2026 paper questions

$f^{-1}(5)$ is the input that $f$ sends to 5. Solve $3x-4=5$: $x=3$. Equivalently $f^{-1}(x)=\frac{x+4}{3}$, so $f^{-1}(5)=\frac{9}{3}=3$.

Option D (11) is the trap: it is $f(5)$, not $f^{-1}(5)$.

By the chain rule $f^{\prime }(x)=\frac{1}{2x}\cdot 2=\frac{1}{x}$ (or write $\ln 2x=\ln 2+\ln x$). So $f^{\prime }(3)=\frac{1}{3}$. Option B forgets that the factor $2$ cancels.

By the Fundamental Theorem, ${\int }_0^{1}x{e}^{x}dx=[e^x(x-1){]}_0^1=e^1(0)-e^0(-1)=0+1=1$.

Option C, $e-1$, is the value of ${\int }_0^1{e}^{x}dx$; it ignores the factor $x$.

Separate the variables: $ydy=xdx$. Integrate: $\frac{y^2}{2}=\frac{x^2}{2}+c$, so $y^2=x^2+C$ with $C=2c$.

Option (B) comes from a sign slip (it solves $\frac{dy}{dx}=-\frac{x}{y}$), and (C) from integrating $x$ as $x^2$ without the factor $\frac{1}{2}$.

$\ln (t-1)$ needs $t-1>0$, so $t>1$ (1 is excluded). $\sqrt{4-t}$ needs $4-t\ge 0$, so $t\le 4$ (4 is included). The domain is the intersection $(1,4]$.

(B) and (D) wrongly include $t=1$, where $\ln 0$ is not defined; (A) wrongly leaves out $t=4$, where $\sqrt{0}=0$ is fine.

Here $a^2=16$, $b^2=9$, so $a=4$, $b=3$. The asymptotes of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are $y=\pm \frac{b}{a}x=\pm \frac{3}{4}x$, so the slopes are $\pm \frac{3}{4}$.

Option B, $\pm \frac{4}{3}$, is $a/b$ (upside down); options A and C use $a^2$ and $b^2$ instead of $a$ and $b$.

The range of an inverse function is the domain of the original function: $\text{Rang}{f}^{-1}=\text{Dom}f=[-1,\infty )$. (Indeed $f^{-1}(x)=x^2-1$ for $x\ge 0$, whose values start at $-1$.)

Option B, $[0,\infty )$, is the range of $f$, which is the domain of $f^{-1}$.

$f^{\prime }(x)=3x^2-6x$, so $f^{\prime }(2)=0$ and $x=2$ is a critical value. $f^{\prime \prime }(x)=6x-6$ gives $f^{\prime \prime }(2)=6>0$, so by the second derivative test $f(2)=-2$ is a local minimum. The inflection point is at $x=1$, where $f^{\prime \prime }=0$.

Disc method: $V=\pi {\int }_0^4(\sqrt{x}{)}^{2}dx=\pi {\int }_0^{4}xdx=\pi {\left[\frac{x^2}{2}\right]}_0^4=\pi \cdot \frac{16}{2}=8\pi$.

Forgetting to square $\sqrt{x}$ gives $\frac{16}{3}\pi$, and forgetting the $\frac{1}{2}$ gives $16\pi$; neither is an option.

The triangle has a right angle at the origin, with legs $4$ (along the $x$-axis) and $6$ (along the $y$-axis). Area $=\frac{1}{2}(4)(6)=12$. The determinant gives the same: $\frac{1}{2}\left|0(0-6)+4(6-0)+0(0-0)\right|=\frac{1}{2}(24)=12$.

(D) $24$ is the determinant before halving: a common slip.

Here $a=3$, $h=-\frac{5}{2}$, $b=2$, so $h^2-ab=\frac{25}{4}-6=\frac{1}{4}>0$: two real, distinct lines through the origin. Indeed $3x^2-5xy+2y^2=(x-y)(3x-2y)$.

Differentiating $x^2+y^2=25$ gives $\frac{dy}{dx}=-\frac{x}{y}$. At $(3,4)$ the slope is $-\frac{3}{4}$.

Option D, $\frac{4}{3}$, is the slope of the radius to $(3,4)$; the tangent is perpendicular to it, so its slope is the negative reciprocal.

Work from the inside: $\sin \frac{5\pi }{6}=\sin \left(\pi -\frac{\pi }{6}\right)=\frac{1}{2}$. The principal value of ${\sin }^{-1}\frac{1}{2}$ is $\frac{\pi }{6}$, which lies in $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$.

Option B is the trap: ${\sin }^{-1}(\sin x)=x$ only for $x\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, and $\frac{5\pi }{6}$ is outside that interval. Option C has the wrong sign ($\sin \frac{5\pi }{6}>0$).

$\mathbf{v}(t)=⟨5,4-2t,0⟩$ and $\mathbf{a}(t)=⟨0,-2,0⟩$, the same at every time, including $t=2$.

(C) keeps the 5 from the first derivative; the constant 5 differentiates to 0.

Length of tangent $=\sqrt{x_1^2+y_1^2-r^2}=\sqrt{100+0-36}=\sqrt{64}=8$. The radius (6), the tangent (8) and the distance to the centre (10) form a 6, 8, 10 right triangle.

$V=x^3$, so $dV=3x^{2}dx=3(4{)}^2(0.1)=4.8\mathrm{c}{\mathrm{m}}^3$. (The exact increase is $4.1^3-4^3=4.921\mathrm{c}{\mathrm{m}}^3$, close to the differential estimate.)

$f_{\text{avg}}=\frac{1}{b-a}{\int }_a^{b}f(x)dx=\frac{1}{3-0}{\int }_0^3{x}^{2}dx=\frac{1}{3}{\left[\frac{x^3}{3}\right]}_0^3=\frac{1}{3}(9)=3$.

The integral alone is 9; remember to divide by the length of the interval, $b-a=3$.

Finding the inverse

Domain and range of the inverse

Check

Model and general solution

(a) Growth constant

(b) Population in 2030

Dividing by y+1 assumed y\neq-1; the constant solution y=-1 is included by c=0.

(a) Pair of lines through the origin

(b) Individual lines

So each focus is 6\,\mathrm{m} from the centre of the room along the floor, that is 4\,\mathrm{m} from each end wall. A whisper at one focus is reflected by the ceiling to the other focus, 12\,\mathrm{m} away.

Finding the inverse

Verification

Condition for continuity

f is continuous at x=c when f(c) is defined, \lim_{x\to c}f(x) exists, and \lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)=f(c). Each piece is a polynomial or a rational function whose denominator is non-zero on its interval, so only the joining points x=1 and x=3 need to be checked.

Continuity at x = 1

Continuity at x = 3

Solving

Check

(a) Critical points

Critical points are the numbers in [0,\pi] where f'(x)=0 or f' does not exist. Here f' exists everywhere, so we solve f'(x)=0.

(b) Second derivative test

(c) Absolute extrema

f is continuous on the closed interval [0,\pi], so it has an absolute maximum and minimum (extreme value theorem). Compare the values at the critical points and at the endpoints.

The largest value in the table is \tfrac32 and the smallest is 1. So the absolute maximum value of f is \tfrac32, reached at both x=\tfrac{\pi}{6} and x=\tfrac{5\pi}{6}, and the absolute minimum value is 1, reached at x=0, x=\tfrac{\pi}{2} and x=\pi. Here the local minimum at \tfrac{\pi}{2} happens to equal the endpoint values.

The printed interval

The question asks for x\in[-2\pi,2\pi], but \sin^{-1}x is the angle whose sine is x, and the sine of every angle lies between -1 and 1. So \sin^{-1}x is defined only for -1\le x\le1; values such as \sin^{-1}2 or \sin^{-1}\pi do not exist. Since 2\pi\approx6.28, most of [-2\pi,2\pi] is outside the domain, and the graph can be drawn only on [-1,1], the part of [-2\pi,2\pi] where the function is defined.

The interval [-2\pi,2\pi] comes from the unit's learning outcome, where the sine curve is drawn from -2\pi to 2\pi and its principal part \left[-\frac{\pi}{2},\frac{\pi}{2}\right] is reflected in the line y=x (second sketch below).

Domain, range and table of values

Scale

x-axis: 1 cm represents 0.25, so the domain from -1 to 1 takes 8 cm. y-axis: 1 cm represents \frac{\pi}{12}, so each step of \frac{\pi}{6} is 2 cm and the range from -\frac{\pi}{2} to \frac{\pi}{2} takes 12 cm.

Graph

Labels and properties

• Domain: [-1,1], shown by the dotted lines x=\pm1; nothing is drawn outside them.
• Range: \left[-\frac{\pi}{2},\frac{\pi}{2}\right], shown by the dotted lines y=\pm\frac{\pi}{2}.
• Intercepts: the graph meets both axes only at the origin O(0,0), since \sin^{-1}0=0.
• End points: A\left(-1,-\frac{\pi}{2}\right) and B\left(1,\frac{\pi}{2}\right), both included (filled dots).
• Shape: strictly increasing and symmetric about the origin (odd function: \sin^{-1}(-x)=-\sin^{-1}x). It is the reflection of y=\sin x, -\frac{\pi}{2}\le x\le\frac{\pi}{2}, in the line y=x.

Sketch from the sine curve on [-2π, 2π]

A median joins a vertex to the mid-point of the opposite side. Plan: find the three mid-points, write the equation of each median, show that the determinant of their coefficients is zero (the condition of concurrency), and then solve two of the equations to get the common point, the centroid.

Mid-points of the sides

Equations of the medians

Concurrency

Centroid

Result. The three medians pass through one point, so they are concurrent, and the point of concurrency is the centroid G\left(\tfrac43,2\right).

Step 1: Show that the equation is homogeneous

Let f(x,y)=\dfrac{x^{2}+y^{2}}{xy}. Then f(tx,ty)=\dfrac{t^{2}\left(x^{2}+y^{2}\right)}{t^{2}xy}=t^{0}f(x,y). Numerator and denominator are both of degree 2, so f is homogeneous of degree 0 and the DE is homogeneous. It can be solved with y=ux.

Step 2: Substitute y = ux

Step 3: Separate the variables

Step 4: Integrate

Step 5: Return to x and y

Verification

f_{\text{avg}}=\dfrac{1}{3-1}\int_{1}^{3}x^{2}\,dx=\dfrac12\left[\dfrac{x^{3}}{3}\right]_{1}^{3}=\dfrac12\left(9-\dfrac13\right)=\dfrac12\cdot\dfrac{26}{3}=\dfrac{13}{3}.

Option B, \tfrac{26}{3}, is the integral itself; it has not been divided by b-a=2.

f^{-1}(13) is the input that gives the output 13: 4x+9=13\Rightarrow x=1. Equivalently f^{-1}(x)=\dfrac{x-9}{4} and f^{-1}(13)=\dfrac{4}{4}=1.

f'(x)=\dfrac{4}{4x+1} (chain rule: derivative of the inside over the inside). At x=2: \dfrac{4}{9}. Option D forgets the factor 4 from the inside.

\Big[\sin^{-1}\dfrac{x}{2}\Big]_{0}^{1}=\sin^{-1}\dfrac12-\sin^{-1}0=\dfrac{\pi}{6}-0=\dfrac{\pi}{6}.

Option C, \tfrac{\pi}{3}, would come from \cos^{-1}\tfrac12; the inverse sine of \tfrac12 is \tfrac{\pi}{6}.

Separate: y\,dy=3x^{2}\,dx. Integrate: \dfrac{y^{2}}{2}=x^{3}+c. Multiply by 2: y^{2}=2x^{3}+C.

Option (C) forgets to multiply x^{3} by 2, and (D) integrates 3x^{2} wrongly.

\sqrt{t+3} needs t\ge-3 (included). \ln(6-t) needs 6-t>0, so t<6 (excluded). The intersection is [-3,6).

Put the right side equal to zero: \dfrac{y^2}{36}=\dfrac{x^2}{49}, so y=\pm\dfrac67x. The slopes are \pm\dfrac67.

For this vertical hyperbola, \dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1 with a=6, b=7, the asymptotes are y=\pm\dfrac{a}{b}x, not \pm\dfrac{b}{a}x. Choosing \pm\tfrac76 (option A) is the common mistake.

The intended model is P(t)=1000e^{0.03t}. At t=10: P=1000e^{0.3}=1000\times1.3499\approx1350.

Option (A) 2000 would need e^{0.3}=2, and (B) 1000 is the starting population.

f'(x)=3x^{2}-6x=3x(x-2) exists everywhere and is zero at x=0 and x=2, both inside [-1,3]. The endpoints -1 and 3 are not critical points; they are checked separately when finding absolute extrema, which is why options A, C and D are tempting but wrong.

V=\pi\int_{0}^{16}(\sqrt{x})^{2}\,dx=\pi\int_{0}^{16}x\,dx=\pi\left[\dfrac{x^{2}}{2}\right]_{0}^{16}=\pi\cdot\dfrac{256}{2}=128\pi.

Option A, 256\pi, forgets the \tfrac12 from \int x\,dx=\tfrac{x^{2}}{2}.

Right angle at the origin with legs 6 and 9: area =\tfrac12(6)(9)=27. (B) 54 forgets the \tfrac12.

a=2, h=-\tfrac32, b=-2: h^2-ab=\tfrac94+4=\tfrac{25}4>0, so two real, distinct lines: (x-2y)(2x+y)=0. Also a+b=0, so the two lines are perpendicular.

\dfrac{dy}{dx}=-\dfrac{x}{y}=-\dfrac{5}{-12}=\dfrac{5}{12}. Check: the radius to (5,-12) has slope -\tfrac{12}{5}, and \left(-\tfrac{12}{5}\right)\left(\tfrac{5}{12}\right)=-1.

Option C forgets that y_1=-12 is negative.

\cos\dfrac{4\pi}{3}=\cos\left(\pi+\dfrac{\pi}{3}\right)=-\dfrac12, and the principal value of \cos^{-1}\left(-\dfrac12\right) is \pi-\dfrac{\pi}{3}=\dfrac{2\pi}{3}, which lies in [0,\pi].

Option B is the trap: \dfrac{4\pi}{3} is greater than \pi, so it can never be a value of \cos^{-1}. Option C, \frac{\pi}{3}, has cosine +\frac12.

\mathbf{v}(t)=\langle8,\,6-4t,\,0\rangle and \mathbf{a}(t)=\langle0,\,-4,\,0\rangle, constant for all t, so at t=3 it is \langle0,-4,0\rangle.

Length =\sqrt{x_1^2+y_1^2-144}=\sqrt{0+169-144}=\sqrt{25}=5 (the 5, 12, 13 right triangle). Option D adds 144 instead of subtracting it.

V=\dfrac43\pi r^{3} gives dV=4\pi r^{2}\,dr=4\pi(25)(0.2)=20\pi\,\mathrm{cm^{3}}.

Finding the inverse

Domain and range of the inverse

Check

Model and general solution

(a) Growth constant

(b) Population in 2030

The solutions are straight lines through the point (-6,2). The constant solution y=2 (lost when dividing by y-2) is included by c=0.

(a) Pair of lines through the origin

(b) Individual lines

Condition for continuity

f is continuous at x=c when \lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)=f(c). The pieces are continuous on their own intervals (the last denominator x+1 is not zero for x\ge5), so we only need the joins at x=2 and x=5.

Continuity at x = 2

Continuity at x = 5

Solving

Check

Step 1: check the point

Version 1: circle x^2+y^2-6x-4y-12=0 (sign of 4y corrected), point P(7,-1)

Version 2: printed circle, point P(7,1) (sign of y_1 corrected)

(a) Critical points

f' exists for every x, so the critical points are the solutions of f'(x)=0 in [0,\pi].

(b) Second derivative test

(c) Absolute extrema

f is continuous on [0,\pi], so compare the critical values with the endpoint values f(0)=0+1=1 and f(\pi)=0+\cos2\pi=1.

So the absolute maximum value is \tfrac{17}{8}, reached twice, at x=\sin^{-1}\tfrac34 and at x=\pi-\sin^{-1}\tfrac34. The absolute minimum value is 1, at the endpoints x=0 and x=\pi. The local minimum 2 at x=\tfrac{\pi}{2} is not the absolute minimum, because the endpoints give the smaller value 1.

The printed interval

The question asks for x\in[-2\pi,2\pi], but \cos^{-1}x is the angle whose cosine is x, and the cosine of every angle lies between -1 and 1. So \cos^{-1}x is defined only for -1\le x\le1; for example \cos^{-1}3 does not exist. Since 2\pi\approx6.28, most of [-2\pi,2\pi] lies outside the domain, and the graph can be drawn only on [-1,1].

The interval [-2\pi,2\pi] comes from the unit's learning outcome, where the cosine curve is drawn from -2\pi to 2\pi and its principal part [0,\pi] is reflected in the line y=x (second sketch below).

Domain, range and table of values

Scale

x-axis: 1 cm represents 0.25, so the domain from -1 to 1 takes 8 cm. y-axis: 1 cm represents \frac{\pi}{12}, so each step of \frac{\pi}{6} is 2 cm and the range from 0 to \pi takes 12 cm.

Graph

Labels and properties

• Domain: [-1,1], shown by the dotted lines x=\pm1.
• Range: [0,\pi]; the curve stays between the x-axis and the dotted line y=\pi.
• Intercepts: y-intercept C\left(0,\frac{\pi}{2}\right) because \cos^{-1}0=\frac{\pi}{2}; x-intercept B(1,0) because \cos^{-1}1=0.
• End points: A(-1,\pi) and B(1,0), both included (filled dots).
• Shape: strictly decreasing; neither even nor odd, but symmetric about the point C\left(0,\frac{\pi}{2}\right) because \cos^{-1}(-x)=\pi-\cos^{-1}x. It is the reflection of y=\cos x, 0\le x\le\pi, in the line y=x.

Sketch from the cosine curve on [-2π, 2π]

Mid-points and slopes of the sides

Equations of the right bisectors

Concurrency

Circumcentre

Step 1: Show that the equation is homogeneous

f(x,y)=\dfrac{x^{2}+3y^{2}}{xy} gives f(tx,ty)=\dfrac{t^{2}\left(x^{2}+3y^{2}\right)}{t^{2}xy}=t^{0}f(x,y). Numerator and denominator are both of degree 2, so the DE is homogeneous.

Step 2: Substitute y = ux

Step 3: Separate the variables

Step 4: Integrate

Step 5: Return to x and y

Verification

f(x)=x^3-x gives f(-x)=-x^3+x=-(x^3-x)=-f(x), so it is odd. x^2+1 and \cos x are even (f(-x)=f(x)), and e^{x} is neither.

f'(x)=4x^{3} and f''(x)=12x^{2}. Option A is only the first derivative; C and D are the third and fourth derivatives.

The derivative of \ln|x| is \dfrac1x (Formula 3.2), so \int\dfrac{dx}{x}=\ln|x|+C. The power rule cannot be used because n=-1.

The median from A goes to the mid-point of BC: \left(\tfrac{3+5}2,\tfrac{4+0}2\right)=(4,2). This is the expected answer.

Note: the median from A is the line y=2, so it also passes through (3,2), which is the centroid \left(\tfrac{1+3+5}3,\tfrac{2+4+0}3\right). Option (C) is therefore also on the median; the question intends the mid-point (D).

The slopes are -1 and 1, which are different, so the lines meet in exactly one point. Adding the equations: 2x=6, x=3, y=2. The unique solution is (3,2). Two straight lines can never meet in exactly two points.

y^2=4ax (with a>0) opens to the right, its axis is the x-axis, the vertex is the origin, the focus is (a,0) and the directrix is x=-a. Option D is the focus of x^2=4ay.

x^2+y^2=9=3^2 is the standard form x^2+y^2=r^2: a circle with centre (0,0) and radius 3. The squared terms have equal coefficients and no xy term.

Chain rule: the derivative of the outer \sin is \cos(3x^{2}), times the derivative of the inside, 6x. Option A forgets the inside derivative.

\int_{0}^{1}x^{2}\,dx=\left[\dfrac{x^{3}}{3}\right]_{0}^{1}=\dfrac13-0=\dfrac13.

m_1=2, m_2=-3: \tan\theta=\left|\dfrac{-3-2}{1+(2)(-3)}\right|=\left|\dfrac{-5}{-5}\right|=1, so \theta=45^\circ.

The larger denominator is a^2=16, so a=4 and the major axis has length 2a=8. Option B is only the semi-major axis a; option C is the minor axis 2b.

(f\circ g)(x)=f\big(g(x)\big)=\big(\sqrt{x}\big)^2=x for x\ge0. Options A and D come from multiplying the functions instead of composing them.

f'(x)=2x-4=0 gives x=2, and f''(x)=2>0, so the minimum is at x=2 (the minimum value is f(2)=-1). x=1 and x=3 are the zeros of f, not its minimum.

Separate: \dfrac{dy}{y}=x\,dx. Integrate: \ln y=\dfrac{x^{2}}{2}+c. Take antilog: y=e^{c}e^{x^{2}/2}=Ce^{x^{2}/2}.

Option (A) is tempting, but the constant multiplies the exponential; it is not added to it.

d=\dfrac{|3(3)+4(4)+5|}{\sqrt{3^2+4^2}}=\dfrac{30}{5}=6.

With u=x^{2}, du=2x\,dx, so \int2xe^{x^{2}}dx=\int e^{u}\,du=e^{u}+C=e^{x^{2}}+C. The factor 2x is exactly the derivative of the power x^{2}.

Option B also happens to work (du=2xe^{x^{2}}dx turns the integral into \int du), but the standard substitution the question expects is the power of e, u=x^{2}.

Substitute: 3^2+4^2-25=9+16-25=0. The value is zero, so P lies on the circle (a negative value would mean inside, a positive value outside).

Use the definitions of the hyperbolic functions (here x\neq0, so that \sinh x\neq0):

(a) Slope-intercept form y=mx+c

(b) Two-intercept form \dfrac xa+\dfrac yb=1