The 20 one-mark questions from the March 2026 12th Standard Mathematics public exam, in paper order, with the correct option marked and a short explanation.
Q1
If A is a \(3\times 3\) non-singular matrix such that \(AA^T=A^TA\) and \(B=A^{-1}A^T\), then \(BB^T=\)
- A. \(I_3\)Correct
- B. A
- C. \(B^T\)
- D. B
Explanation. \(BB^T=A^{-1}A^T\,A\,(A^{-1})^T=A^{-1}(AA^T)(A^T)^{-1}=(A^{-1}A)(A^T(A^T)^{-1})=I_3\), using \(A^TA=AA^T\).
Q2
If \(\rho(A)=\rho([A|B])\), then the system \(AX=B\) of linear equations is :
- A. consistent and has infinitely many solutions
- B. consistent and has a unique solution
- C. inconsistent
- D. consistentCorrect
Explanation. By the Rouché–Capelli theorem, equal ranks mean the system is consistent. Whether the solution is unique or infinite depends on the number of unknowns, which is not given.
Q3
If z is a complex number such that \(z\in\mathbb{C}\setminus\mathbb{R}\) and \(z+\frac{1}{z}\in\mathbb{R}\), then \(|z|\) is :
- A. 2
- B. 0
- C. 3
- D. 1Correct
Explanation. Let \(z=re^{i\theta}\). The imaginary part of \(z+\frac{1}{z}\) is \((r-\frac{1}{r})\sin\theta=0\). Since z is not real, \(\sin\theta\neq0\), so \(r=1\).
Q4
The product of all four values of \(\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)^{\frac{3}{4}}\) is :
- A. 1Correct
- B. \(-2\)
- C. 2
- D. \(-1\)
Explanation. \((\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})^3=\cos\pi+i\sin\pi=-1\), so the four values are the roots of \(w^4+1=0\). Their product is the constant term 1.
Q5
If f and g are polynomials of degrees m and n respectively and if \(h(x)=(f\circ g)(x)\), then the degree of h is :
- A. \(m^n\)
- B. mnCorrect
- C. \(n^m\)
- D. m + n
Explanation. The leading term of f is \(ax^m\); substituting g (leading term \(bx^n\)) gives leading term \(ab^m x^{mn}\), so the degree is mn.
Q6
If \(x<0\), then \(\tan^{-1}\left(\frac{1}{x}\right)\) is equal to :
- A. \(-\pi+\cot^{-1}(x)\)Correct
- B. \(\tan^{-1}(x)\)
- C. \(-\pi+\tan^{-1}x\)
- D. \(\cot^{-1}(x)\)
Explanation. For \(x<0\), \(\cot^{-1}x\in(\frac{\pi}{2},\pi)\) while \(\tan^{-1}\frac{1}{x}\in(-\frac{\pi}{2},0)\), so \(\tan^{-1}\frac{1}{x}=\cot^{-1}x-\pi\). Check x = −1: \(-\frac{\pi}{4}=\frac{3\pi}{4}-\pi\).
Q7
The eccentricity of the circle is :
- A. \(\frac{1}{2}\)
- B. 0Correct
- C. 2
- D. 1
Explanation. A circle is an ellipse with a = b, so \(e=\sqrt{1-\frac{b^2}{a^2}}=0\).
Q8
If a vector \(\vec{\alpha}\) lies in the plane of \(\vec{\beta}\) and \(\vec{\gamma}\), then
- A. \([\vec{\alpha},\vec{\beta},\vec{\gamma}]=0\)Correct
- B. \([\vec{\alpha},\vec{\beta},\vec{\gamma}]=1\)
- C. \([\vec{\alpha},\vec{\beta},\vec{\gamma}]=2\)
- D. \([\vec{\alpha},\vec{\beta},\vec{\gamma}]=-1\)
Explanation. Three coplanar vectors have scalar triple product zero, so \([\vec{\alpha},\vec{\beta},\vec{\gamma}]=0\).
Q9
If the image of the point A(1, 2, 3) with respect to the plane \(\vec{r}\cdot(\hat{i}+2\hat{j}+4\hat{k})=38\) is A'(3, 6, 11), then the foot of the perpendicular from the point A to the given plane is :
- A. (2, 5, 7)
- B. (2, 3, 7)
- C. (2, −4, 7)
- D. (2, 4, 7)Correct
Explanation. The foot of the perpendicular is the midpoint of AA': \(\left(\frac{1+3}{2},\frac{2+6}{2},\frac{3+11}{2}\right)=(2,4,7)\). Check: 2 + 8 + 28 = 38.
Q10
One of the closest points on the curve \(x^2-y^2=4\) to the point (6, 0) is :
- A. \((3,\sqrt{5})\)Correct
- B. (2, 0)
- C. \((\sqrt{13},-\sqrt{3})\)
- D. \((\sqrt{5},1)\)
Explanation. \(D^2=(x-6)^2+y^2=(x-6)^2+x^2-4\). \(\frac{d}{dx}D^2=2(x-6)+2x=0\) gives x = 3, so \(y^2=5\); the point is \((3,\sqrt{5})\).
Q11
The value of 'c' satisfied by the Rolle's theorem for the function \(f(x)=x^3-3x^2,\ x\in[0,3]\) is :
- A. \(\frac{3}{2}\)
- B. 1
- C. 2Correct
- D. \(\sqrt{2}\)
Explanation. f(0) = f(3) = 0. \(f'(c)=3c^2-6c=0\) gives c = 0 or 2; the value in (0, 3) is c = 2.
Q12
The percentage error of fifth root of 31 is approximately how many times the percentage error in 31?
- A. 5
- B. \(\frac{1}{31}\)
- C. 31
- D. \(\frac{1}{5}\)Correct
Explanation. If \(y=x^{1/5}\), then \(\frac{dy}{y}=\frac{1}{5}\frac{dx}{x}\), so the percentage error of the fifth root is \(\frac{1}{5}\) times that of x.
Q13
Let \(A=\{(x,y)\mid a<x<b,\ c<y<d\}\subset\mathbb{R}^2\). If the function \(u:A\to\mathbb{R}^2\) is harmonic in A, then :
- A. \(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0\;\forall (x,y)\in A\)Correct
- B. \(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=1\;\forall (x,y)\in A\)
- C. \(\frac{\partial^2 u}{\partial x^2}-\frac{\partial^2 u}{\partial y^2}=0\;\forall (x,y)\in A\)
- D. \(\frac{\partial^2 u}{\partial x^2}-\frac{\partial^2 u}{\partial y^2}=1\;\forall (x,y)\in A\)
Explanation. A harmonic function satisfies Laplace's equation \(u_{xx}+u_{yy}=0\) at every point of A. (The paper prints \(u:A\to\mathbb{R}^2\); the textbook has \(\mathbb{R}\), which does not change the answer.)
Q14
If \(f(x)=\int_0^x t\cos t\,dt\), then \(\frac{df}{dx}=\)
- A. \(x\cos x\)Correct
- B. \(\cos x-x\sin x\)
- C. \(x\sin x\)
- D. \(\sin x+x\cos x\)
Explanation. By the fundamental theorem of calculus, \(\frac{d}{dx}\int_0^x t\cos t\,dt=x\cos x\).
Q15
If \(f(x)=\int_1^x\frac{e^{\sin u}}{u}\,du,\ x>1\) and \(\int_1^3\frac{e^{\sin x^2}}{x}\,dx=\frac{1}{2}[f(a)-f(1)]\), then one of the possible values of a is :
- A. 9Correct
- B. 3
- C. 5
- D. 6
Explanation. Put \(u=x^2\), \(du=2x\,dx\), so \(\frac{dx}{x}=\frac{du}{2u}\); limits 1 to 9. The integral becomes \(\frac{1}{2}\int_1^9\frac{e^{\sin u}}{u}du=\frac{1}{2}[f(9)-f(1)]\), so a = 9.
Q16
The solution of the differential equation \(2x\frac{dy}{dx}-y=3\) represents :
- A. ParabolaCorrect
- B. Straight lines
- C. Ellipse
- D. Circles
Explanation. Separate variables: \(\frac{2\,dy}{y+3}=\frac{dx}{x}\), so \(2\log(y+3)=\log x+\log C\), i.e. \((y+3)^2=Cx\), a family of parabolas.
Q17
P is the amount of certain substance left after time t. If the rate of evaporation of the substance is proportional to the amount remaining, then :
- A. \(P=Ckt\)
- B. \(P=Ce^{kt}\)
- C. \(Pt=C\)
- D. \(P=Ce^{-kt}\)Correct
Explanation. The amount decreases: \(\frac{dP}{dt}=-kP\) (k > 0), so \(\frac{dP}{P}=-k\,dt\) and \(P=Ce^{-kt}\).
Q18
If the function \(f(x)=\frac{1}{12}\) for \(a<x<b\), represents a probability density function of a continuous random variable X, then which of the following cannot be the value of a and b?
- A. 7 and 19
- B. 0 and 12
- C. 16 and 24Correct
- D. 5 and 17
Explanation. Total probability \(\frac{b-a}{12}=1\) needs b − a = 12. For 16 and 24, b − a = 8, so it cannot be a pdf.
Q19
A rod of length \(2l\) is broken into two pieces at random. The probability density function of the shorter of the two pieces is \(f(x)=\begin{cases}\frac{1}{l}, & 0<x<l\\ 0, & l\le x<2l\end{cases}\). The mean and variance of the shorter of the two pieces are respectively :
- A. \(l,\ \frac{l^2}{12}\)
- B. \(\frac{l}{2},\ \frac{l^2}{3}\)
- C. \(\frac{l}{2},\ \frac{l^2}{12}\)Correct
- D. \(\frac{l}{2},\ \frac{l^2}{6}\)
Explanation. X is uniform on (0, l): \(E(X)=\frac{l}{2}\), \(E(X^2)=\frac{l^2}{3}\), so \(Var(X)=\frac{l^2}{3}-\frac{l^2}{4}=\frac{l^2}{12}\).
Q20
The dual of \(\neg(p\vee q)\vee[p\vee(p\wedge\neg r)]\) is :
- A. \(\neg(p\wedge q)\wedge[p\wedge(p\wedge r)]\)
- B. \(\neg(p\wedge q)\wedge[p\vee(p\wedge\neg r)]\)
- C. \(\neg(p\wedge q)\wedge[p\wedge(p\vee\neg r)]\)Correct
- D. \((p\wedge q)\wedge[p\wedge(p\vee\neg r)]\)
Explanation. The dual interchanges \(\vee\) and \(\wedge\) and keeps \(\neg\): \(\neg(p\wedge q)\wedge[p\wedge(p\vee\neg r)]\).
