Past paper practice · Mathematics Extension 2

2025 HSC
worked solutions

A clear method for every question. Find a topic, try it on paper, then open the working when you’re ready.

16 questions47 parts100 marks

Use these independently written solutions alongside the official NESA questions. This is an unofficial study resource, not endorsed by NESA.

Section I Multiple choice

Question 11 mark

A line through two points

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VectorsCoordinate geometry

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The direction from $(-3,1)$ to $(1,4)$ is $(4,3)$ . Using $(-3,1)$ as a point gives the line equation in option D.

Answer

\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}-3\\1\end{pmatrix}+\lambda\begin{pmatrix}4\\3\end{pmatrix}

Question 21 mark

Negating an existential statement

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ProofLogic

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Negating “there exists an integer whose square is odd” gives “for every integer, its square is not odd”. An integer square is either odd or even.

Answer

\forall x\in\mathbb Z,\ x^2\text{ is even}.

Question 31 mark

Square roots of a complex number

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Complex numbers

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Squaring $2-i$ gives $4-4i+i^2=3-4i$ . Its negative is the other square root.

Answer

2-i

and

-2+i

Question 41 mark

Amplitude from the speed equation

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MechanicsSimple harmonic motion

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The speed is zero at the turning points: $-x^2+2x+8=0$ , so $x=-2$ or $x=4$ . The midpoint is $1$ , and the distance from it to either turning point is $3$ .

Answer

3\text{ m}.

Question 51 mark

Writing a contrapositive

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ProofLogic

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For $P\Rightarrow Q$ , the contrapositive is $\neg Q\Rightarrow\neg P$ . Here $\neg Q$ is $x>0$ , and $\neg P$ is $x^2-2x<0$ .

Answer

A: If

x>0

, then

x^2-2x<0

Question 61 mark

Moduli of sums and differences

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Complex numbers

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For unit complex numbers, $|z+w|^2+|z-w|^2=2|z|^2+2|w|^2=4$ . Therefore $|z-w|^2=4-\frac94=\frac74$ .

Answer

|z-w|=\frac{\sqrt7}{2}.

Question 71 mark

Arguments subtended by a circle

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Complex numbersGeometry

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The points $z-2i$ form the circle of radius $1$ centred at $-2i$ . From the origin, its tangent rays make angles $\arcsin(1/2)=\pi/6$ on either side of the direction $-\pi/2$ .

Answer

-\frac{2\pi}{3}\leq\operatorname{Arg}(z-2i)\leq-\frac{\pi}{3}.

Question 81 mark

Acceleration as a function of displacement

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MechanicsCalculus

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Since $a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$ , acceleration is zero where the velocity is zero or has a stationary point. The displayed velocity curve has three zeros and two turning points, with acceleration negative at the far left and positive at the far right; graph A has this pattern.

Answer

Question 91 mark

Classifying a complex quadrilateral

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Complex numbersGeometry

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The relation $v+z=u+w$ says the diagonals have the same midpoint, so $UVWZ$ is a parallelogram. Also $u-w=ki(v-z)$ , so its diagonals are perpendicular. A parallelogram with perpendicular diagonals is a rhombus.

Answer

C: Rhombus.

Question 101 mark

Reparametrising a space curve

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VectorsParametric curves

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Set the parameter in option D to $s=2t+\pi/2$ . It ranges over all real values, and the coordinates become $(\cos s,s,-\sin s)$ , the same helix as $(\cos t,-t,\sin t)$ after taking $s=-t$ .

Answer

Section II Written responses

Question 11(a)2 marks

Conjugate and rotated complex numbers

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Complex numbers

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Write $z=a+ib$ , with $a,b>0$ as shown. Then $\bar z=a-ib$ , the reflection of $z$ in the real axis, while $i\bar z=b+ia$ , at coordinates $(b,a)$ . Mark both points on the diagram.

Solution sketch for Question 11(a); key values and curve behaviour are described in the steps above. — Original solution sketch; see the steps for key values.

Answer

\bar z

is the reflection of

z

across the real axis;

i\bar z

is the point

(b,a)

, in the first quadrant.

Question 11(b)2 marks

Multiplying complex numbers in polar form

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Complex numbers

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Multiply the moduli and add the arguments: $wz=(2\cdot3)e^{i(\pi/6+\pi/6)}=6e^{i\pi/3}$ .

Answer

Modulus

6

; argument

\frac{\pi}{3}

Question 11(c)(i)1 mark

Squaring a complex number

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Complex numbers

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Expand $(x+iy)^2$ and use $i^2=-1$ .

Answer

z^2=(x^2-y^2)+2xyi.

Question 11(c)(ii)2 marks

Taking a complex reciprocal

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Complex numbers

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Multiply numerator and denominator by the conjugate $x-iy$ to make the denominator real.

Answer

\displaystyle\frac1z=\frac{x-iy}{x^2+y^2}=\frac{x}{x^2+y^2}-\frac{y}{x^2+y^2}i.

This requires

z\ne0

Question 11(d)(i)1 mark

Resolving a force along a vector

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VectorsMechanics

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The direction vector has magnitude $\sqrt{2^2+(-2)^2+1^2}=3$ . Scale it by $12/3=4$ to obtain the force vector.

Answer

\mathbf F_1=8\mathbf i-8\mathbf j+4\mathbf k.

Question 11(d)(ii)1 mark

Adding force vectors

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VectorsMechanics

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Add the force components: $(8,-8,4)+(-6,12,4)=(2,4,8)$ .

Answer

\mathbf F_3=2\mathbf i+4\mathbf j+8\mathbf k.

Question 11(d)(iii)1 mark

Dot product of force and displacement

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VectorsMechanics

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Take the componentwise dot product with $\mathbf d=(1,1,2)$ : $2+4+8\cdot2=22$ .

Answer

\mathbf F_3\cdot\mathbf d=22.

Question 11(e)2 marks

Proof by contradiction with surds

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ProofInequalities

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Assume the negation, $\sqrt3+\sqrt5\leq\sqrt{11}$ . Squaring positive sides gives $8+2\sqrt{15}\leq11$ , hence $2\sqrt{15}\leq3$ . Squaring again gives $60\leq9$ , a contradiction.

Answer

Therefore

\sqrt3+\sqrt5>\sqrt{11}

Question 11(f)2 marks

Completing the square in an integral

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Integration

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Complete the square: $7-x^2-6x=16-(x+3)^2$ . Apply the standard inverse-sine integral with the substitution $u=(x+3)/4$ .

Answer

\displaystyle\int\frac{5}{\sqrt{7-x^2-6x}}\,dx=5\sin^{-1}\!\left(\frac{x+3}{4}\right)+C.

Question 12(a)3 marks

A definite integral by parts

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Integration

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Choose $u=x$ and $dv=\sin x\,dx$ , so $du=dx$ and $v=-\cos x$ . Integration by parts gives $\int_0^{\pi/2}x\sin x\,dx=[-x\cos x]_0^{\pi/2}+\int_0^{\pi/2}\cos x\,dx$ .
The product term is zero at both endpoints. The remaining integral is $[\sin x]_0^{\pi/2}=1$ .

Answer

1

Question 12(b)3 marks

Induction for higher derivatives

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ProofInductionCalculus

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For $n=1$ , the product rule gives $y^{\prime}=(2x+1)e^{2x}$ , agreeing with the proposed expression.
Assume $y^{(k)}=(2^kx+k2^{k-1})e^{2x}$ for some integer $k\ge1$ . Differentiating gives $y^{(k+1)}=[2^k+2(2^kx+k2^{k-1})]e^{2x}$ .
Simplifying yields $[2^{k+1}x+(k+1)2^k]e^{2x}$ , exactly the statement for $k+1$ . Therefore induction proves the formula for every positive integer.

Answer

y^{(n)}=(2^nx+n2^{n-1})e^{2x}

, for all integers

n\ge1

Question 12(c)3 marks

A half-plane from complex distances

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Complex numbers

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Write $z=x+iy$ . Squaring both non-negative distances gives $(x+5)^2+(y-1)^2>(x-3)^2+(y+3)^2$ .
After cancellation, $16x-8y+8>0$ , or $y<2x+1$ . Sketch the line through $(-1/2,0)$ and $(0,1)$ as a dashed boundary, then shade the side below it. The origin is a useful test point in the shaded half-plane.

Solution sketch for Question 12(c); key values and curve behaviour are described in the steps above. — Original solution sketch; see the steps for key values.

Answer

The open half-plane

y<2x+1

; its boundary is excluded.

Question 12(d)4 marks

Partial fractions with a quadratic factor

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Integration

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Write the fraction as $A/(4-x)+(Bx+C)/(x^2+1)$ . Multiplication by the denominator gives $x^2-2x+9=A(x^2+1)+(Bx+C)(4-x)$ .
Substituting $x=4$ gives $A=1$ . Setting $x=0$ then gives $C=2$ ; comparing the squared terms gives $B=0$ . Thus the integrand is $1/(4-x)+2/(x^2+1)$ .
Integrate each term, allowing for the negative derivative of $4-x$ .

Answer

-\ln|4-x|+2\tan^{-1}x+C

Question 12(e)3 marks

Stopping distance under resistance

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MechanicsDifferential equations

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Take distance $s$ along the initial direction of travel. Before the particle stops, Newton’s law and $a=v\,dv/ds$ give $v\,dv/ds=-(k+v^2)$ , after cancelling the mass.
Separate variables and integrate from initial speed $V_0$ to zero: $S=-\int_{V_0}^{0}v/(k+v^2)\,dv=-\tfrac12[\ln(k+v^2)]_{V_0}^{0}$ .
The logarithms combine into a ratio. Since $k>0$ , the result is non-negative.

Answer

S=\frac12\ln\left(\frac{k+V_0^2}{k}\right)\text{ m}

Question 13(a)3 marks

Reflecting the limits of an integral

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Integration

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Use $u=m-x$ , so $dx=-du$ . The lower endpoint $x=m-4$ becomes $u=4$ , and the upper endpoint $x=m-2$ becomes $u=2$ .
Also $e^{-x}=e^{-m}e^u$ and $x-m+1=1-u$ . Consequently the transformed integral is $e^{-m}\int_4^2 e^u/(u-1)\,du$ .
Reversing the endpoints introduces a minus sign, leaving the integral defining $A$ .

Answer

The integral equals

-e^{-m}A

, so

k=-e^{-m}

Question 13(b)4 marks

A unit vector perpendicular to two vectors

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Vectors

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Perpendicularity gives $2x+4y-3z=0$ and $-4x-5y+3z=0$ . Adding them gives $y=-2x$ ; substituting into the first equation gives $z=-2x$ .
The unit-length condition is $x^2+y^2+z^2=1$ . Hence $9x^2=1$ , so $x=\pm1/3$ , with $y$ and $z$ carrying the opposite signs.

Answer

\mathbf c=\pm\frac13(1,-2,-2)

Question 13(c)(i)2 marks

Proving the arithmetic-geometric mean inequality

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ProofInequalities

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Since $a,b>0$ , the square $(\sqrt a-\sqrt b)^2$ is non-negative. Expanding gives $a+b-2\sqrt{ab}\ge0$ . Divide by 2.
Equality holds exactly when $a=b$ .

Answer

\frac{a+b}{2}\ge\sqrt{ab}

Question 13(c)(ii)2 marks

A strict inequality from distinct inputs

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ProofInequalities

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Apply part (i) to $a=2n+1$ and $b=2n+3$ . Both are positive for $n\ge0$ , and they differ, so $\sqrt{(2n+1)(2n+3)}<2n+2$ .
Divide by the positive quantity $(2n+2)\sqrt{2n+3}$ , then multiply by $\sqrt{2n+1}$ , to obtain the required comparison.

Answer

\frac{2n+1}{2n+2}<\frac{\sqrt{2n+1}}{\sqrt{2n+3}}

, for every integer

n\ge0

Question 13(d)4 marks

Symmetry followed by a half-angle substitution

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IntegrationTrigonometry

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Call the given integral $J$ . First substitute $u=\pi/2-x$ . Since sine and cosine exchange, $J=\int_0^{\pi/2}(\pi/2-x)/(1+\sin x+\cos x)\,dx$ .
Adding this expression to the original gives $2J=(\pi/2)K$ , where $K=\int_0^{\pi/2}dx/(1+\sin x+\cos x)$ . Thus $J=\pi K/4$ .
For $K$ , put $t=\tan(x/2)$ . The endpoints become 0 and 1, and the half-angle identities reduce the integrand to $dt/(1+t)$ . Hence $K=[\ln(1+t)]_0^1=\ln2$ .

Answer

J=\frac{\pi\ln2}{4}

Question 14(a)(i)3 marks

Reducing an integral of an even cotangent power

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IntegrationReduction formulasTrigonometry

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Separate the final factor of $\cot^2\theta$ , and replace it by $\csc^2\theta-1$ :
$I_n=\int_{\pi/4}^{\pi/2}\cot^{2n-2}\theta\,\csc^2\theta\,d\theta-I_{n-1}.$
The substitution $u=\cot\theta$ integrates the first term as $-\cot^{2n-1}\theta/(2n-1)$ . Its upper value is zero and its lower value is $-1/(2n-1)$ .

Answer

\displaystyle I_n=\frac1{2n-1}-I_{n-1},\quad n>0

Question 14(a)(ii)1 mark

Applying a reduction formula twice

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IntegrationReduction formulas

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The starting integral is $I_0=\pi/4$ . Thus $I_1=1-\pi/4$ , and $I_2=1/3-I_1$ .

Answer

I_2=\pi/4-2/3

Question 14(b)(i)2 marks

Velocity from acceleration as a function of position

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MotionDifferential equations

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Use $a=v\,dv/dx$ : $v\,dv=32x(x^2+3)\,dx$ . Integration gives $v^2=16(x^2+3)^2+C$ .
The initial conditions $x=0$ , $v=-12$ give $C=0$ . The velocity starts negative and its magnitude never vanishes, so retain the negative branch throughout.

Answer

v=-4(x^2+3)\text{ m s}^{-1}

Question 14(b)(ii)2 marks

Travel time with position-dependent velocity

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MotionIntegration

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The velocity is always negative, so travelling 3 metres from the origin means reaching $x=-3$ . Since $dt=dx/v$ ,
$t=-\frac14\int_0^{-3}\frac{dx}{x^2+3}=-\frac1{4\sqrt3}\left[\tan^{-1}(x/\sqrt3)\right]_0^{-3}.$
The inverse tangent at the lower displacement is $-\pi/3$ .

Answer

t=\dfrac{\pi}{12\sqrt3}\text{ s}

Question 14(c)(i)1 mark

Identifying a root of unity

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Complex numbersRoots of unity

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Multiplying the given sum by $w-1$ makes all intermediate powers cancel, leaving $w^7-1=0$ . Also $w\ne1$ , because substituting 1 into the original sum gives 7.

Answer

w^7=1

, so

w

is a non-trivial seventh root of unity.

Question 14(c)(ii)2 marks

Conjugate roots written as powers

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Complex numbersRoots of unityPolynomials

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A non-real root of a polynomial with real coefficients comes with its complex conjugate. Hence the second root is $\overline{w+w^2+w^4}$ .
Every root of unity has modulus 1, so $\overline w=1/w=w^6$ . Reduce exponents modulo 7: $\overline w^{\,2}=w^{12}=w^5$ and $\overline w^{\,4}=w^{24}=w^3$ .

Answer

The other root is

w^3+w^5+w^6

Question 14(c)(iii)1 mark

The product of conjugate roots

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Complex numbersRoots of unityPolynomials

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The constant term is the product of the roots. Expanding $(w+w^2+w^4)(w^3+w^5+w^6)$ and using $w^7=1$ gives $3+w+w^2+\cdots+w^6$ . The original relation says the six non-constant powers sum to $-1$ .

Answer

c=2

Question 14(d)3 marks

Comparing sums of reciprocal arithmetic terms

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ProofInequalities

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Let $u=1/a>0$ . The four reciprocal terms are $u,u+k,u+2k,u+3k$ , so
$a+d-b-c=\left(\frac1u-\frac1{u+k}\right)-\left(\frac1{u+2k}-\frac1{u+3k}\right).$
This is $\dfrac{k}{u(u+k)}-\dfrac{k}{(u+2k)(u+3k)}$ . The second denominator is strictly larger than the first because $u,k>0$ . The difference is therefore positive.

Answer

b+c<a+d

Question 15(a)4 marks

Area of a parallelogram using a scalar product

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VectorsGeometry

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The side lengths satisfy $|\mathbf a|^2=4^2+3^2+(-1)^2=26$ and $|\mathbf b|^2=2^2+(-1)^2+3^2=14$ . Their scalar product is $8-3-3=2$ .
If $\phi$ is the angle between them, $\cos\phi=2/\sqrt{26\cdot14}$ . Hence
$\text{Area}^2=|\mathbf a|^2|\mathbf b|^2\sin^2\phi=26\cdot14-2^2=360.$
Take the positive square root for an area.

Answer

The area is

6\sqrt{10}

square units.

Question 15(b)4 marks

Maximum acceleration in simple harmonic motion

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Simple harmonic motionMotion

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Two cycles per second means frequency $f=2$ , so the angular frequency is $\omega=2\pi f=4\pi$ .
For SHM, $v^2=\omega^2(A^2-x^2)$ and $v_{\max}=\omega A$ . At $|x|=1/4$ , the given half-speed condition gives $\omega^2A^2/4=\omega^2(A^2-1/16)$ .
Therefore $3A^2/4=1/16$ , so $A=1/(2\sqrt3)$ . The acceleration is $-\omega^2x$ , whose largest positive value occurs at $x=-A$ .

Answer

a_{\max}=\omega^2A=\dfrac{8\pi^2}{\sqrt3}\text{ m s}^{-2}

Question 15(c)(i)3 marks

A complex ratio and the half-angle formula

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Complex numbersTrigonometry

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Write $\cos\theta+i\sin\theta=e^{i\theta}$ . Factoring out $e^{i\theta/2}$ from numerator and denominator gives
$\frac{1+e^{i\theta}}{1-e^{i\theta}}=\frac{e^{-i\theta/2}+e^{i\theta/2}}{e^{-i\theta/2}-e^{i\theta/2}}=\frac{2\cos(\theta/2)}{-2i\sin(\theta/2)}.$
Since $1/(-i)=i$ , the result is $i\cot(\theta/2)$ . Both expressions require $\theta\notin2\pi\mathbb Z$ .

Answer

\displaystyle\frac{1+\cos\theta+i\sin\theta}{1-\cos\theta-i\sin\theta}=i\cot(\theta/2)

Question 15(c)(ii)2 marks

Sixth roots by De Moivre’s theorem

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Complex numbersRoots of unity

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If $w^6=-1$ , then $|w|^6=1$ , so $|w|=1$ . Write $w=\cos\phi+i\sin\phi$ .
De Moivre’s theorem gives $w^6=\cos6\phi+i\sin6\phi$ . To equal $-1$ , the angle must satisfy $6\phi=(2j+1)\pi$ . The six values $j=0,\ldots,5$ give all distinct roots.

Answer

\displaystyle w_j=\cos\frac{(2j+1)\pi}{6}+i\sin\frac{(2j+1)\pi}{6},\quad j=0,\ldots,5

Question 15(c)(iii)2 marks

Mapping the six roots to the required solutions

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Complex numbersRoots of unity

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Put $w=(z-1)/(z+1)$ . Then $w^6=-1$ , so use one of the six roots in part (ii). Rearranging gives $z=(1+w)/(1-w)$ . No sixth root of $-1$ equals 1, so this denominator is non-zero.
Apply part (i) with $\theta=(2j+1)\pi/6$ . The half-angle is $(2j+1)\pi/12$ , producing six distinct values.

Answer

\displaystyle z=i\cot\frac{(2j+1)\pi}{12},\quad j=0,1,2,3,4,5

Question 16(a)4 marks

Bounding complex roots with the triangle inequality

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Complex numbersProofInequalities

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Suppose, for a contradiction, that a solution has $|z|\le1/2$ . Write the left side as $\sum_{j=1}^{n}z^j\cos(j\theta)$ .
The triangle inequality and $|\cos(j\theta)|\le1$ imply
$1=\left|\sum_{j=1}^{n}z^j\cos(j\theta)\right|\le\sum_{j=1}^{n}|z|^j\le\sum_{j=1}^{n}2^{-j}=1-2^{-n}<1.$
This is impossible. The finite geometric sum is strictly less than 1, so the contradiction excludes the boundary as well as the interior.

Answer

Every solution satisfies

|z|>1/2

Question 16(b)5 marks

Projectile motion with linear resistance

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Projectile motionDifferential equations

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The launch components are $\dot x(0)=50(4/5)=40$ and $\dot y(0)=-50(3/5)=-30$ . Solving $\ddot x=-k\dot x$ and using $x(0)=0$ gives
$\dot x=40e^{-kt},\qquad x=\frac{40}{k}(1-e^{-kt}).$
For the vertical component, $\dot y+10/k$ decays exponentially. The initial velocity and $y(0)=0$ therefore give
$\dot y=(10/k-30)e^{-kt}-10/k,\quad y=\frac{10/k-30}{k}(1-e^{-kt})-\frac{10t}{k}.$
Eliminate time using $e^{-kt}=1-kx/40$ and $t=-\ln(1-kx/40)/k$ .

Answer

\displaystyle y=\frac{1-3k}{4k}x+\frac{10}{k^2}\ln\left(1-\frac{kx}{40}\right),\quad0\le x<40/k

Question 16(c)(i)2 marks

The closest point on a line

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VectorsOptimisation

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Minimising distance is equivalent to minimising its square. Since $|\mathbf d|=1$ ,
$f(\lambda)^2=|\mathbf a-\mathbf b+\lambda\mathbf d|^2=|\mathbf a-\mathbf b|^2+2\lambda(\mathbf a-\mathbf b)\cdot\mathbf d+\lambda^2.$
This quadratic opens upwards. Its derivative vanishes at $2(\mathbf a-\mathbf b)\cdot\mathbf d+2\lambda=0$ , giving the unique minimum.

Answer

\lambda_0=(\mathbf b-\mathbf a)\cdot\mathbf d

Question 16(c)(ii)1 mark

The shortest connecting segment is perpendicular

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VectorsGeometry

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The connecting vector is $\overrightarrow{PB}=\mathbf b-\mathbf a-\lambda_0\mathbf d$ . Its scalar product with $\mathbf d$ is $(\mathbf b-\mathbf a)\cdot\mathbf d-\lambda_0|\mathbf d|^2=\lambda_0-\lambda_0=0$ .

Answer

\overrightarrow{PB}\cdot\mathbf d=0

, so PB is perpendicular to the line direction.

Question 16(c)(iii)3 marks

Distance between a line and a sphere

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VectorsGeometry

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First find the line’s distance from the origin. Using part (i) with $\mathbf b=\mathbf0$ , the closest point has vector $\mathbf p_0=\mathbf a-(\mathbf a\cdot\mathbf d)\mathbf d$ .
Because $|\mathbf d|=1$ , its distance from the origin is $D=\sqrt{|\mathbf a|^2-(\mathbf a\cdot\mathbf d)^2}$ .
If $D>1$ , move one unit from the origin towards this closest point to reach the sphere; the remaining gap is $D-1$ . If $D\le1$ , the line intersects or touches the sphere, so the shortest distance is zero.

Answer

\displaystyle\max\left\{0,\sqrt{|\mathbf a|^2-(\mathbf a\cdot\mathbf d)^2}-1\right\}

units.

2025 HSCworked solutions

Choose your topics

Section I Multiple choice

A line through two points

Negating an existential statement

Square roots of a complex number

Amplitude from the speed equation

Writing a contrapositive

Moduli of sums and differences

Arguments subtended by a circle

Acceleration as a function of displacement

Classifying a complex quadrilateral

Reparametrising a space curve

Section II Written responses

Conjugate and rotated complex numbers

Multiplying complex numbers in polar form

Squaring a complex number

Taking a complex reciprocal

Resolving a force along a vector

Adding force vectors

Dot product of force and displacement

Proof by contradiction with surds

Completing the square in an integral

A definite integral by parts

Induction for higher derivatives

A half-plane from complex distances

Partial fractions with a quadratic factor

Stopping distance under resistance

Reflecting the limits of an integral

A unit vector perpendicular to two vectors

Proving the arithmetic-geometric mean inequality

A strict inequality from distinct inputs

Symmetry followed by a half-angle substitution

Reducing an integral of an even cotangent power

Applying a reduction formula twice

Velocity from acceleration as a function of position

Travel time with position-dependent velocity

Identifying a root of unity

Conjugate roots written as powers

The product of conjugate roots

Comparing sums of reciprocal arithmetic terms

Area of a parallelogram using a scalar product

Maximum acceleration in simple harmonic motion

A complex ratio and the half-angle formula

Sixth roots by De Moivre’s theorem

Mapping the six roots to the required solutions

Bounding complex roots with the triangle inequality

Projectile motion with linear resistance

The closest point on a line

The shortest connecting segment is perpendicular

Distance between a line and a sphere

No matching question parts

2025 HSC
worked solutions