Free UPSC Mathematics Optional Paper-1 2024 Solutions: View Online | UPSC Maths Solution | IAS Maths Solution

July 13, 2025ByAbstract Classes

Question:-1(a)

Let $H$ be a subspace of $R^{4}$ spanned by the vectors $v_{1} = (1, - 2, 5, - 3)$ , $v_{2} = (2, 3, 1, - 4), v_{3} = (3, 8, - 3, - 5)$ . Then find a basis and dimension of $H$ , and extend the basis of $H$ to a basis of $R^{4}$ .

Answer:

We are given a subspace

H \subseteq R^{4}

spanned by:

v_{1} = (1, - 2, 5, - 3), v_{2} = (2, 3, 1, - 4), v_{3} = (3, 8, - 3, - 5)

Step 1: Find the Basis and Dimension of $H$

We check the linear independence of

v_{1}, v_{2}, v_{3}

by row reducing the matrix formed by these vectors as rows or columns. Let’s form a matrix with vectors as rows:

A = [\begin{matrix} 1 & - 2 & 5 & - 3 \\ 2 & 3 & 1 & - 4 \\ 3 & 8 & - 3 & - 5 \end{matrix}]

Perform row reduction:

Step 1: Make the first pivot 1 (already done)

R_{1} = (1, - 2, 5, - 3)

Step 2: Eliminate below using $R_{1}$

R_{2} = R_{2} - 2 R_{1} = (2, 3, 1, - 4) - 2 (1, - 2, 5, - 3) = (0, 7, - 9, 2)

R_{3} = R_{3} - 3 R_{1} = (3, 8, - 3, - 5) - 3 (1, - 2, 5, - 3) = (0, 14, - 18, 4)

Matrix becomes:

[\begin{matrix} 1 & - 2 & 5 & - 3 \\ 0 & 7 & - 9 & 2 \\ 0 & 14 & - 18 & 4 \end{matrix}]

Step 3: Eliminate below using $R_{2}$

R_{3} = R_{3} - 2 R_{2} = (0, 14, - 18, 4) - 2 (0, 7, - 9, 2) = (0, 0, 0, 0)

Now the matrix is:

[\begin{matrix} 1 & - 2 & 5 & - 3 \\ 0 & 7 & - 9 & 2 \\ 0 & 0 & 0 & 0 \end{matrix}]

This matrix has 2 non-zero rows, so the rank is 2.

✅ Conclusion:

Dimension of $H$ = 2
Basis of $H$ = Any two linearly independent vectors. We can take:
${v_{1} = (1, - 2, 5, - 3), v_{2} = (2, 3, 1, - 4)}$

Step 2: Extend this Basis to a Basis for $R^{4}$

To extend this basis to

R^{4}

, two additional vectors are needed. The standard basis vectors

e_{1} =

(1, 0, 0, 0)

and

e_{2} = (0, 1, 0, 0)

are checked for linear independence from the basis of

H

$e_{1}$ is not in the span of $v_{1}$ and $v_{2}$ (since $a v_{1} + b v_{2} = e_{1}$ leads to a contradiction).
The set ${v_{1}, v_{2}, e_{1}}$ is linearly independent.
$e_{2}$ is not in the span of ${v_{1}, v_{2}, e_{1}}$ (since $a v_{1} + b v_{2} + c e_{1} = e_{2}$ leads to a contradiction).
The set ${v_{1}, v_{2}, e_{1}, e_{2}}$ is linearly independent.

Thus, a basis for

R^{4}

{(1, - 2, 5, - 3), (2, 3, 1, - 4), (1, 0, 0, 0), (0, 1, 0, 0)}

Basis of H : {(\begin{matrix} 1 \\ - 2 \\ 5 \\ - 3 \end{matrix}), (\begin{matrix} 2 \\ 3 \\ 1 \\ - 4 \end{matrix})}

Dimension of

H : 2

Extended basis of

R^{4} : {(\begin{matrix} 1 \\ - 2 \\ 5 \\ - 3 \end{matrix}), (\begin{matrix} 2 \\ 3 \\ 1 \\ - 4 \end{matrix}), (\begin{array}{l} 1 \\ 0 \\ 0 \\ 0 \end{array}), (\begin{array}{l} 0 \\ 1 \\ 0 \\ 0 \end{array})}

Question:-1(b)

Let $T : R^{3} \to R^{3}$ be a linear operator and $B = {v_{1}, v_{2}, v_{3}}$ be a basis of $R^{3}$ over $R$ . Suppose that $T v_{1} = (1, 1, 0), T v_{2} = (1, 0, - 1), T v_{3} = (2, 1, - 1)$ . Find a basis for the range space and null space of $T$ .

Answer:

To find a basis for the range space and null space of the linear operator

T : R^{3} \to R^{3}

, given the action of

T

on the basis

B = {v_{1}, v_{2}, v_{3}}

R^{3}

, where:

T v_{1} = (1, 1, 0), T v_{2} = (1, 0, - 1), T v_{3} = (2, 1, - 1),

we proceed as follows.

Step 1: Find the Matrix Representation of $T$

Since

B = {v_{1}, v_{2}, v_{3}}

is a basis for

R^{3}

, and the vectors

T v_{1}, T v_{2}, T v_{3}

are given in standard coordinates, we assume the outputs are expressed in the standard basis

{e_{1}, e_{2}, e_{3}}

, where

e_{1} = (1, 0, 0)

e_{2} = (0, 1, 0)

e_{3} = (0, 0, 1)

. The matrix of

T

with respect to the basis

B

(for the domain) and the standard basis (for the codomain) is formed by taking the coordinates of

T v_{i}

$T v_{1} = (1, 1, 0)$
$T v_{2} = (1, 0, - 1)$
$T v_{3} = (2, 1, - 1)$

The matrix

A

T

has columns given by

T v_{1}, T v_{2}, T v_{3}

A = [\begin{matrix} 1 & 1 & 2 \\ 1 & 0 & 1 \\ 0 & - 1 & - 1 \end{matrix}]

Step 2: Basis for the Range Space

The range space of

T

, denoted

Range (T)

, is the span of the images of the basis vectors, i.e.,

span {T v_{1}, T v_{2}, T v_{3}}

, which corresponds to the column space of

A

. We need a basis for the column space, so we check if the columns are linearly independent by row reducing

A

Row Reduction of $A$ :

A = [\begin{matrix} 1 & 1 & 2 \\ 1 & 0 & 1 \\ 0 & - 1 & - 1 \end{matrix}]

R2 = R2 – R1:

[\begin{matrix} 1 & 1 & 2 \\ 0 & - 1 & - 1 \\ 0 & - 1 & - 1 \end{matrix}]

R3 = R3 – R2 (not necessary, as R3 = R2), set R3 = 0:

[\begin{matrix} 1 & 1 & 2 \\ 0 & - 1 & - 1 \\ 0 & 0 & 0 \end{matrix}]

R2 = -R2:

[\begin{matrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{matrix}]

R1 = R1 – R2:

[\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{matrix}]

The row-reduced echelon form shows two pivot columns (columns 1 and 2), indicating that the first two columns of

A

are linearly independent, and the third column is a linear combination of the first two. Check the dependency:

Column 3: $(2, 1, - 1)$
Columns 1 and 2: $(1, 1, 0)$ , $(1, 0, - 1)$

Solve

a (1, 1, 0) + b (1, 0, - 1) = (2, 1, - 1)

{\begin{cases} a + b = 2 \\ a = 1 \\ - b = - 1 \end{cases}

From the third equation,

b = 1

. From the second,

a = 1

. Check the first:

1 + 1 = 2

, which holds. Thus:

(2, 1, - 1) = (1, 1, 0) + (1, 0, - 1)

So,

Range (T) = span {(1, 1, 0), (1, 0, - 1)}

, and since the first two columns are independent (as confirmed by the pivots),

{(1, 1, 0), (1, 0, - 1)}

is a basis for the range space.

Dimension Check: The rank of

A

is 2 (two pivots), so

\dim (Range (T)) = 2

Step 3: Basis for the Null Space

The null space of

T

, denoted

Null (T)

, consists of vectors

x = (x_{1}, x_{2}, x_{3})

in the basis

B

, such that

T (x) = 0

. Express

x = x_{1} v_{1} + x_{2} v_{2} + x_{3} v_{3}

, so:

T (x) = x_{1} T v_{1} + x_{2} T v_{2} + x_{3} T v_{3} = x_{1} (1, 1, 0) + x_{2} (1, 0, - 1) + x_{3} (2, 1, - 1) = 0

Compute:

(x_{1} + x_{2} + 2 x_{3}, x_{1} + x_{3}, - x_{2} - x_{3}) = (0, 0, 0)

This gives the system:

{\begin{cases} x_{1} + x_{2} + 2 x_{3} = 0 \\ x_{1} + x_{3} = 0 \\ - x_{2} - x_{3} = 0 \end{cases}

Solve:

From the third equation: $x_{2} = - x_{3}$ .
From the second: $x_{1} = - x_{3}$ .
Substitute into the first: $(- x_{3}) + (- x_{3}) + 2 x_{3} = - x_{3} - x_{3} + 2 x_{3} = 0$ , which is satisfied.

Let

x_{3} = t

, then

x_{1} = - t

x_{2} = - t

. The solution is:

x = (- t, - t, t) = t (- 1, - 1, 1)

Thus, the null space is spanned by

(- 1, - 1, 1)

(in coordinates relative to basis

B

). To confirm, check if the vector is non-zero and satisfies the equation. Test

(- 1, - 1, 1)

T (- v_{1} - v_{2} + v_{3}) = - T v_{1} - T v_{2} + T v_{3} = - (1, 1, 0) - (1, 0, - 1) + (2, 1, - 1) = (- 1 - 1 + 2, - 1 - 0 + 1, 0 + 1 - 1) = (0, 0, 0)

This confirms the vector is in the null space. Since

\dim (Null (T)) = 3 - rank (A) = 3 - 2 = 1

, the vector

(- 1, - 1, 1)

forms a basis for the null space (in coordinates relative to

B

Step 4: Express Null Space Basis in Standard Coordinates (if needed)

The null space basis is given in coordinates

(x_{1}, x_{2}, x_{3})

relative to

B

, meaning the vector is

- v_{1} - v_{2} + v_{3}

. If the basis vectors

v_{1}, v_{2}, v_{3}

were given in standard coordinates, we would express

- v_{1} - v_{2} + v_{3}

accordingly, but since

B

is an arbitrary basis, we keep the null space basis as

{- v_{1} - v_{2} + v_{3}}

Final Answer

Basis for the range space: ${(1, 1, 0), (1, 0, - 1)}$
Basis for the null space: ${- v_{1} - v_{2} + v_{3}}$

Range space basis: {(1, 1, 0), (1, 0, - 1)}, Null space basis: {- v_{1} - v_{2} + v_{3}}

Question:-1(c)

Discuss the continuity of the function

f (x) = {\begin{cases} \frac{1}{1 - e^{- 1 / x}}, & x \neq 0 \\ 0, & x = 0 \end{cases}

for all values of

x

Answer:

To determine the continuity of the function

f (x) = {\begin{cases} \frac{1}{1 - e^{- 1 / x}}, & x \neq 0 \\ 0, & x = 0 \end{cases}

for all values of

x

, we analyze its behavior at different points, particularly focusing on

x = 0

1. Continuity for $x \neq 0$ :

For

x \neq 0

, the function

f (x) = \frac{1}{1 - e^{- 1 / x}}

is a composition of continuous functions:

$e^{- 1 / x}$ is continuous for all $x \neq 0$ .
The denominator $1 - e^{- 1 / x}$ is continuous and non-zero for $x \neq 0$ , since $e^{- 1 / x} \neq 1$ when $x \neq 0$ .

Thus,

f (x)

is continuous for all

x \neq 0

2. Continuity at $x = 0$ :

To check continuity at

x = 0

, we need to verify:

lim_{x \to 0} f (x) = f (0) = 0.

Case 1: $x \to 0^{+}$ (Right-Hand Limit)

As $x \to 0^{+}$ , $\frac{1}{x} \to + \infty$ , so $e^{- 1 / x} \to 0$ .
Thus: $lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} \frac{1}{1 - e^{- 1 / x}} = \frac{1}{1 - 0} = 1.$

Case 2: $x \to 0^{-}$ (Left-Hand Limit)

As $x \to 0^{-}$ , $\frac{1}{x} \to - \infty$ , so $e^{- 1 / x} \to + \infty$ .
Thus: $lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} \frac{1}{1 - e^{- 1 / x}} = \frac{1}{1 - \infty} = 0.$

Conclusion at $x = 0$ :

Since the left-hand limit (

0

) and the right-hand limit (

1

) are not equal, the limit

lim_{x \to 0} f (x)

does not exist. Therefore,

f (x)

is not continuous at

x = 0

3. Summary of Continuity:

Continuous for all $x \neq 0$ .
Discontinuous at $x = 0$ because the limit does not exist (left and right limits disagree).

Final Answer:

{\begin{cases} Continuous for all x \neq 0, \\ Discontinuous at x = 0 (limit does not exist). \end{cases}

Question:-1(d)

Expand $\ln (x)$ in powers of $(x - 1)$ by Taylor’s theorem and hence find the value of $\ln (1 \cdot 1)$ correct up to four decimal places.

Answer:

To solve the problem of expanding

\ln (x)

in powers of

(x - 1)

using Taylor’s theorem and finding the value of

\ln (1.1)

correct to four decimal places, we proceed step by step.

Step 1: Understand Taylor’s Theorem

Taylor’s theorem allows us to expand a function

f (x)

around a point

a

as an infinite series:

f (x) = f (a) + f^{'} (a) (x - a) + \frac{f^{″} (a)}{2!} (x - a)^{2} + \frac{f^{‴} (a)}{3!} (x - a)^{3} + \dots

Here, we need to expand

\ln (x)

around

x = 1

, so

a = 1

, and the series will be in powers of

(x - 1)

Step 2: Compute the Function and Its Derivatives at $x = 1$

Let’s compute the value of

\ln (x)

and its derivatives at

x = 1

Function value:
$f (x) = \ln (x), f (1) = \ln (1) = 0$
First derivative:
$f^{'} (x) = \frac{d}{d x} \ln (x) = \frac{1}{x}, f^{'} (1) = \frac{1}{1} = 1$
Second derivative:
$f^{″} (x) = \frac{d}{d x} (\frac{1}{x}) = - \frac{1}{x^{2}}, f^{″} (1) = - \frac{1}{1^{2}} = - 1$
Third derivative:
$f^{‴} (x) = \frac{d}{d x} (- \frac{1}{x^{2}}) = \frac{2}{x^{3}}, f^{‴} (1) = \frac{2}{1^{3}} = 2$
Fourth derivative:
$f^{(4)} (x) = \frac{d}{d x} (\frac{2}{x^{3}}) = - \frac{6}{x^{4}}, f^{(4)} (1) = - \frac{6}{1^{4}} = - 6$
Fifth derivative:
$f^{(5)} (x) = \frac{d}{d x} (- \frac{6}{x^{4}}) = \frac{24}{x^{5}}, f^{(5)} (1) = \frac{24}{1^{5}} = 24$

To generalize, let’s find the pattern for the

n

-th derivative. For

n \geq 1

First derivative: $f^{'} (x) = x^{- 1}$
Second derivative: $f^{″} (x) = - x^{- 2}$
Third derivative: $f^{‴} (x) = 2 x^{- 3} = 2 \cdot 1 \cdot x^{- 3}$
Fourth derivative: $f^{(4)} (x) = - 6 x^{- 4} = - 3 \cdot 2 \cdot 1 \cdot x^{- 4}$
Fifth derivative: $f^{(5)} (x) = 24 x^{- 5} = 4 \cdot 3 \cdot 2 \cdot 1 \cdot x^{- 5}$

The

n

-th derivative appears to be:

f^{(n)} (x) = (- 1)^{n - 1} (n - 1)! x^{- n}

Evaluate at

x = 1

f^{(n)} (1) = (- 1)^{n - 1} (n - 1)! \cdot 1^{- n} = (- 1)^{n - 1} (n - 1)!

Step 3: Construct the Taylor Series

The Taylor series for

\ln (x)

around

x = 1

is:

\ln (x) = f (1) + f^{'} (1) (x - 1) + \frac{f^{″} (1)}{2!} (x - 1)^{2} + \frac{f^{‴} (1)}{3!} (x - 1)^{3} + \frac{f^{(4)} (1)}{4!} (x - 1)^{4} + \dots

Substitute the derivatives:

$f (1) = 0$
$f^{'} (1) = 1$
$f^{″} (1) = - 1$ , so $\frac{f^{″} (1)}{2!} = \frac{- 1}{2} = - \frac{1}{2}$
$f^{‴} (1) = 2$ , so $\frac{f^{‴} (1)}{3!} = \frac{2}{6} = \frac{1}{3}$
$f^{(4)} (1) = - 6$ , so $\frac{f^{(4)} (1)}{4!} = \frac{- 6}{24} = - \frac{1}{4}$
$f^{(5)} (1) = 24$ , so $\frac{f^{(5)} (1)}{5!} = \frac{24}{120} = \frac{1}{5}$

For the

n

-th term (

n \geq 1

\frac{f^{(n)} (1)}{n!} = \frac{(- 1)^{n - 1} (n - 1)!}{n!} = \frac{(- 1)^{n - 1}}{n}

Thus, the series is:

\ln (x) = 0 + (x - 1) - \frac{(x - 1)^{2}}{2} + \frac{(x - 1)^{3}}{3} - \frac{(x - 1)^{4}}{4} + \dots

\ln (x) = \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1}}{n} (x - 1)^{n}

This is the Taylor series expansion of

\ln (x)

in powers of

(x - 1)

, valid for

| x - 1 | < 1

, and at

x = 2

by checking the radius of convergence and endpoint behavior.

Step 4: Evaluate $\ln (1.1)$

To find

\ln (1.1)

correct to four decimal places, set

x = 1.1

, so

x - 1 = 1.1 - 1 = 0.1

. Substitute into the series:

\ln (1.1) = (0.1) - \frac{(0.1)^{2}}{2} + \frac{(0.1)^{3}}{3} - \frac{(0.1)^{4}}{4} + \frac{(0.1)^{5}}{5} - \dots

This is an alternating series:

\ln (1.1) = \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1}}{n} (0.1)^{n}

Compute the terms:

$n = 1$ : $\frac{(0.1)^{1}}{1} = 0.1$
$n = 2$ : $- \frac{(0.1)^{2}}{2} = - \frac{0.01}{2} = - 0.005$
$n = 3$ : $\frac{(0.1)^{3}}{3} = \frac{0.001}{3} \approx 0.000333333$
$n = 4$ : $- \frac{(0.1)^{4}}{4} = - \frac{0.0001}{4} = - 0.000025$
$n = 5$ : $\frac{(0.1)^{5}}{5} = \frac{0.00001}{5} = 0.000002$
$n = 6$ : $- \frac{(0.1)^{6}}{6} = - \frac{0.000001}{6} \approx - 0.00000016667$

Sum the first few terms:

First term: $0.1$
First two terms: $0.1 - 0.005 = 0.095$
First three terms: $0.095 + 0.000333333 \approx 0.095333333$
First four terms: $0.095333333 - 0.000025 = 0.095308333$
First five terms: $0.095308333 + 0.000002 = 0.095310333$
First six terms: $0.095310333 - 0.00000016667 \approx 0.09531016667$

Step 5: Determine Terms Needed for Four Decimal Places

To ensure accuracy to four decimal places (error less than

0.00005

), use the alternating series error bound. For an alternating series

\sum (- 1)^{n - 1} b_{n}

, the error after

k

terms is less than the absolute value of the

(k + 1)

-th term:

Error < b_{k + 1} = \frac{(0.1)^{k + 1}}{k + 1}

We need:

\frac{(0.1)^{k + 1}}{k + 1} < 0.00005 = 5 \times 10^{- 5}

Test

k = 4

(after four terms, error is bounded by the fifth term):

b_{5} = \frac{(0.1)^{5}}{5} = \frac{0.00001}{5} = 0.000002 < 0.00005

This satisfies the requirement. The partial sum after four terms is:

S_{4} \approx 0.095308333

Round to four decimal places:

0.095308333 \approx 0.0953

The fifth term (

0.000002

) suggests the true value is slightly higher, but let’s check with one more term:

S_{5} \approx 0.095310333 \approx 0.0953

The sixth term is

- 0.00000016667

, which affects the fifth decimal place, confirming

S_{5} \approx 0.0953

to four decimal places.

Step 6: Verify the Result

The actual value of

\ln (1.1) \approx 0.09531017980432493

. Rounding to four decimal places:

0.09531017980432493 \approx 0.0953

Our series approximation

S_{5} \approx 0.095310333

matches this when rounded, confirming correctness.

Final Answer

The Taylor series expansion of

\ln (x)

in powers of

(x - 1)

is:

\ln (x) = \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1}}{n} (x - 1)^{n}

The value of

\ln (1.1)

correct to four decimal places is:

0.0953

Question:-1(e)

Find the equation of the right circular cylinder which passes through the circle $x^{2} + y^{2} + z^{2} = 9, x - y + z = 3$ .

Answer:

To find the equation of the right circular cylinder that passes through the given circle, we’ll follow these steps:

Given:

The circle is the intersection of the sphere

x^{2} + y^{2} + z^{2} = 9

and the plane

x - y + z = 3

Step 1: Find the Center and Radius of the Given Circle

Center of the Sphere:
The sphere $x^{2} + y^{2} + z^{2} = 9$ has its center at the origin $(0, 0, 0)$ and radius $3$ .
Distance from the Center to the Plane:
The plane is $x - y + z = 3$ . The distance $d$ from the center $(0, 0, 0)$ to the plane is:
$d = \frac{| 0 - 0 + 0 - 3 |}{\sqrt{1^{2} + (- 1)^{2} + 1^{2}}} = \frac{3}{\sqrt{3}} = \sqrt{3} .$
Radius of the Circle:
Using the Pythagorean theorem for the sphere and plane intersection:
$Radius of the circle = \sqrt{R^{2} - d^{2}} = \sqrt{9 - 3} = \sqrt{6} .$
Center of the Circle:
The center of the circle lies along the normal from the sphere’s center to the plane. The normal vector to the plane $x - y + z = 3$ is $n = (1, - 1, 1)$ .
The parametric equations for the line from the origin in the direction of $n$ are:
$x = t, y = - t, z = t .$
Substituting into the plane equation to find the foot of the perpendicular:
$t - (- t) + t = 3 \Rightarrow 3 t = 3 \Rightarrow t = 1.$
So, the center of the circle is $(1, - 1, 1)$ .

Step 2: Determine the Axis of the Cylinder

The axis of the right circular cylinder is parallel to the normal vector of the given plane

n = (1, - 1, 1)

Step 3: Find the Equation of the Cylinder

A right circular cylinder with axis parallel to

n = (1, - 1, 1)

and passing through the circle centered at

(1, - 1, 1)

with radius

\sqrt{6}

can be described as follows:

Distance from a Point $(x, y, z)$ to the Axis:
The distance $D$ from a point $(x, y, z)$ to the line (axis) passing through $(1, - 1, 1)$ in the direction $n = (1, - 1, 1)$ is given by:
$D = \frac{‖ n \times v ‖}{‖ n ‖},$
where $v = (x - 1, y + 1, z - 1)$ .
Compute the cross product:
$n \times v = | \begin{matrix} i & j & k \\ 1 & - 1 & 1 \\ x - 1 & y + 1 & z - 1 \end{matrix} | = i ((- 1) (z - 1) - 1 (y + 1)) - j (1 (z - 1) - 1 (x - 1)) + k (1 (y + 1) - (- 1) (x - 1)) .$
Simplifying:
$n \times v = (- (z - 1) - (y + 1)) i - ((z - 1) - (x - 1)) j + ((y + 1) + (x - 1)) k .$
$= (- y - z) i - (z - x) j + (x + y) k .$
The magnitude is:
$‖ n \times v ‖ = \sqrt{(- y - z)^{2} + (- z + x)^{2} + (x + y)^{2}} .$
$= \sqrt{(y + z)^{2} + (z - x)^{2} + (x + y)^{2}} .$
The magnitude of $n$ is:
$‖ n ‖ = \sqrt{1^{2} + (- 1)^{2} + 1^{2}} = \sqrt{3} .$
Therefore:
$D = \frac{\sqrt{(y + z)^{2} + (z - x)^{2} + (x + y)^{2}}}{\sqrt{3}} .$
Equation of the Cylinder:
The cylinder consists of all points $(x, y, z)$ such that the distance $D$ to the axis is equal to the radius $\sqrt{6}$ :
$\frac{\sqrt{(y + z)^{2} + (z - x)^{2} + (x + y)^{2}}}{\sqrt{3}} = \sqrt{6} .$
Squaring both sides:
$\frac{(y + z)^{2} + (z - x)^{2} + (x + y)^{2}}{3} = 6.$
Multiply by 3:
$(y + z)^{2} + (z - x)^{2} + (x + y)^{2} = 18.$
Expand each square:
$(y^{2} + 2 y z + z^{2}) + (z^{2} - 2 x z + x^{2}) + (x^{2} + 2 x y + y^{2}) = 18.$
Combine like terms:
$2 x^{2} + 2 y^{2} + 2 z^{2} + 2 x y + 2 y z - 2 x z = 18.$
Divide by 2:
$x^{2} + y^{2} + z^{2} + x y + y z - x z = 9.$

Final Answer:

The equation of the right circular cylinder is:

x^{2} + y^{2} + z^{2} + x y + y z - x z = 9

Question:-2(a)

Consider a linear operator $T$ on $R^{3}$ over $R$ defined by $T (x, y, z) = (2 x, 4 x - y, 2 x + 3 y - z)$ . Is $T$ invertible? If yes, justify your answer and find $T^{- 1}$ .

Answer:

To determine if the linear operator

T

is invertible and to find its inverse

T^{- 1}

if it exists, we follow these steps:

1. Definition of $T$ :

The linear operator

T

R^{3}

is defined by:

T (x, y, z) = (2 x, 4 x - y, 2 x + 3 y - z) .

2. Matrix Representation of $T$ :

To represent

T

as a matrix, we apply

T

to the standard basis vectors

e_{1} = (1, 0, 0)

e_{2} = (0, 1, 0)

, and

e_{3} = (0, 0, 1)

T (e_{1}) = (2, 4, 2), T (e_{2}) = (0, - 1, 3), T (e_{3}) = (0, 0, - 1) .

Thus, the matrix

A

representing

T

is:

A = [\begin{matrix} 2 & 0 & 0 \\ 4 & - 1 & 0 \\ 2 & 3 & - 1 \end{matrix}] .

3. Checking Invertibility:

A matrix (and hence the operator) is invertible if its determinant is non-zero. Let’s compute

det (A)

det (A) = 2 \cdot | \begin{matrix} - 1 & 0 \\ 3 & - 1 \end{matrix} | - 0 \cdot | \begin{matrix} 4 & 0 \\ 2 & - 1 \end{matrix} | + 0 \cdot | \begin{matrix} 4 & - 1 \\ 2 & 3 \end{matrix} | = 2 \cdot ((- 1) (- 1) - 0 \cdot 3) = 2 \cdot 1 = 2.

Since

det (A) = 2 \neq 0

T

is invertible.

4. Finding the Inverse $T^{- 1}$ :

To find

A^{- 1}

, we use the formula for the inverse of a

3 \times 3

matrix:

A^{- 1} = \frac{1}{det (A)} \cdot adj (A),

where

adj (A)

is the adjugate of

A

First, compute the cofactor matrix

C

A

C = [\begin{matrix} | \begin{matrix} - 1 & 0 \\ 3 & - 1 \end{matrix} | & - | \begin{matrix} 4 & 0 \\ 2 & - 1 \end{matrix} | & | \begin{matrix} 4 & - 1 \\ 2 & 3 \end{matrix} | \\ - | \begin{matrix} 0 & 0 \\ 3 & - 1 \end{matrix} | & | \begin{matrix} 2 & 0 \\ 2 & - 1 \end{matrix} | & - | \begin{matrix} 2 & 0 \\ 2 & 3 \end{matrix} | \\ | \begin{matrix} 0 & 0 \\ - 1 & 0 \end{matrix} | & - | \begin{matrix} 2 & 0 \\ 4 & 0 \end{matrix} | & | \begin{matrix} 2 & 0 \\ 4 & - 1 \end{matrix} | \end{matrix}] = [\begin{matrix} 1 & 4 & 14 \\ 0 & - 2 & - 6 \\ 0 & 0 & - 2 \end{matrix}] .

The adjugate

adj (A)

is the transpose of

C

adj (A) = [\begin{matrix} 1 & 0 & 0 \\ 4 & - 2 & 0 \\ 14 & - 6 & - 2 \end{matrix}] .

Thus, the inverse matrix is:

A^{- 1} = \frac{1}{2} \cdot [\begin{matrix} 1 & 0 & 0 \\ 4 & - 2 & 0 \\ 14 & - 6 & - 2 \end{matrix}] = [\begin{matrix} \frac{1}{2} & 0 & 0 \\ 2 & - 1 & 0 \\ 7 & - 3 & - 1 \end{matrix}] .

5. Expression for $T^{- 1}$ :

Using

A^{- 1}

, the inverse operator

T^{- 1}

is given by:

T^{- 1} (x, y, z) = (\frac{1}{2} x, 2 x - y, 7 x - 3 y - z) .

Verification:

To ensure correctness, let’s verify

T \circ T^{- 1} = I

T (\frac{1}{2} x, 2 x - y, 7 x - 3 y - z) = (2 \cdot \frac{1}{2} x, 4 \cdot \frac{1}{2} x - (2 x - y), 2 \cdot \frac{1}{2} x + 3 (2 x - y) - (7 x - 3 y - z)) .

Simplifying:

= (x, 2 x - 2 x + y, x + 6 x - 3 y - 7 x + 3 y + z) = (x, y, z) .

Thus,

T^{- 1}

is indeed the inverse of

T

Final Answer:

T^{- 1} (x, y, z) = (\frac{1}{2} x, 2 x - y, 7 x - 3 y - z)

Question:-2(b)

Answer:

To solve the problem, we’ll follow these steps:

Given:

u = \frac{x + y}{1 - x y}, v = \tan^{- 1} x + \tan^{- 1} y .

Step 1: Compute the Jacobian $\frac{\partial (u, v)}{\partial (x, y)}$

The Jacobian determinant is given by:

\frac{\partial (u, v)}{\partial (x, y)} = | \begin{matrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{matrix} | .

Partial Derivatives of $u$ :

\frac{\partial u}{\partial x} = \frac{(1) (1 - x y) - (x + y) (- y)}{(1 - x y)^{2}} = \frac{1 - x y + x y + y^{2}}{(1 - x y)^{2}} = \frac{1 + y^{2}}{(1 - x y)^{2}},

\frac{\partial u}{\partial y} = \frac{(1) (1 - x y) - (x + y) (- x)}{(1 - x y)^{2}} = \frac{1 - x y + x^{2} + x y}{(1 - x y)^{2}} = \frac{1 + x^{2}}{(1 - x y)^{2}} .

Partial Derivatives of $v$ :

\frac{\partial v}{\partial x} = \frac{1}{1 + x^{2}}, \frac{\partial v}{\partial y} = \frac{1}{1 + y^{2}} .

Compute the Jacobian Determinant:

\frac{\partial (u, v)}{\partial (x, y)} = | \begin{matrix} \frac{1 + y^{2}}{(1 - x y)^{2}} & \frac{1 + x^{2}}{(1 - x y)^{2}} \\ \frac{1}{1 + x^{2}} & \frac{1}{1 + y^{2}} \end{matrix} | .

= (\frac{1 + y^{2}}{(1 - x y)^{2}}) (\frac{1}{1 + y^{2}}) - (\frac{1 + x^{2}}{(1 - x y)^{2}}) (\frac{1}{1 + x^{2}}),

= \frac{1}{(1 - x y)^{2}} - \frac{1}{(1 - x y)^{2}} = 0.

Step 2: Determine Functional Relationship

Since the Jacobian determinant is zero,

u

and

v

are functionally dependent. This means there exists a relationship between

u

and

v

Find the Relationship:

Recall that:

v = \tan^{- 1} x + \tan^{- 1} y .

Using the tangent addition formula:

\tan (v) = \tan (\tan^{- 1} x + \tan^{- 1} y) = \frac{x + y}{1 - x y} = u .

Thus, the relationship is:

u = \tan (v) .

Final Answer:

\begin{aligned} \frac{\partial (u, v)}{\partial (x, y)} & = 0, \\ Relationship: u & = \tan (v) . \end{aligned}

Question:-2(c)

Find the image of the line $x = 3 - 6 t, y = 2 t, z = 3 + 2 t$ in the plane $3 x + 4 y - 5 z + 26 = 0$ .

Answer:

To find the image of the line

x = 3 - 6 t

y = 2 t

z = 3 + 2 t

in the plane

3 x + 4 y - 5 z + 26 = 0

, we interpret the "image" as the reflection of the line over the plane. The reflection of a line in a plane is the set of points obtained by reflecting each point on the line across the plane. Since the line is in

R^{3}

and the plane is a two-dimensional subspace, the reflection is another line. We will find the parametric equations of the reflected line by reflecting a point on the line and determining the direction of the reflected line.

Step 1: Understand the Line and Plane

The line is given parametrically as:

x = 3 - 6 t, y = 2 t, z = 3 + 2 t

A point on the line at parameter

t

(3 - 6 t, 2 t, 3 + 2 t)

. The direction vector of the line is obtained from the coefficients of

t

\vec{d} = (- 6, 2, 2)

The plane is:

3 x + 4 y - 5 z + 26 = 0

The normal vector to the plane is:

\vec{n} = (3, 4, - 5)

Step 2: Reflection of a Point

To find the reflected line, we can reflect a specific point on the line and determine the direction of the reflected line. Choose a point on the line, say at

t = 0

(3, 0, 3)

Verify if this point lies on the plane (though not necessary, it helps understand the geometry):

3 \cdot 3 + 4 \cdot 0 - 5 \cdot 3 + 26 = 9 + 0 - 15 + 26 = 20 \neq 0

The point is not on the plane, so we proceed to find its reflection.

The reflection of a point

P = (x_{0}, y_{0}, z_{0})

across the plane involves finding the foot of the perpendicular from

P

to the plane (the midpoint of

P

and its image

P^{'}

) and then extending to

P^{'}

Line from $P (3, 0, 3)$ to the plane: The line through $P$ parallel to the normal $\vec{n} = (3, 4, - 5)$ is:
$x = 3 + 3 s, y = 0 + 4 s, z = 3 - 5 s$
Intersection with the plane: Substitute into the plane equation:
$3 (3 + 3 s) + 4 (4 s) - 5 (3 - 5 s) + 26 = 0$
$9 + 9 s + 16 s - 15 + 25 s + 26 = 0$
$50 s + 20 = 0 ⟹ s = - \frac{20}{50} = - \frac{2}{5}$
Foot of the perpendicular (point $Q$ ):
$x = 3 + 3 \cdot (- \frac{2}{5}) = 3 - \frac{6}{5} = \frac{9}{5}$
$y = 4 \cdot (- \frac{2}{5}) = - \frac{8}{5}$
$z = 3 - 5 \cdot (- \frac{2}{5}) = 3 + 2 = 5$
So, $Q = (\frac{9}{5}, - \frac{8}{5}, 5)$ .
Find the reflected point $P^{'}$ : The foot $Q$ is the midpoint of $P (3, 0, 3)$ and $P^{'} (x^{'}, y^{'}, z^{'})$ . Using the midpoint formula:
$\frac{3 + x^{'}}{2} = \frac{9}{5} ⟹ 3 + x^{'} = \frac{18}{5} ⟹ x^{'} = \frac{18}{5} - 3 = \frac{18}{5} - \frac{15}{5} = \frac{3}{5}$
$\frac{0 + y^{'}}{2} = - \frac{8}{5} ⟹ y^{'} = - \frac{16}{5}$
$\frac{3 + z^{'}}{2} = 5 ⟹ 3 + z^{'} = 10 ⟹ z^{'} = 7$
So, the reflected point is:
$P^{'} = (\frac{3}{5}, - \frac{16}{5}, 7)$

Step 3: Direction of the Reflected Line

The reflected line has a direction vector obtained by reflecting the original line’s direction vector

\vec{d} = (- 6, 2, 2)

across the plane. The reflection of a vector

\vec{v}

across a plane with normal

\vec{n}

is given by:

{\vec{v}}^{'} = \vec{v} - 2 \frac{\vec{v} \cdot \vec{n}}{‖ \vec{n} ‖^{2}} \vec{n}

Compute $\vec{v} \cdot \vec{n}$ : $\vec{d} \cdot \vec{n} = (- 6) \cdot 3 + 2 \cdot 4 + 2 \cdot (- 5) = - 18 + 8 - 10 = - 20$
Compute $‖ \vec{n} ‖^{2}$ : $‖ \vec{n} ‖^{2} = 3^{2} + 4^{2} + (- 5)^{2} = 9 + 16 + 25 = 50$
Compute the projection term: $2 \frac{\vec{d} \cdot \vec{n}}{‖ \vec{n} ‖^{2}} \vec{n} = 2 \cdot \frac{- 20}{50} \cdot (3, 4, - 5) = - \frac{40}{50} \cdot (3, 4, - 5) = - \frac{4}{5} \cdot (3, 4, - 5) = (- \frac{12}{5}, - \frac{16}{5}, 4)$
Reflected direction vector: ${\vec{d}}^{'} = (- 6, 2, 2) - (- \frac{12}{5}, - \frac{16}{5}, 4) = (- 6 + \frac{12}{5}, 2 + \frac{16}{5}, 2 - 4)$ $= (- \frac{30}{5} + \frac{12}{5}, \frac{10}{5} + \frac{16}{5}, - 2) = (- \frac{18}{5}, \frac{26}{5}, - 2)$

Simplify by multiplying by 5 (direction vectors are scalable):

{\vec{d}}^{'} = (- 18, 26, - 10)

Step 4: Parametric Equations of the Reflected Line

The reflected line passes through

P^{'} = (\frac{3}{5}, - \frac{16}{5}, 7)

with direction vector

(- 18, 26, - 10)

. Parametric equations are:

x = \frac{3}{5} - 18 u, y = - \frac{16}{5} + 26 u, z = 7 - 10 u

To make coefficients integers, use a parameter

s = 5 u

x = \frac{3}{5} - \frac{18}{5} s, y = - \frac{16}{5} + \frac{26}{5} s, z = 7 - \frac{10}{5} s = 7 - 2 s

x = \frac{3 - 18 s}{5}, y = \frac{- 16 + 26 s}{5}, z = 7 - 2 s

Alternatively, write as:

x = \frac{3}{5} - \frac{18}{5} s, y = - \frac{16}{5} + \frac{26}{5} s, z = 7 - 2 s

Step 5: Verify the Reflection

Check if $P^{'}$ is the reflection: The midpoint $Q$ is correct, and the vector from $Q$ to $P^{'}$ should be opposite to $Q$ to $P$ , adjusted by the normal. This was computed correctly.
Check if the reflected line is consistent: Reflect another point, e.g., at $t = 1$ : $P_{1} = (3 - 6, 2, 3 + 2) = (- 3, 2, 5)$ This computation is complex, so instead, verify the direction vector by ensuring the reflected line is consistent with the plane’s geometry. The reflected direction should satisfy symmetry across the plane, which we’ve computed.

Final Answer

The image (reflection) of the line is:

x = \frac{3 - 18 s}{5}, y = \frac{- 16 + 26 s}{5}, z = 7 - 2 s

Question:-3(a)

Let $V = M_{2 \times 2} (R)$ denote a vector space over the field of real numbers. Find the matrix of the linear mapping $ϕ : V \to V$ given by

ϕ (v) = (\begin{matrix} 1 & 2 \\ 3 & - 1 \end{matrix}) v

with respect to the standard basis of

M_{2 \times 2} (R)

, and hence find the rank of

ϕ

. Is

ϕ

invertible? Justify your answer.

Answer:

To solve the problem, we need to find the matrix representation of the linear mapping

ϕ : M_{2 \times 2} (R) \to M_{2 \times 2} (R)

defined by

ϕ (v) = (\begin{matrix} 1 & 2 \\ 3 & - 1 \end{matrix}) v

, where

v \in M_{2 \times 2} (R)

is a 2×2 matrix, with respect to the standard basis of

M_{2 \times 2} (R)

. We then determine the rank of

ϕ

and check if

ϕ

is invertible. Let’s proceed step by step.

Step 1: Understand the Vector Space and Basis

The vector space

M_{2 \times 2} (R)

consists of all 2×2 matrices over

R

, and it has dimension 4 (since each matrix has 4 entries). The standard basis for

M_{2 \times 2} (R)

is:

E_{11} = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}), E_{12} = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}), E_{21} = (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}), E_{22} = (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix})

A general matrix

v = (\begin{matrix} a & b \\ c & d \end{matrix})

can be written as:

v = a E_{11} + b E_{12} + c E_{21} + d E_{22}

The mapping

ϕ (v) = A v

, where

A = (\begin{matrix} 1 & 2 \\ 3 & - 1 \end{matrix})

, is a linear transformation from

M_{2 \times 2} (R)

to itself.

Step 2: Compute the Matrix of $ϕ$

To find the matrix of

ϕ

with respect to the standard basis

{E_{11}, E_{12}, E_{21}, E_{22}}

, we apply

ϕ

to each basis element and express the result as a linear combination of the basis elements. The coefficients form the columns of the matrix representation. Since

M_{2 \times 2} (R)

is isomorphic to

R^{4}

, we can represent matrices as vectors by stacking their entries, but we’ll work directly with matrices for clarity.

Let’s compute

ϕ (E_{i j}) = A E_{i j}

For $E_{11} = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})$ :
$ϕ (E_{11}) = (\begin{matrix} 1 & 2 \\ 3 & - 1 \end{matrix}) (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) = (\begin{matrix} 1 \cdot 1 + 2 \cdot 0 & 1 \cdot 0 + 2 \cdot 0 \\ 3 \cdot 1 + (- 1) \cdot 0 & 3 \cdot 0 + (- 1) \cdot 0 \end{matrix}) = (\begin{matrix} 1 & 0 \\ 3 & 0 \end{matrix})$
Express in the basis:
$(\begin{matrix} 1 & 0 \\ 3 & 0 \end{matrix}) = 1 \cdot E_{11} + 0 \cdot E_{12} + 3 \cdot E_{21} + 0 \cdot E_{22}$
Column 1: $(1, 0, 3, 0)$ .
For $E_{12} = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ :
$ϕ (E_{12}) = (\begin{matrix} 1 & 2 \\ 3 & - 1 \end{matrix}) (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}) = (\begin{matrix} 1 \cdot 0 + 2 \cdot 0 & 1 \cdot 1 + 2 \cdot 0 \\ 3 \cdot 0 + (- 1) \cdot 0 & 3 \cdot 1 + (- 1) \cdot 0 \end{matrix}) = (\begin{matrix} 0 & 1 \\ 0 & 3 \end{matrix})$
$(\begin{matrix} 0 & 1 \\ 0 & 3 \end{matrix}) = 0 \cdot E_{11} + 1 \cdot E_{12} + 0 \cdot E_{21} + 3 \cdot E_{22}$
Column 2: $(0, 1, 0, 3)$ .
For $E_{21} = (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix})$ :
$ϕ (E_{21}) = (\begin{matrix} 1 & 2 \\ 3 & - 1 \end{matrix}) (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}) = (\begin{matrix} 1 \cdot 0 + 2 \cdot 1 & 1 \cdot 0 + 2 \cdot 0 \\ 3 \cdot 0 + (- 1) \cdot 1 & 3 \cdot 0 + (- 1) \cdot 0 \end{matrix}) = (\begin{matrix} 2 & 0 \\ - 1 & 0 \end{matrix})$
$(\begin{matrix} 2 & 0 \\ - 1 & 0 \end{matrix}) = 2 \cdot E_{11} + 0 \cdot E_{12} + (- 1) \cdot E_{21} + 0 \cdot E_{22}$
Column 3: $(2, 0, - 1, 0)$ .
For $E_{22} = (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix})$ :
$ϕ (E_{22}) = (\begin{matrix} 1 & 2 \\ 3 & - 1 \end{matrix}) (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}) = (\begin{matrix} 1 \cdot 0 + 2 \cdot 0 & 1 \cdot 0 + 2 \cdot 1 \\ 3 \cdot 0 + (- 1) \cdot 0 & 3 \cdot 0 + (- 1) \cdot 1 \end{matrix}) = (\begin{matrix} 0 & 2 \\ 0 & - 1 \end{matrix})$
$(\begin{matrix} 0 & 2 \\ 0 & - 1 \end{matrix}) = 0 \cdot E_{11} + 2 \cdot E_{12} + 0 \cdot E_{21} + (- 1) \cdot E_{22}$
Column 4: $(0, 2, 0, - 1)$ .

The matrix of

ϕ

with respect to the standard basis (ordered as

E_{11}, E_{12}, E_{21}, E_{22}

) is:

[ϕ] = [\begin{matrix} 1 & 0 & 2 & 0 \\ 0 & 1 & 0 & 2 \\ 3 & 0 & - 1 & 0 \\ 0 & 3 & 0 & - 1 \end{matrix}]

Step 3: Alternative Approach Using Vector Representation

To confirm, we can represent matrices in

M_{2 \times 2} (R)

as vectors in

R^{4}

. Map a matrix

v = (\begin{matrix} a & b \\ c & d \end{matrix})

to the vector

(a, b, c, d)

. Then:

ϕ (v) = (\begin{matrix} 1 & 2 \\ 3 & - 1 \end{matrix}) (\begin{matrix} a & b \\ c & d \end{matrix}) = (\begin{matrix} a + 2 c & b + 2 d \\ 3 a - c & 3 b - d \end{matrix})

The output vector is:

(a + 2 c, b + 2 d, 3 a - c, 3 b - d)

The matrix acting on

(a, b, c, d)

is exactly the one derived above, confirming:

[\begin{matrix} a + 2 c \\ b + 2 d \\ 3 a - c \\ 3 b - d \end{matrix}] = [\begin{matrix} 1 & 0 & 2 & 0 \\ 0 & 1 & 0 & 2 \\ 3 & 0 & - 1 & 0 \\ 0 & 3 & 0 & - 1 \end{matrix}] [\begin{matrix} a \\ b \\ c \\ d \end{matrix}]

Step 4: Find the Rank of $ϕ$

The rank of

ϕ

is the rank of its matrix

[ϕ]

, which is the dimension of the image of

ϕ

. Compute the rank by row reducing the matrix:

[ϕ] = [\begin{matrix} 1 & 0 & 2 & 0 \\ 0 & 1 & 0 & 2 \\ 3 & 0 & - 1 & 0 \\ 0 & 3 & 0 & - 1 \end{matrix}]

$R_{3} \leftarrow R_{3} - 3 R_{1}$ : $R_{3} = (3, 0, - 1, 0) - 3 (1, 0, 2, 0) = (0, 0, - 7, 0)$
$R_{4} \leftarrow R_{4} - 3 R_{2}$ : $R_{4} = (0, 3, 0, - 1) - 3 (0, 1, 0, 2) = (0, 0, 0, - 7)$

[\begin{matrix} 1 & 0 & 2 & 0 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & - 7 & 0 \\ 0 & 0 & 0 & - 7 \end{matrix}]

The matrix is in row echelon form with 4 non-zero rows, so the rank is:

rank (ϕ) = 4

Step 5: Determine Invertibility

A linear operator on a finite-dimensional vector space is invertible if and only if its matrix is invertible, which occurs when the rank equals the dimension of the space (here, 4) or equivalently, the determinant is non-zero. Since

rank (ϕ) = 4

ϕ

is invertible (it is bijective).

Alternatively, compute the determinant of

[ϕ]

. The matrix is block diagonal:

[ϕ] = [\begin{matrix} A & B \\ C & D \end{matrix}], A = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}), B = (\begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix}), C = (\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix}), D = (\begin{matrix} - 1 & 0 \\ 0 & - 1 \end{matrix})

However, computing the determinant directly is complex. Since the rank is 4, the matrix is full rank, implying

det ([ϕ]) \neq 0

, so

ϕ

is invertible.

Step 6: Optional Verification

To confirm invertibility, we could compute the determinant explicitly or check if the kernel is trivial, but the rank being 4 is sufficient. For completeness, the kernel of

ϕ

is:

ϕ (v) = 0 ⟹ A v = 0

Since

A = (\begin{matrix} 1 & 2 \\ 3 & - 1 \end{matrix})

has

det (A) = (1) (- 1) - (2) (3) = - 1 - 6 = - 7 \neq 0

, for

v \neq 0

A v \neq 0

, so the kernel of

ϕ

is trivial, confirming invertibility.

Final Answer

The matrix of

ϕ

with respect to the standard basis

{E_{11}, E_{12}, E_{21}, E_{22}}

is:

[\begin{matrix} 1 & 0 & 2 & 0 \\ 0 & 1 & 0 & 2 \\ 3 & 0 & - 1 & 0 \\ 0 & 3 & 0 & - 1 \end{matrix}]

The rank of

ϕ

is:

4

Since the rank is 4, equal to the dimension of

M_{2 \times 2} (R)

ϕ

is invertible.

Question:-3(b)

Find the volume of the greatest cylinder which can be inscribed in a cone of height $h$ and semi-vertical angle $α$ .

Answer:

To find the volume of the largest cylinder that can be inscribed in a cone with height

h

and semi-vertical angle

α

, we need to maximize the volume of the cylinder subject to the constraint that it is fully contained within the cone. Let’s proceed step by step, using geometric insights and calculus to optimize the volume.

Step 1: Set Up the Geometry of the Cone

Consider a right circular cone with its vertex at the origin

O (0, 0, 0)

, axis along the positive

z

-axis, and base at

z = h

. The semi-vertical angle

α

is the angle between the cone’s axis (the

z

-axis) and its slant surface. In the

x z

-plane, the cone’s surface satisfies the equation derived from the angle

α

Place the cone’s base in the plane

z = h

. At

z = h

, the radius of the base is

r = h \tan α

, since the line from the vertex to the base edge makes an angle

α

with the

z

-axis. The equation of the cone’s surface can be derived as follows:

For a point

(x, y, z)

on the cone’s surface, the distance from the

z

-axis is

\sqrt{x^{2} + y^{2}}

. The slope of the line from the origin to a point on the cone’s surface at height

z

is:

\tan α = \frac{radius at height z}{z} = \frac{\sqrt{x^{2} + y^{2}}}{z}

Thus:

\sqrt{x^{2} + y^{2}} = z \tan α

x^{2} + y^{2} = (z \tan α)^{2} = z^{2} \tan^{2} α

Since

0 \leq z \leq h

, and at

z = h

, the radius is

x^{2} + y^{2} = h^{2} \tan^{2} α

, this equation describes the cone.

Step 2: Define the Inscribed Cylinder

Assume the cylinder is right circular with its axis coincident with the cone’s axis (the

z

-axis) to maximize symmetry and volume. Let the cylinder have radius

r

and extend from height

z = a

z = b

(where

0 \leq a < b \leq h

). The radius of the cone at height

z

z \tan α

. For the cylinder to be inscribed, its radius

r

must not exceed the cone’s radius at any height between

z = a

and

z = b

. The cone’s radius is smallest at the lower height

z = a

r \leq a \tan α

To maximize the cylinder’s volume, assume the cylinder touches the cone’s surface along its lateral surface, so at

z = a

, the cylinder’s radius equals the cone’s radius:

r = a \tan α

However, the cylinder extends to height

b

, where the cone’s radius is

b \tan α

. For the cylinder to be inside the cone, we need

r \leq b \tan α

. Since

r = a \tan α

, and

a < b

, this is satisfied:

a \tan α < b \tan α

Let’s reconsider the cylinder’s configuration. To maximize volume, the cylinder’s top and bottom circular faces typically touch the cone’s surface. Suppose the cylinder’s top is at height

z = H

, and its base is at height

z = H - L

, where

L

is the cylinder’s height, and

0 < H - L < H \leq h

. At

z = H

, the cone’s radius is

H \tan α

, and at

z = H - L

, it’s

(H - L) \tan α

. For the cylinder to touch the cone’s surface at both ends, its radius

r

satisfies:

r = (H - L) \tan α (at the base)

z = H

, the cone’s radius is

H \tan α

, so:

r \leq H \tan α

Since

H - L < H

, we have

(H - L) \tan α < H \tan α

, which is consistent. For points between

z = H - L

and

z = H

, the cone’s radius

z \tan α

satisfies:

(H - L) \tan α \leq z \tan α \leq H \tan α

Thus, the cylinder’s radius

r = (H - L) \tan α

is valid if it doesn’t exceed the cone’s radius at all points, which it does since the cone widens as

z

increases.

Step 3: Volume of the Cylinder

The volume of the cylinder is:

V = π r^{2} L = π ((H - L) \tan α)^{2} L = π \tan^{2} α (H - L)^{2} L

We need to maximize:

V (L, H) = π \tan^{2} α (H - L)^{2} L

subject to constraints:

0 < H - L < H \leq h

Since

π \tan^{2} α

is constant, maximize:

f (L, H) = (H - L)^{2} L

Constraints:

0 < L < H \leq h

Step 4: Optimize the Volume

To find the maximum, compute partial derivatives and find critical points.

Partial derivative with respect to $L$ :
$\frac{\partial f}{\partial L} = \frac{\partial}{\partial L} [(H - L)^{2} L] = (H - L)^{2} \cdot 1 + L \cdot 2 (H - L) \cdot (- 1)$
$= (H - L)^{2} - 2 L (H - L) = (H - L) [(H - L) - 2 L] = (H - L) (H - 3 L)$
Set to zero:
$(H - L) (H - 3 L) = 0$
$H - L = 0 or H - 3 L = 0$
$H = L or H = 3 L$
Since $L < H$ , discard $H = L$ . Thus:
$H = 3 L$
Substitute $H = 3 L$ into the volume:
$f (L, H) = (3 L - L)^{2} L = (2 L)^{2} L = 4 L^{2} \cdot L = 4 L^{3}$
Constraints: $0 < L < H = 3 L \leq h$ , so:
$3 L \leq h ⟹ L \leq \frac{h}{3}$
Maximize:
$g (L) = 4 L^{3}, 0 < L \leq \frac{h}{3}$
Since $g (L) = 4 L^{3}$ is increasing ( $g^{'} (L) = 12 L^{2} > 0$ ), the maximum occurs at the boundary $L = \frac{h}{3}$ .
Compute $H$ and $r$ :
$H = 3 L = 3 \cdot \frac{h}{3} = h$
Base at:
$H - L = h - \frac{h}{3} = \frac{2 h}{3}$
Radius:
$r = (H - L) \tan α = \frac{2 h}{3} \tan α$
Volume:
$V = π {(\frac{2 h}{3} \tan α)}^{2} \cdot \frac{h}{3} = π \cdot \frac{4 h^{2} \tan^{2} α}{9} \cdot \frac{h}{3} = π \cdot \frac{4 h^{3} \tan^{2} α}{27} = \frac{4 π h^{3} \tan^{2} α}{27}$

Step 5: Verify the Critical Point

To ensure the maximum, consider the second derivatives or test boundaries. Using the single-variable function:

V (L) = 4 π \tan^{2} α L^{3}

V^{'} (L) = 12 π \tan^{2} α L^{2}

The critical point is at

L = 0

, but since

L > 0

, and

V^{'} (L) > 0

, the function increases up to

L = \frac{h}{3}

. Alternatively, use the two-variable function and check the Hessian at

H = 3 L

, but the boundary

H = h

simplifies the process.

Step 6: Geometric Verification

The cylinder’s base is at

z = \frac{2 h}{3}

, top at

z = h

, with radius

r = \frac{2 h}{3} \tan α

. At

z = h

, the cone’s radius is

h \tan α

, which is larger, ensuring the cylinder fits inside. The height

L = \frac{h}{3}

and radius are consistent with the cone’s geometry.

Final Answer

The volume of the greatest cylinder inscribed in the cone is:

\frac{4 π h^{3} \tan^{2} α}{27}

Question:-3(c)

Find the vertex of the cone $4 x^{2} - y^{2} + 2 z^{2} + 2 x y - 3 y z + 12 x - 11 y + 6 z + 4 = 0$ .

Answer:

To find the vertex of the given cone, we’ll follow these steps:

Given Equation:

4 x^{2} - y^{2} + 2 z^{2} + 2 x y - 3 y z + 12 x - 11 y + 6 z + 4 = 0

Step 1: Rewrite the Equation in Matrix Form

The general equation of a quadric surface is:

X^{T} A X + B X + C = 0

where:

$X = [\begin{matrix} x \\ y \\ z \end{matrix}]$
$A$ is the symmetric matrix of the quadratic terms.
$B$ is the row matrix of the linear terms.
$C$ is the constant term.

For the given equation:

A = [\begin{matrix} 4 & 1 & 0 \\ 1 & - 1 & - \frac{3}{2} \\ 0 & - \frac{3}{2} & 2 \end{matrix}], B = [\begin{matrix} 12 & - 11 & 6 \end{matrix}], C = 4

Step 2: Find the Vertex

The vertex

V = [\begin{matrix} x_{0} \\ y_{0} \\ z_{0} \end{matrix}]

of the cone satisfies:

2 A V + B^{T} = 0

A V = - \frac{1}{2} B^{T}

Substituting $A$ and $B$ :
[
$[\begin{matrix} 4 & 1 & 0 \\ 1 & - 1 & - \frac{3}{2} \\ 0 & - \frac{3}{2} & 2 \end{matrix}]$
$[\begin{matrix} x_{0} \\ y_{0} \\ z_{0} \end{matrix}]$
= -\frac{1}{2}
$[\begin{matrix} 12 \\ - 11 \\ 6 \end{matrix}]$

[\begin{matrix} - 6 \\ \frac{11}{2} \\ - 3 \end{matrix}]

]

This gives us the system of equations:

$4 x_{0} + y_{0} = - 6$

Free UPSC Mathematics Optional Paper-1 2024 Solutions: View Online | UPSC Maths Solution | IAS Maths Solution

Question:-1(a)

Answer:

Step 1: Find the Basis and Dimension of H H HHH

✅ Conclusion:

Step 2: Extend this Basis to a Basis for R 4 R 4 R^(4)\mathbb{R}^4R4

Question:-1(b)

Answer:

Step 1: Find the Matrix Representation of T T TTT

Step 2: Basis for the Range Space

Step 3: Basis for the Null Space

Step 4: Express Null Space Basis in Standard Coordinates (if needed)

Final Answer

Question:-1(c)

Discuss the continuity of the function

Answer:

1. Continuity for x ≠ 0 x ≠ 0 x!=0x \neq 0x≠0:

2. Continuity at x = 0 x = 0 x=0x = 0x=0:

Case 1: x → 0 + x → 0 + x rarr0^(+)x \to 0^+x→0+ (Right-Hand Limit)

Case 2: x → 0 − x → 0 − x rarr0^(-)x \to 0^-x→0− (Left-Hand Limit)

Conclusion at x = 0 x = 0 x=0x = 0x=0:

3. Summary of Continuity:

Final Answer:

Question:-1(d)

Expand ln ⁡ ( x ) ln ⁡ ( x ) ln(x)\ln(x)ln⁡(x) in powers of ( x − 1 ) ( x − 1 ) (x-1)(x-1)(x−1) by Taylor’s theorem and hence find the value of ln ⁡ ( 1 ⋅ 1 ) ln ⁡ ( 1 ⋅ 1 ) ln(1*1)\ln (1 \cdot 1)ln⁡(1⋅1) correct up to four decimal places.

Answer:

Step 1: Understand Taylor’s Theorem

Step 2: Compute the Function and Its Derivatives at x = 1 x = 1 x=1x = 1x=1

Step 3: Construct the Taylor Series

Step 4: Evaluate ln ⁡ ( 1.1 ) ln ⁡ ( 1.1 ) ln(1.1)\ln(1.1)ln⁡(1.1)

Step 5: Determine Terms Needed for Four Decimal Places

Step 6: Verify the Result

Final Answer

Question:-1(e)

Find the equation of the right circular cylinder which passes through the circle x 2 + y 2 + z 2 = 9 , x − y + z = 3 x 2 + y 2 + z 2 = 9 , x − y + z = 3 x^(2)+y^(2)+z^(2)=9,x-y+z=3x^{2}+y^{2}+z^{2}=9,\; x-y+z=3x2+y2+z2=9,x−y+z=3.

Answer:

Given:

Step 1: Find the Center and Radius of the Given Circle

Step 2: Determine the Axis of the Cylinder

Step 3: Find the Equation of the Cylinder

Final Answer:

Question:-2(a)

Answer:

1. Definition of T T TTT:

2. Matrix Representation of T T TTT:

3. Checking Invertibility:

4. Finding the Inverse T − 1 T − 1 T^(-1)T^{-1}T−1:

5. Expression for T − 1 T − 1 T^(-1)T^{-1}T−1:

Verification:

Final Answer:

Question:-2(b)

Answer:

Given:

Step 1: Compute the Jacobian ∂ ( u , v ) ∂ ( x , y ) ∂ ( u , v ) ∂ ( x , y ) (del(u,v))/(del(x,y))\frac{\partial(u, v)}{\partial(x, y)}∂(u,v)∂(x,y)

Partial Derivatives of u u uuu:

Partial Derivatives of v v vvv:

Compute the Jacobian Determinant:

Step 2: Determine Functional Relationship

Find the Relationship:

Final Answer:

Question:-2(c)

Find the image of the line x = 3 − 6 t , y = 2 t , z = 3 + 2 t x = 3 − 6 t , y = 2 t , z = 3 + 2 t x=3-6t,y=2t,z=3+2tx=3-6t,\; y=2t,\; z=3+2tx=3−6t,y=2t,z=3+2t in the plane 3 x + 4 y − 5 z + 26 = 0 3 x + 4 y − 5 z + 26 = 0 3x+4y-5z+26=03x+4y-5z+26=03x+4y−5z+26=0.

Answer:

Step 1: Understand the Line and Plane

Step 2: Reflection of a Point

Step 3: Direction of the Reflected Line

Step 4: Parametric Equations of the Reflected Line

Step 5: Verify the Reflection

Final Answer

Question:-3(a)

Let V = M 2 × 2 ( R ) V = M 2 × 2 ( R ) V=M_(2xx2)(R)V=M_{2 \times 2}(\mathbb{R})V=M2×2(R) denote a vector space over the field of real numbers. Find the matrix of the linear mapping ϕ : V → V ϕ : V → V phi:V rarr V\phi: V \rightarrow Vϕ:V→V given by

Answer:

Step 1: Understand the Vector Space and Basis

Step 2: Compute the Matrix of ϕ ϕ phi\phiϕ

Step 3: Alternative Approach Using Vector Representation

Step 4: Find the Rank of ϕ ϕ phi\phiϕ

Step 5: Determine Invertibility

Step 6: Optional Verification

Final Answer

Question:-3(b)

Step 1: Find the Basis and Dimension of $H$

Step 2: Extend this Basis to a Basis for $R^{4}$

Step 1: Find the Matrix Representation of $T$

1. Continuity for $x \neq 0$ :

2. Continuity at $x = 0$ :

Case 1: $x \to 0^{+}$ (Right-Hand Limit)

Case 2: $x \to 0^{-}$ (Left-Hand Limit)

Conclusion at $x = 0$ :

Expand $\ln (x)$ in powers of $(x - 1)$ by Taylor’s theorem and hence find the value of $\ln (1 \cdot 1)$ correct up to four decimal places.

Step 2: Compute the Function and Its Derivatives at $x = 1$

Step 4: Evaluate $\ln (1.1)$

Find the equation of the right circular cylinder which passes through the circle $x^{2} + y^{2} + z^{2} = 9, x - y + z = 3$ .

1. Definition of $T$ :

2. Matrix Representation of $T$ :

4. Finding the Inverse $T^{- 1}$ :

5. Expression for $T^{- 1}$ :

Step 1: Compute the Jacobian $\frac{\partial (u, v)}{\partial (x, y)}$

Partial Derivatives of $u$ :

Partial Derivatives of $v$ :

Find the image of the line $x = 3 - 6 t, y = 2 t, z = 3 + 2 t$ in the plane $3 x + 4 y - 5 z + 26 = 0$ .

Let $V = M_{2 \times 2} (R)$ denote a vector space over the field of real numbers. Find the matrix of the linear mapping $ϕ : V \to V$ given by

Step 2: Compute the Matrix of $ϕ$

Step 4: Find the Rank of $ϕ$

Find the volume of the greatest cylinder which can be inscribed in a cone of height $h$ and semi-vertical angle $α$ .

Find the vertex of the cone $4 x^{2} - y^{2} + 2 z^{2} + 2 x y - 3 y z + 12 x - 11 y + 6 z + 4 = 0$ .