + 49)\ dx`	`= 1/7 int 7/(x^2 + 7^2)\ dx`
	`= 1/7\ tan^-1\ x/7 + c`

Calculus, EXT1 C2 2015 HSC 13d

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

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

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

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`text(See Worked Solutions.)`
`text(See Worked Solutions.)`

Show Worked Solution

i. `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%.

ii.	`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`
`sqrt3`
`2`
`2sqrt3`

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`B`

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`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 2008 HSC 1c

Evaluate `int_-1^1 1/sqrt(4 - x^2)\ dx`. (2 marks)

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`pi/3`

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`int_-1^1 1/sqrt(4 – x^2)\ dx`

`= [sin^(-1) (x/2)]_(-1)^1`

`= sin^(-1) (1/2) – sin^(-1) (-1/2)`

`= pi/6 – (- pi/6)`

`= pi/3`

Calculus, EXT1 C2 2008 HSC 1b

Differentiate `cos^(–1) (3x)` with respect to `x`. (2 marks)

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`(-3)/sqrt(1 – 9x^2)`

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`y`	`= cos^(-1) (3x)`
`dy/dx`	`= – 1/sqrt(1 – (3x)^2) * d/(dx) (3x)`
	`= (-3)/sqrt(1 – 9x^2)`

Calculus, EXT1 C2 2014 HSC 6 MC

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

`6/sqrt(4 - x^2)`
`3/sqrt(4 - x^2)`
`3/(2sqrt(4 - x^2))`
`3/(4sqrt(4 - x^2))`

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`B`

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`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`

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`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`

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`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`.

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

--- 6 WORK AREA LINES (style=lined) ---
Given that `f(x)` is an odd function, sketch the graph `y = f(x)`. (1 mark)

--- 6 WORK AREA LINES (style=lined) ---

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

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♦ 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`.

i.	`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%

ii.	`text(Given)\ \ f(x)\ \ text(is odd)`
	`f(–x) = -f(x)`

Calculus, EXT1 C2 2010 HSC 1a

Use the table of standard integrals to find `int 1/sqrt(4 - x^2)\ dx`. (1 mark)

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

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`int 1/sqrt(4\ – x^2)\ dx`

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

Calculus, EXT1 C2 2012 HSC 11a

Evaluate `int_0^3 1/(9 + x^2)\ dx`. (3 marks)

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`pi/12`

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`int_0^3 1/(9 + x^2)\ dx`	`= int_0^3 1/(3^2 + x^2)\ dx`
	`= [1/3 tan^(-1) (x/3)]_0^3`
	`= [(1/3 tan^(-1) 1) – 0]`
	`= 1/3 xx pi/4`
	`= pi/12`

Calculus, EXT1 C2 2012 HSC 9 MC

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

`1/(3 sqrt(1 - 9x^2))`
`(-1)/(3 sqrt(1 - 9x^2))`
`3/sqrt(1 - 9x^2)`
`(-3)/sqrt(1 - 9x^2)`

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`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`