Q: A variable plane passes through a fixed point $(a, b, c)$ and meets the axes at points $A, B$ and $C$ respectively. Find the locus of the centre of the sphere passing through the points $O, A, B$ and $C$, $O$ being the origin.

Analytical Geometry of the Sphere in Three Dimensions We are tasked with determining the locus of the center of a sphere passing through the origin $O(0, 0, 0)$ and three intercept points $A, B, C$ formed by a variable plane passing through a given fixed point $(a, b, c)$. --- Step 1: Equation of the Variable Plane Let the variable plane intercept the coordinate axes at: - $A$ on the $x$-axis: $(alpha, 0, 0)$ - $B$ on the $y$-axis: $(0, beta, 0)$ - $C$ on the $z$-axis: $(0, 0, gamma)$ where $alpha eq 0, beta eq 0, gamma eq 0$ are the non-zero variable intercepts. The equation of the variable plane in intercept form is: $$frac{x}{alpha} + frac{y}{beta} + frac{z}{gamma} = 1 tag{1}$$ Since this plane is given to pass through the fixed point $(a, b, c)$, the coordinates $(a, b, c)$ must satisfy equation (1): $$frac{a}{alpha} + frac{b}{beta} + frac{c}{gamma} = 1 tag{2}$$ --- Step 2: Equation of the Sphere Passing through $O, A, B, C$ The general equation of a sphere in $mathbb{R}^3$ is: $$x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0 tag{3}$$ with center at $(-u, -v, -w)$ and radius $R = sqrt{u^2 + v^2 + w^2 - d}$. Applying the given four points lying on the sphere: 1. Passes through origin $O(0, 0, 0)$: $$0^2 + 0^2 + 0^2 + 0 + 0 + 0 + d = 0 implies d = 0 tag{4}$$ 2. Passes through $A(alpha, 0, 0)$: $$alpha^2 + 2ualpha = 0 implies alpha(alpha + 2u) = 0$$ Since $alpha eq 0$: $$2u = -alpha implies u = -frac{alpha}{2} tag{5a}$$ 3. Passes through $B(0, beta, 0)$: $$beta^2 + 2vbeta = 0 implies 2v = -beta implies v = -frac{beta}{2} tag{5b}$$ 4. Passes through $C(0, 0, gamma)$: $$gamma^2 + 2wgamma = 0 implies 2w = -gamma implies w = -frac{gamma}{2} tag{5c}$$ Substituting $d = 0$, $2u = -alpha$, $2v = -beta$, and $2w = -gamma$ into (3), the explicit equation of the circumscribing sphere $OABC$ is: $$x^2 + y^2 + z^2 - alpha x - beta y - gamma z = 0 tag{6}$$ --- Step 3: Coordinates of the Sphere's Center Let $(xi, eta, zeta)$ denote the coordinates of the center of the sphere (6). Com

Q: A wire of length $l$ is cut into two parts which are bent in the form of a square and a circle respectively. Using Lagrange's method of undetermined multipliers, find the least value of the sum of the areas so formed.

Mathematical Modeling & Formulation Let a piece of wire of total length $l$ be cut into two portions of lengths $x$ and $y$ such that: $$x + y = l tag{1}$$ where $x ge 0$ is the length of wire used to form a square, and $y ge 0$ is the length of wire used to form a circle. 1. Geometry of the Square: - Perimeter of square $= 4s = x implies text{side length } s = frac{x}{4}$ - Area of the square: $$A1 = s^2 = left( frac{x}{4} right)^2 = frac{x^2}{16} tag{2}$$ 2. Geometry of the Circle: - Circumference of circle $= 2pi r = y implies text{radius } r = frac{y}{2pi}$ - Area of the circle: $$A2 = pi r^2 = pi left( frac{y}{2pi} right)^2 = frac{y^2}{4pi} tag{3}$$ 3. Objective Function: We seek to minimize the total combined area: $$f(x, y) = A1 + A2 = frac{x^2}{16} + frac{y^2}{4pi} tag{4}$$ subject to the equality constraint: $$g(x, y) = x + y - l = 0 tag{5}$$ --- Step 1: Formulation of the Lagrange Function According to Lagrange's method of undetermined multipliers, we introduce a scalar multiplier $lambda$ and construct the Lagrangian function: $$mathcal{L}(x, y, lambda) = f(x, y) - lambda cdot g(x, y)$$ $$mathcal{L}(x, y, lambda) = frac{x^2}{16} + frac{y^2}{4pi} - lambda (x + y - l) tag{6}$$ --- Step 2: First-Order Necessary Conditions (Stationary Points) For stationary points, the partial derivatives of $mathcal{L}$ with respect to $x, y,$ and $lambda$ must vanish: 1. Differentiating with respect to $x$: $$frac{partial mathcal{L}}{partial x} = frac{2x}{16} - lambda = 0 implies frac{x}{8} - lambda = 0 implies x = 8lambda tag{7a}$$ 2. Differentiating with respect to $y$: $$frac{partial mathcal{L}}{partial y} = frac{2y}{4pi} - lambda = 0 implies frac{y}{2pi} - lambda = 0 implies y = 2pilambda tag{7b}$$ 3. Differentiating with respect to $lambda$: $$frac{partial mathcal{L}}{partial lambda} = -(x + y - l) = 0 implies x + y = l tag{7c}$$ --- Step 3: Determining the Optimal Multiplier $lambda$ Substitute equations (7a) and (7b) into the constraint equation (7c): $$8lambda + 2pi

Question 1

Prove that any set of $n$ linearly independent vectors in a vector space $V$ of dimension $n$ constitutes a basis for $V$.

Accepted Answer

Theoretical Principles and Fundamental Definitions Let $V$ be a finite-dimensional vector space over a field $mathbb{F}$ (such as $mathbb{R}$ or $mathbb{C}$). 1. Linear Independence: A finite set of vectors $S = {v1, v2, dots, vn} subset V$ is said to be linearly independent if the only linear combination of these vectors that equals the zero vector is the trivial one: $$sum{i=1}^n ci vi = c1 v1 + c2 v2 + dots + cn vn = mathbf{0} implies c1 = c2 = dots = cn = 0 tag{1}$$ 2. Spanning Set: The set $S$ is said to span $V$ (denoted $operatorname{span}(S) = V$) if every arbitrary vector $v in V$ can be expressed as a linear combination of vectors in $S$: $$forall v in V, quad exists alpha1, alpha2, dots, alphan in mathbb{F} quad text{such that} quad v = sum{i=1}^n alphai vi tag{2}$$ 3. Basis: A subset $B subset V$ is a basis of $V$ if and only if: - $B$ is linearly independent, and - $B$ spans $V$ ($operatorname{span}(B) = V$). 4. Dimension: The dimension of a finite-dimensional vector space $V$, denoted $dim(V) = n$, is defined as the cardinal number of vectors in any basis of $V$. --- Statement of the Theorem to be Proven Theorem: Let $V$ be a vector space of dimension $n$ over a field $mathbb{F}$ ($dim(V) = n$). If $S = {v1, v2, dots, vn}$ is a set of $n$ linearly independent vectors in $V$, then $S$ constitutes a basis for $V$. --- Lemma Required for the Proof (Steinitz Extension / Linear Dependence) Lemma: If $S = {v1, v2, dots, vn}$ is a linearly independent subset of $V$ and $v in V$ is any vector such that $v otin operatorname{span}(S)$, then the extended set: $$S' = S cup {v} = {v1, v2, dots, vn, v}$$ is also linearly independent. Proof of Lemma: Suppose there exist scalars $c1, c2, dots, cn, c in mathbb{F}$ such that: $$c1 v1 + c2 v2 + dots + cn vn + c v = mathbf{0} tag{3}$$ We show that $c = 0$ and $c1 = c2 = dots = cn = 0$: - If $c eq 0$, then its multiplicative inverse $c^{-1}$ exists in the field $mathbb{F}$. Multiplying (3) by $-c^{-1}$ yields: $$v = -frac{

Question 2

Let $T : mathbb{R}^2 to mathbb{R}^3$ be a linear transformation such that $Tbegin{pmatrix} 1 0 end{pmatrix} = begin{pmatrix} 1 2 3 end{pmatrix}$ and $Tbegin{pmatrix} 1 1 end{pmatrix} = begin{pmatrix} -3 2 8 end{pmatrix}$. Find $Tbegin{pmatrix} 2 4 end{pmatrix}$.

Accepted Answer

Theoretical Principles of Linear Transformations A mapping $T : V to W$ between two real vector spaces is a linear transformation if and only if it preserves vector addition and scalar multiplication: $$T(alpha u + beta v) = alpha T(u) + beta T(v) quad forall u, v in V, ; forall alpha, beta in mathbb{R} tag{1}$$ By the fundamental theorem of linear transformations, if $B = {v1, v2}$ is an ordered basis of a finite-dimensional domain space $V$, then a linear transformation $T$ is uniquely and completely determined by its action on the basis vectors. --- Step 1: Verification of Linear Independence of Given Vectors Let: $$v1 = begin{pmatrix} 1 0 end{pmatrix}, quad v2 = begin{pmatrix} 1 1 end{pmatrix}$$ Consider the determinant formed by $v1$ and $v2$: $$begin{vmatrix} 1 & 1 0 & 1 end{vmatrix} = (1)(1) - (0)(1) = 1 eq 0$$ Since the determinant is non-zero, ${v1, v2}$ is linearly independent in $mathbb{R}^2$. Since $dim(mathbb{R}^2) = 2$, the set $B = {v1, v2}$ forms a basis for $mathbb{R}^2$. We are given the images under $T$: $$T(v1) = Tbegin{pmatrix} 1 0 end{pmatrix} = begin{pmatrix} 1 2 3 end{pmatrix} tag{2a}$$ $$T(v2) = Tbegin{pmatrix} 1 1 end{pmatrix} = begin{pmatrix} -3 2 8 end{pmatrix} tag{2b}$$ --- Step 2: Method 1 — Linear Combination of Basis Vectors We are required to compute $T(w)$ where: $$w = begin{pmatrix} 2 4 end{pmatrix}$$ Express $w$ as a unique linear combination of the basis vectors $v1$ and $v2$: $$w = c1 v1 + c2 v2$$ $$begin{pmatrix} 2 4 end{pmatrix} = c1 begin{pmatrix} 1 0 end{pmatrix} + c2 begin{pmatrix} 1 1 end{pmatrix} = begin{pmatrix} c1 + c2 c2 end{pmatrix} tag{3}$$ Equating corresponding components: 1. From the second row: $$c2 = 4 tag{4a}$$ 2. From the first row: $$c1 + c2 = 2 implies c1 + 4 = 2 implies c1 = 2 - 4 = -2 tag{4b}$$ Thus, the vector $w$ is represented as: $$begin{pmatrix} 2 4 end{pmatrix} = -2 begin{pmatrix} 1 0 end{pmatrix} + 4 begin{pmatrix} 1 1 end{pmatrix} tag{5}$$ Now, apply the linearity property (1) of $T$: $$Tbegin{pmat

Question 3

Evaluate $lim{x to infty} (e^x + x)^{frac{1}{x}}$.

Accepted Answer

Mathematical Analysis of Indeterminate Forms We are required to evaluate the limiting value of the real-valued function: $$L = lim{x to infty} f(x) = lim{x to infty} (e^x + x)^{frac{1}{x}} tag{1}$$ Initial Form Inspection: As $x to infty$: - Base: $e^x + x to infty$ - Exponent: $frac{1}{x} to 0$ Thus, the limit assumes the classical indeterminate form of type: $$boldsymbol{infty^0}$$ To evaluate indeterminate powers of the form $infty^0$, we employ the logarithmic transformation technique combined with asymptotic expansions and the Squeeze Theorem. --- Method 1: Logarithmic Transformation and L'Hôpital's Rule Let: $$y = (e^x + x)^{frac{1}{x}} tag{2}$$ Taking the natural logarithm of both sides for $x 0$: $$ln y = lnleft[ (e^x + x)^{frac{1}{x}} right] = frac{1}{x} ln(e^x + x) = frac{ln(e^x + x)}{x} tag{3}$$ Now evaluate the limit of $ln y$ as $x to infty$: $$lim{x to infty} ln y = lim{x to infty} frac{ln(e^x + x)}{x} tag{4}$$ Dominant Term Factoring: In the argument of the logarithm, $e^x$ grows exponentially faster than the linear term $x$. We factor out the dominant term $e^x$: $$e^x + x = e^x left( 1 + frac{x}{e^x} right) tag{5}$$ Using the logarithmic identity $ln(A cdot B) = ln A + ln B$: $$ln(e^x + x) = lnleft[ e^x left( 1 + x e^{-x} right) right] = ln(e^x) + lnleft( 1 + frac{x}{e^x} right)$$ $$ln(e^x + x) = x + lnleft( 1 + frac{x}{e^x} right) tag{6}$$ Substitute equation (6) back into the numerator of (4): $$lim{x to infty} ln y = lim{x to infty} frac{x + lnleft( 1 + frac{x}{e^x} right)}{x}$$ $$lim{x to infty} ln y = lim{x to infty} left[ 1 + frac{lnleft( 1 + frac{x}{e^x} right)}{x} right] tag{7}$$ Evaluating the Component Limits: First, examine the growth of $x/e^x$ as $x to infty$ using L'Hôpital's rule ($frac{infty}{infty}$ form): $$lim{x to infty} frac{x}{e^x} overset{text{H}}{=} lim{x to infty} frac{frac{d}{dx}[x]}{frac{d}{dx}[e^x]} = lim{x to infty} frac{1}{e^x} = 0 tag{8}$$ Consequently, by the continuity of the natural logarithm: $$lim{x to infty} lnle

Question 4

Examine the convergence of $int0^2 frac{dx}{(2x - x^2)}$.

Accepted Answer

Classification of the Improper Integral We are required to examine the convergence of the definite integral: $$I = int0^2 frac{dx}{2x - x^2} tag{1}$$ Let the integrand function be: $$f(x) = frac{1}{2x - x^2} = frac{1}{x(2 - x)} tag{2}$$ Singularity Analysis: The integrand $f(x)$ is continuous on the open interval $(0, 2)$, but it becomes unbounded at the boundary points: - As $x to 0^+$: $$lim{x to 0^+} f(x) = lim{x to 0^+} frac{1}{x(2 - x)} = frac{1}{0^+ cdot 2} = +infty$$ - As $x to 2^-$: $$lim{x to 2^-} f(x) = lim{x to 2^-} frac{1}{x(2 - x)} = frac{1}{2 cdot 0^+} = +infty$$ Since the integrand has infinite discontinuities at both the lower limit $x = 0$ and the upper limit $x = 2$, integral (1) is an improper integral of the second kind. --- Step 1: Decomposition at an Interior Point According to the Cauchy definition for improper integrals with multiple singularities, we choose an arbitrary interior point $c in (0, 2)$, conveniently chosen as $c = 1$: $$I = int0^2 frac{dx}{x(2 - x)} = int0^1 frac{dx}{x(2 - x)} + int1^2 frac{dx}{x(2 - x)} = I1 + I2 tag{3}$$ By definition, the integral $I$ converges if and only if both $I1$ and $I2$ converge independently. If either $I1$ or $I2$ diverges, the original integral $I$ is divergent. --- Step 2: Testing Convergence of $I1 = int0^1 frac{dx}{x(2 - x)}$ at $x = 0$ In $I1$, the point of infinite discontinuity is $x = 0$. Limit Comparison Test: Compare $f(x)$ with the standard $p$-integral test function: $$g(x) = frac{1}{x^p} = frac{1}{x} quad (text{taking } p = 1)$$ Compute the limit of the ratio as $x to 0^+$: $$lim{x to 0^+} frac{f(x)}{g(x)} = lim{x to 0^+} frac{frac{1}{x(2 - x)}}{frac{1}{x}} = lim{x to 0^+} frac{1}{2 - x} = frac{1}{2 - 0} = frac{1}{2} tag{4}$$ Since $0 < frac{1}{2} < infty$ (a non-zero finite positive real number), the Limit Comparison Test asserts that the two integrals $int0^1 f(x) dx$ and $int0^1 g(x) dx$ either both converge or both diverge. We know from standard analysis that the reference $p$-integ

Question 5

A variable plane passes through a fixed point $(a, b, c)$ and meets the axes at points $A, B$ and $C$ respectively. Find the locus of the centre of the sphere passing through the points $O, A, B$ and $C$, $O$ being the origin.

Accepted Answer

Analytical Geometry of the Sphere in Three Dimensions We are tasked with determining the locus of the center of a sphere passing through the origin $O(0, 0, 0)$ and three intercept points $A, B, C$ formed by a variable plane passing through a given fixed point $(a, b, c)$. --- Step 1: Equation of the Variable Plane Let the variable plane intercept the coordinate axes at: - $A$ on the $x$-axis: $(alpha, 0, 0)$ - $B$ on the $y$-axis: $(0, beta, 0)$ - $C$ on the $z$-axis: $(0, 0, gamma)$ where $alpha eq 0, beta eq 0, gamma eq 0$ are the non-zero variable intercepts. The equation of the variable plane in intercept form is: $$frac{x}{alpha} + frac{y}{beta} + frac{z}{gamma} = 1 tag{1}$$ Since this plane is given to pass through the fixed point $(a, b, c)$, the coordinates $(a, b, c)$ must satisfy equation (1): $$frac{a}{alpha} + frac{b}{beta} + frac{c}{gamma} = 1 tag{2}$$ --- Step 2: Equation of the Sphere Passing through $O, A, B, C$ The general equation of a sphere in $mathbb{R}^3$ is: $$x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0 tag{3}$$ with center at $(-u, -v, -w)$ and radius $R = sqrt{u^2 + v^2 + w^2 - d}$. Applying the given four points lying on the sphere: 1. Passes through origin $O(0, 0, 0)$: $$0^2 + 0^2 + 0^2 + 0 + 0 + 0 + d = 0 implies d = 0 tag{4}$$ 2. Passes through $A(alpha, 0, 0)$: $$alpha^2 + 2ualpha = 0 implies alpha(alpha + 2u) = 0$$ Since $alpha eq 0$: $$2u = -alpha implies u = -frac{alpha}{2} tag{5a}$$ 3. Passes through $B(0, beta, 0)$: $$beta^2 + 2vbeta = 0 implies 2v = -beta implies v = -frac{beta}{2} tag{5b}$$ 4. Passes through $C(0, 0, gamma)$: $$gamma^2 + 2wgamma = 0 implies 2w = -gamma implies w = -frac{gamma}{2} tag{5c}$$ Substituting $d = 0$, $2u = -alpha$, $2v = -beta$, and $2w = -gamma$ into (3), the explicit equation of the circumscribing sphere $OABC$ is: $$x^2 + y^2 + z^2 - alpha x - beta y - gamma z = 0 tag{6}$$ --- Step 3: Coordinates of the Sphere's Center Let $(xi, eta, zeta)$ denote the coordinates of the center of the sphere (6). Com

Question 6

Find all solutions to the following system of equations by row-reduced method: $$begin{aligned} x1 + 2x2 - x3 &= 2 2x1 + 3x2 + 5x3 &= 5 -x1 - 3x2 + 8x3 &= -1 end{aligned}$$

Accepted Answer

Systematic Solution via Gaussian Elimination (Row-Reduced Method) We consider the non-homogeneous system of three linear algebraic equations in three unknowns $x1, x2, x3$: $$begin{aligned} x1 + 2x2 - x3 &= 2 tag{1} 2x1 + 3x2 + 5x3 &= 5 tag{2} -x1 - 3x2 + 8x3 &= -1 tag{3} end{aligned}$$ In matrix notation $A x = b$: $$begin{pmatrix} 1 & 2 & -1 2 & 3 & 5 -1 & -3 & 8 end{pmatrix} begin{pmatrix} x1 x2 x3 end{pmatrix} = begin{pmatrix} 2 5 -1 end{pmatrix} tag{4}$$ The augmented matrix $[A mid b]$ of the system is: $$[A mid b] = begin{pmatrix} 1 & 2 & -1 & bigm| & 2 2 & 3 & 5 & bigm| & 5 -1 & -3 & 8 & bigm| & -1 end{pmatrix} tag{5}$$ We now apply elementary row operations to reduce $[A mid b]$ to row echelon form and reduced row echelon form (RREF). --- Step 1: Forward Elimination to Row Echelon Form Operation 1: Eliminate $x1$ from Rows 2 and 3 The leading pivot in Row 1 is already $a{11} = 1$. - Perform $R2 to R2 - 2R1$: $$begin{aligned} R2 - 2R1 &= (2 - 2(1), ; 3 - 2(2), ; 5 - 2(-1) mid 5 - 2(2)) &= (0, ; -1, ; 7 mid 1) end{aligned}$$ - Perform $R3 to R3 + R1$: $$begin{aligned} R3 + R1 &= (-1 + 1, ; -3 + 2, ; 8 + (-1) mid -1 + 2) &= (0, ; -1, ; 7 mid 1) end{aligned}$$ The augmented matrix becomes: $$begin{pmatrix} 1 & 2 & -1 & bigm| & 2 0 & -1 & 7 & bigm| & 1 0 & -1 & 7 & bigm| & 1 end{pmatrix} tag{6}$$ --- Operation 2: Eliminate $x2$ from Row 3 Notice that Row 2 and Row 3 are identical. - Perform $R3 to R3 - R2$: $$begin{aligned} R3 - R2 &= (0 - 0, ; -1 - (-1), ; 7 - 7 mid 1 - 1) &= (0, ; 0, ; 0 mid 0) end{aligned}$$ The augmented matrix is now in Row Echelon Form: $$begin{pmatrix} 1 & 2 & -1 & bigm| & 2 0 & -1 & 7 & bigm| & 1 0 & 0 & 0 & bigm| & 0 end{pmatrix} tag{7}$$ --- Step 2: Rank Analysis & Consistency (Rouché-Capelli Theorem) From the echelon matrix (7): - Rank of coefficient matrix $A$: Number of non-zero rows in $A$ is $operatorname{rank}(A) = 2$. - Rank of augmented matrix $[A mid b]$: Number of non-zero rows in $[A mid b]$ is $operatorname{rank}([A mid b])

Question 7

A wire of length $l$ is cut into two parts which are bent in the form of a square and a circle respectively. Using Lagrange's method of undetermined multipliers, find the least value of the sum of the areas so formed.

Accepted Answer

Mathematical Modeling & Formulation Let a piece of wire of total length $l$ be cut into two portions of lengths $x$ and $y$ such that: $$x + y = l tag{1}$$ where $x ge 0$ is the length of wire used to form a square, and $y ge 0$ is the length of wire used to form a circle. 1. Geometry of the Square: - Perimeter of square $= 4s = x implies text{side length } s = frac{x}{4}$ - Area of the square: $$A1 = s^2 = left( frac{x}{4} right)^2 = frac{x^2}{16} tag{2}$$ 2. Geometry of the Circle: - Circumference of circle $= 2pi r = y implies text{radius } r = frac{y}{2pi}$ - Area of the circle: $$A2 = pi r^2 = pi left( frac{y}{2pi} right)^2 = frac{y^2}{4pi} tag{3}$$ 3. Objective Function: We seek to minimize the total combined area: $$f(x, y) = A1 + A2 = frac{x^2}{16} + frac{y^2}{4pi} tag{4}$$ subject to the equality constraint: $$g(x, y) = x + y - l = 0 tag{5}$$ --- Step 1: Formulation of the Lagrange Function According to Lagrange's method of undetermined multipliers, we introduce a scalar multiplier $lambda$ and construct the Lagrangian function: $$mathcal{L}(x, y, lambda) = f(x, y) - lambda cdot g(x, y)$$ $$mathcal{L}(x, y, lambda) = frac{x^2}{16} + frac{y^2}{4pi} - lambda (x + y - l) tag{6}$$ --- Step 2: First-Order Necessary Conditions (Stationary Points) For stationary points, the partial derivatives of $mathcal{L}$ with respect to $x, y,$ and $lambda$ must vanish: 1. Differentiating with respect to $x$: $$frac{partial mathcal{L}}{partial x} = frac{2x}{16} - lambda = 0 implies frac{x}{8} - lambda = 0 implies x = 8lambda tag{7a}$$ 2. Differentiating with respect to $y$: $$frac{partial mathcal{L}}{partial y} = frac{2y}{4pi} - lambda = 0 implies frac{y}{2pi} - lambda = 0 implies y = 2pilambda tag{7b}$$ 3. Differentiating with respect to $lambda$: $$frac{partial mathcal{L}}{partial lambda} = -(x + y - l) = 0 implies x + y = l tag{7c}$$ --- Step 3: Determining the Optimal Multiplier $lambda$ Substitute equations (7a) and (7b) into the constraint equation (7c): $$8lambda + 2pi

Maths Optional 2022 Paper I

7 Questions Available

Maths Optional 2022 Paper I

7 Questions Available

Question 1: Prove that any set of $n$ linearly independent vectors in a vector space $V$ of dimension $n$ constitutes a basis for $V$.

Question 2: Let $T : mathbb{R}^2 to mathbb{R}^3$ be a linear transformation such that $Tbegin{pmatrix} 1 0 end{pmatrix} = begin{pmatrix} 1 2 3 end{pmatrix}$ and $

Question 3: Evaluate $lim{x to infty} (e^x + x)^{frac{1}{x}}$.

Question 4: Examine the convergence of $int0^2 frac{dx}{(2x - x^2)}$.

Question 5: A variable plane passes through a fixed point $(a, b, c)$ and meets the axes at points $A, B$ and $C$ respectively. Find the locus of the centre of th

Question 6: Find all solutions to the following system of equations by row-reduced method: $$begin{aligned} x1 + 2x2 - x3 &= 2 2x1 + 3x2 + 5x3 &= 5 -x1 - 3x2 + 8x

Question 7: A wire of length $l$ is cut into two parts which are bent in the form of a square and a circle respectively. Using Lagrange's method of undetermined m