Free UPSC Mathematics Optional Paper-1 2019 Solutions: View Online | UPSC Maths Solution | IAS Maths Solution

July 12, 2025ByAbstract Classes

खण्ड-A / SECTION-A

(a) माना कि $f : [0, \frac{π}{2}] \to R$ एक संतत फलन है, जैसा कि

f (x) = \frac{\cos^{2} x}{4 x^{2} - π^{2}}, 0 \leq x < \frac{π}{2}

f (\frac{π}{2})

का मान ज्ञात कीजिए।

Let

f : [0, \frac{π}{2}] \to R

be a continuous function such that

f (x) = \frac{\cos^{2} x}{4 x^{2} - π^{2}}, 0 \leq x < \frac{π}{2}

Find the value of

f (\frac{π}{2})

Answer:

Introduction

The problem asks us to find the value of

f (\frac{π}{2})

for a given function

f (x) = \frac{\cos^{2} x}{4 x^{2} - π^{2}}

defined on the interval

[0, \frac{π}{2}]

. The function is continuous except at

x = \frac{π}{2}

because the denominator becomes zero at that point. To find

f (\frac{π}{2})

, we’ll need to evaluate the limit of

f (x)

x

approaches

\frac{π}{2}

Work/Calculations

Step 1: Define the Function and the Limit

The function

f (x)

is given as:

f (x) = \frac{\cos^{2} x}{4 x^{2} - π^{2}}

We need to find:

f (\frac{π}{2}) = lim_{x \to \frac{π}{2}} f (x)

Step 2: Simplify the Function

To find the limit, we can use L’Hôpital’s Rule, which states that if

lim_{x \to a} \frac{f (x)}{g (x)}

is an indeterminate form

\frac{0}{0}

\frac{\infty}{\infty}

, then:

lim_{x \to a} \frac{f (x)}{g (x)} = lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}

provided the limits on the right-hand side exist.

First, let’s check if the function

f (x)

is in indeterminate form at

x = \frac{π}{2}

Numerator at x = \frac{π}{2} : \cos^{2} (\frac{π}{2}) = 0

Denominator at x = \frac{π}{2} : 4 {(\frac{π}{2})}^{2} - π^{2} = 0

Both the numerator and the denominator are zero, so we have an indeterminate form

\frac{0}{0}

Step 3: Apply L’Hôpital’s Rule

Let’s differentiate the numerator and the denominator with respect to

x

$f^{'} (x)$ (Derivative of $\cos^{2} x$ ) = $- 2 \cos (x) \sin (x)$
$g^{'} (x)$ (Derivative of $4 x^{2} - π^{2}$ ) = $8 x$

Now, we can find the limit:

lim_{x \to \frac{π}{2}} \frac{f^{'} (x)}{g^{'} (x)} = lim_{x \to \frac{π}{2}} \frac{- 2 \cos (x) \sin (x)}{8 x}

Let’s substitute the values into the formula and calculate the limit.

After substituting the values, we get:

lim_{x \to \frac{π}{2}} \frac{- 2 \cos (x) \sin (x)}{8 x} = \frac{- 2 \cos (\frac{π}{2}) \sin (\frac{π}{2})}{8 \times \frac{π}{2}}

After calculating, we get:

lim_{x \to \frac{π}{2}} \frac{- 2 \cos (x) \sin (x)}{8 x} = 0

Conclusion

The value of

f (\frac{π}{2})

0

. We used L’Hôpital’s Rule to evaluate the limit of the function

f (x)

x

approaches

\frac{π}{2}

, and found that the limit is

0

. Therefore,

f (\frac{π}{2}) = 0

(b) माना कि

f : D (\subseteq R^{2}) \to R

एक फलन है और

(a, b) \in D

. अगर

f (x, y)

बिंदु

(a, b)

पर संतत है, तो दर्शाइए कि फलन

f (x, b)

और

f (a, y)

क्रमशः

x = a

और

y = b

पर संतत हैं।

Let

f : D (\subseteq R^{2}) \to R

be a function and

(a, b) \in D

. If

f (x, y)

is continuous at

(a, b)

, then show that the functions

f (x, b)

and

f (a, y)

are continuous at

x = a

and at

y = b

respectively.

Answer:

Introduction

The problem asks us to prove that if a function

f (x, y)

is continuous at a point

(a, b)

, then the functions

f (x, b)

and

f (a, y)

are continuous at

x = a

and

y = b

respectively. To prove this, we will use the definition of continuity and manipulate the mathematical expressions accordingly.

Work/Calculations

Step 1: Definition of Continuity

A function

f (x, y)

is said to be continuous at

(a, b)

if for every

ϵ > 0

, there exists

δ > 0

such that:

\sqrt{(x - a)^{2} + (y - b)^{2}} < δ ⟹ | f (x, y) - f (a, b) | < ϵ

Step 2: Prove Continuity for $f (x, b)$ at $x = a$

We need to show that for every

ϵ > 0

, there exists

δ > 0

such that:

| x - a | < δ ⟹ | f (x, b) - f (a, b) | < ϵ

To prove this, let’s consider

ϵ > 0

. Since

f (x, y)

is continuous at

(a, b)

, there exists

δ > 0

such that:

\sqrt{(x - a)^{2} + (y - b)^{2}} < δ ⟹ | f (x, y) - f (a, b) | < ϵ

Now, consider

| x - a | < δ

. We can rewrite this as

\sqrt{(x - a)^{2} + (b - b)^{2}} < δ

By the continuity of

f (x, y)

(a, b)

, this implies:

| f (x, b) - f (a, b) | < ϵ

Thus, we have shown that

f (x, b)

is continuous at

x = a

Step 3: Prove Continuity for $f (a, y)$ at $y = b$

Similarly, we need to show that for every

ϵ > 0

, there exists

δ > 0

such that:

| y - b | < δ ⟹ | f (a, y) - f (a, b) | < ϵ

Again, consider

ϵ > 0

. By the continuity of

f (x, y)

(a, b)

, there exists

δ > 0

such that:

\sqrt{(x - a)^{2} + (y - b)^{2}} < δ ⟹ | f (x, y) - f (a, b) | < ϵ

Now, consider

| y - b | < δ

. We can rewrite this as

\sqrt{(a - a)^{2} + (y - b)^{2}} < δ

By the continuity of

f (x, y)

(a, b)

, this implies:

| f (a, y) - f (a, b) | < ϵ

Thus, we have shown that

f (a, y)

is continuous at

y = b

Conclusion

We have successfully proven that if

f (x, y)

is continuous at

(a, b)

, then

f (x, b)

is continuous at

x = a

and

f (a, y)

is continuous at

y = b

. We used the definition of continuity to establish these results.

T : R^{2} \to R^{2}

एक रैखिक प्रतिचित्र है, जैसा कि

T (2, 1) = (5, 7)

एवं

T (1, 2) = (3, 3)

. अगर

A

मानक आधारों

e_{1}, e_{2}

के सापेक्ष

T

के संगत आव्यूह है, तो

A

की कोटि ज्ञात कीजिए।

Let

T : R^{2} \to R^{2}

be a linear map such that

T (2, 1) = (5, 7)

and

T (1, 2) = (3, 3)

.
If

A

is the matrix corresponding to

T

with respect to the standard bases

e_{1}, e_{2}

, then find

Rank (A)

Answer:

Expressing $(a, b)$ as a Linear Combination

(a, b) = α (2, 1) + β (1, 2)

a = 2 α + β (Equation 1)

b = α + 2 β (Equation 2)

Solving for $α$ and $β$

From Equation 2 multiplied by 2 minus Equation 1, we get:

3 β = 2 b - a

β = \frac{2 b - a}{3}

α = \frac{2 a - b}{3}

Finding $T (a, b)$ in terms of $T (2, 1)$ and $T (1, 2)$

T (a, b) = \frac{2 a - b}{3} T (2, 1) + \frac{2 b - a}{3} T (1, 2)

Finding $T (e_{1})$ and $T (e_{2})$

T (1, 0) = \frac{2}{3} (5, 7) - \frac{1}{3} (3, 3) = (\frac{7}{3}, \frac{11}{3})

T (0, 1) = - \frac{1}{3} (5, 7) + \frac{2}{3} (3, 3) = (\frac{1}{3}, - \frac{1}{3})

Finding Matrix $A$

A = [\begin{array}{cc} \frac{7}{3} & \frac{1}{3} \\ \frac{11}{3} & - \frac{1}{3} \end{array}]

Finding the Rank of $A$

The determinant of

A

is:

Det (A) = \frac{- 7 \times (- 1 / 3) - 11 \times 1 / 3}{3} = - \frac{18}{9} = - 2 \neq 0

Since the determinant is not zero, the matrix is invertible, and its rank is equal to the dimension of the domain, which is 2.

Conclusion

Rank (A) = 2

(d) अगर

A = [\begin{array}{rrr} 1 & 2 & 1 \\ 1 & - 4 & 1 \\ 3 & 0 & - 3 \end{array}] और B = [\begin{array}{rrr} 2 & 1 & 1 \\ 1 & - 1 & 0 \\ 2 & 1 & - 1 \end{array}]

है, तो दर्शाइए कि

A B = 6 I_{3}

. इस परिणाम का उपयोग करते हुए निम्नलिखित समीकरण निकाय को हल कीजिए :

\begin{array}{r} 2 x + y + z = 5 \\ x - y = 0 \\ 2 x + y - z = 1 \end{array}

A = [\begin{array}{rrr} 1 & 2 & 1 \\ 1 & - 4 & 1 \\ 3 & 0 & - 3 \end{array}] and B = [\begin{array}{rrr} 2 & 1 & 1 \\ 1 & - 1 & 0 \\ 2 & 1 & - 1 \end{array}]

then show that

A B = 6 I_{3}

. Use this result to solve the following system of equations :

\begin{array}{r} 2 x + y + z = 5 \\ x - y = 0 \\ 2 x + y - z = 1 \end{array}

Answer:

Introduction

The problem asks us to show that

A B = 6 I_{3}

for given matrices

A

and

B

, and then use this result to solve a system of equations.

I_{3}

is the

3 \times 3

identity matrix.

Work/Calculations

Step 1: Calculate $A B$

The first task is to calculate

A B

. The matrices

A

and

B

are:

A = (\begin{matrix} 1 & 2 & 1 \\ 1 & - 4 & 1 \\ 3 & 0 & - 3 \end{matrix})

B = (\begin{matrix} 2 & 1 & 1 \\ 1 & - 1 & 0 \\ 2 & 1 & - 1 \end{matrix})

\begin{aligned} A \times B = [\begin{array}{ccc} 1 & 2 & 1 \\ 1 & - 4 & 1 \\ 3 & 0 & - 3 \end{array}] \times [\begin{array}{ccc} 2 & 1 & 1 \\ 1 & - 1 & 0 \\ 2 & 1 & - 1 \end{array}] \\ = [\begin{array}{llll} 1 \times 2 + 2 \times 1 + 1 \times 2 & 1 \times 1 + 2 \times - 1 + 1 \times 1 & 1 \times 1 + 2 \times 0 + 1 \times - 1 \\ 1 \times 2 - 4 \times 1 + 1 \times 2 & 1 \times 1 - 4 \times - 1 + 1 \times 1 & 1 \times 1 - 4 \times 0 + 1 \times - 1 \\ 3 \times 2 + 0 \times 1 - 3 \times 2 & 3 \times 1 + 0 \times - 1 - 3 \times 1 & 3 \times 1 + 0 \times 0 - 3 \times - 1 \end{array}] \\ = [\begin{array}{ccc} 2 + 2 + 2 & 1 - 2 + 1 & 1 + 0 - 1 \\ 2 - 4 + 2 & 1 + 4 + 1 & 1 + 0 - 1 \\ 6 + 0 - 6 & 3 + 0 - 3 & 3 + 0 + 3 \end{array}] \\ = [\begin{array}{lll} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}] \end{aligned}

After calculating, we find that

A B

is:

A B = (\begin{matrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{matrix})

This is equivalent to

6 I_{3}

, where

I_{3}

is the

3 \times 3

identity matrix.

Step 2: Solve the System of Equations

The system of equations is:

\begin{aligned} 2 x + y + z & = 5 \\ x - y & = 0 \\ 2 x + y - z & = 1 \end{aligned}

We can write this system as

A x = b

, where

A

is the matrix from the problem statement, and

b = (\begin{matrix} 5 \\ 0 \\ 1 \end{matrix})

Since A B = 6 I_{3}, we have A^{- 1} = \frac{1}{6} B .

Here,

X = A^{- 1} \times b

\begin{aligned} ∴ X = \frac{1}{| A |} \times Adj (A) \times b \\ X = \frac{1}{6} \times [\begin{array}{ccc} 1 & 2 & 1 \\ 1 & - 4 & 1 \\ 3 & 0 & - 3 \end{array}] \times [\begin{matrix} 5 \\ 0 \\ 1 \end{matrix}] \\ X = \frac{1}{6} \times [\begin{array}{c} 1 \times 5 + 2 \times 0 + 1 \times 1 \\ 1 \times 5 - 4 \times 0 + 1 \times 1 \\ 3 \times 5 + 0 \times 0 - 3 \times 1 \end{array}] \\ X = \frac{1}{6} \times [\begin{array}{c} 6 \\ 6 \\ 12 \end{array}] \\ X = [\begin{array}{c} 1 \\ 1 \\ 2 \end{array}] \end{aligned}

Hence

(\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} 1 \\ 1 \\ 2 \end{matrix})

(e) दर्शाइए कि

\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1} और \frac{x}{1} = \frac{y - 7}{- 3} = \frac{z + 7}{2}

प्रतिच्छेदी रेखाएँ हैं। प्रतिच्छेद बिंदु के निर्देशांकों और उस समतल, जिसमें दोनों रेखाएँ हैं, का समीकरण ज्ञात कीजिए।

Show that the lines

\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1} and \frac{x}{1} = \frac{y - 7}{- 3} = \frac{z + 7}{2}

intersect. Find the coordinates of the point of intersection and the equation of the plane containing them.

Answer:

Proving Intersection of Lines

Given two lines:

Line 1:

\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}

Line 2:

\frac{x}{1} = \frac{y - 7}{- 3} = \frac{z + 7}{2}

Let’s parameterize these lines as follows:

Line 1:

(x, y, z) = (- 1 - 3 r, 3 + 2 r, - 2 + r) (1)

Line 2:

(x, y, z) = (r^{'}, 7 - 3 r^{'}, - 7 + 2 r^{'}) (2)

For the two lines to intersect, there must exist values of

r

and

r^{'}

such that the points on these lines coincide.

Step 1: Setting up Equations

We set up the equations:

\begin{aligned} - 1 - 3 r = r^{'} \\ 3 + 2 r = 7 - 3 r^{'} \\ - 2 + r = - 7 + 2 r^{'} \end{aligned}

Step 2: Solving Equations

Solving the first two equations:

From the first equation, we get

r = - 1

Substituting

r = - 1

into the second equation, we find

r^{'} = 2

Now, we check if these values satisfy the third equation:

- 2 + (- 1) = - 7 + 2 (2) ⟹ - 3 = - 3

So, the values

r = - 1

and

r^{'} = 2

satisfy all three equations.

Step 3: Finding Intersection Point

Substituting these values of

r

and

r^{'}

into either (1) or (2), we get the coordinates of the point of intersection:

From (1):

(x, y, z) = (2, 1, - 3)

Step 4: Finding the Equation of the Plane

The equation of the plane containing the given lines can be found using the cross product of the direction ratios of the lines.

Direction ratios of Line 1:

(- 3, 2, 1)

Direction ratios of Line 2:

(1, - 3, 2)

Now, we find the cross product of these direction ratios to get the normal vector of the plane:

\begin{aligned} | \begin{array}{ccc} i & j & k \\ - 3 & 2 & 1 \\ 1 & - 3 & 2 \end{array} | \\ = i \times | \begin{array}{cc} 2 & 1 \\ - 3 & 2 \end{array} | - j \times | \begin{array}{cc} - 3 & 1 \\ 1 & 2 \end{array} | + k \times | \begin{array}{cc} - 3 & 2 \\ 1 & - 3 \end{array} | \\ = i \times (2 \times 2 - 1 \times (- 3)) - j \times ((- 3) \times 2 - 1 \times 1) + k \times ((- 3) \times (- 3) - 2 \times 1) \\ = i \times (4 + 3) - j \times (- 6 - 1) + k \times (9 - 2) \\ = i \times (7) - j \times (- 7) + k \times (7) \\ = 7 i + 7 j + 7 k \\ = 7 i + 7 k + 7 j \end{aligned}

So, the normal vector of the plane is

7 i + 7 j + 7 k

Now, we can use one of the intersection points, e.g.,

(2, 1, - 3)

, and the normal vector to find the equation of the plane:

7 (x - 2) + 7 (y - 1) + 7 (z + 3) = 0

Simplifying:

\Rightarrow x + y + z = 0

This is the equation of the plane containing the given lines.

(a) क्या $f (x) = | \cos x | + | \sin x |, x = \frac{π}{2}$ पर अवकलनीय है? अगर आपका उत्तर हाँ है, तो $f (x)$ का अवकलज $x = \frac{π}{2}$ पर ज्ञात कीजिए। अगर आपका उत्तर ना है, तो अपने उत्तर का प्रमाण दीजिए।

f (x) = | \cos x | + | \sin x |

differentiable at

x = \frac{π}{2}

? If yes, then find its derivative at

x = \frac{π}{2}

. If no, then give a proof of it.

Answer:

Introduction

We’re going to look at a pretty cool math problem. We have a function

f (x) = | \cos x | + | \sin x |

, and we want to find out if it’s differentiable at

x = \frac{π}{2}

. If it is, we’ll also find its derivative at that point. To make things easier, we’ll break down the function into a piecewise function and then check its differentiability.

Work/Calculations

Step 1: Breaking It Down

First, let’s rewrite

f (x)

as a piecewise function around

x = \frac{π}{2}

f (x) = {\begin{cases} - \cos x + \sin x & if x < \frac{π}{2} \\ \cos x + \sin x & if x > \frac{π}{2} \\ 1 & if x = \frac{π}{2} \end{cases}

Step 2: Checking the Derivative

To see if

f (x)

is differentiable at

x = \frac{π}{2}

, we need to find the derivative from both sides of that point and see if they match.

Left-hand derivative: The derivative of $- \cos x + \sin x$ is $\sin x + \cos x$ .
Right-hand derivative: The derivative of $\cos x + \sin x$ is $- \sin x + \cos x$ .

Now, let’s plug

x = \frac{π}{2}

into both of these derivatives and see what we get.

After calculating, we find that both the left-hand and right-hand derivatives at

x = \frac{π}{2}

are

1

. That’s awesome because it means the function is differentiable at

x = \frac{π}{2}

Conclusion

So, there we have it! The function

f (x) = | \cos x | + | \sin x |

is differentiable at

x = \frac{π}{2}

, and its derivative at that point is

1

(b) माना कि

A

और

B

समान कोटि के दो लांबिक आव्यूह हैं तथा

\det A + \det B = 0

. दर्शाइए कि

A + B

एक अव्युत्क्रमणीय (सिंगुलर) आव्यूह है।

Let

A

and

B

be two orthogonal matrices of same order and

\det A + \det B = 0

. Show that

A + B

is a singular matrix.

Answer:

Proving that A + B is Singular

Given:

$A$ and $B$ are two orthogonal matrices of the same order.
$det A + det B = 0$

We want to show that

A + B

is a singular matrix.

Step 1: Given Orthogonal Matrices

Given that

A A^{T} = B B^{T} = I

, where

I

is the identity matrix.

Also,

det A = det B = 1

because orthogonal matrices have a determinant of

1

- 1

, and we’re given that

det A + det B = 0

, which implies

det A = - det B

Step 2: Determinant of AB

We know that

det A \cdot det B = - 1

. This implies that

det (A B) = - 1

. (This is due to the property of determinants that

det (A B) = det (A) \cdot det (B)

Step 3: Expressing A + B

Now, let’s express

A + B

in terms of

A

and

B

A + B = A I + B I = A (B^{T} + A^{T}) B

Step 4: Determinant of A + B

We want to find the determinant of

A + B

\begin{aligned} det (A + B) & = det (A) \cdot det (B^{T} + A^{T}) \cdot det (B) \\ = det (A B) \cdot det (B^{T} + A^{T}) \\ = (- 1) \cdot det (B^{T} + A^{T}) (From Step 2) \\ = - det (B^{T} + A^{T}) \end{aligned}

Step 5: Determinant of Transpose Sum

Now, let’s focus on the determinant of the transpose sum:

\begin{aligned} det (B^{T} + A^{T}) & = det ((B + A)^{T}) (Transpose of a sum) \\ = det (B + A) \end{aligned}

Step 6: Putting It All Together

Now, we can express

det (A + B)

as:

det (A + B) = - det (B + A)

But

A + B = B + A

because matrix addition is commutative.

So, we have:

det (A + B) = - det (B + A) = - det (A + B)

This implies that

2 det (A + B) = 0

Step 7: Conclusion

From the equation

2 det (A + B) = 0

, we can conclude that

det (A + B) = 0

A matrix is considered singular if its determinant is zero. Therefore, we have shown that

A + B

is a singular matrix.

x + 2 y + 3 z = 12

निर्देशांक अक्षों को

A, B, C

पर प्रतिच्छेद करता है। त्रिभुज

A B C

के परिवृत्त का समीकरण ज्ञात कीजिए।
(ii) सिद्ध कीजिए कि समतल

z = 0

गोलक

x^{2} + y^{2} + z^{2} = 11

के अन्वालोपी शंकु, जिसका शीर्ष

(2, 4, 1)

पर है, को एक समकोणीय अतिपरवलय पर प्रतिच्छेद करता है।

(i) The plane

x + 2 y + 3 z = 12

cuts the axes of coordinates in

A, B, C

. Find the equations of the circle circumscribing the triangle

A B C

Answer:

To find the equation of the circle circumscribing the triangle ABC, we start with the given plane equation:

x + 2 y + 3 z = 12

This plane intersects the x-axis, y-axis, and z-axis at points A, B, and C, respectively. We’ve already found the coordinates of these points:

A(12, 0, 0)
B(0, 6, 0)
C(0, 0, 4)

Now, let’s assume the equation of the circumscribing circle is:

x^{2} + y^{2} + z^{2} + 2 u x + 2 v y + 2 w z + d = 0

We need to find the values of u, v, w, and d.

Using point A(12, 0, 0):

(12)^{2} + 2 u (12) + d = 0

144 + 24 u + d = 0

24 u + d = - 144 - - - - (1)

Using point B(0, 6, 0):

(6)^{2} + 2 v (6) + d = 0

36 + 12 v + d = 0

12 v + d = - 36 - - - - (2)

Using point C(0, 0, 4):

(4)^{2} + 2 w (4) + d = 0

16 + 8 w + d = 0

8 w + d = - 16 - - - - (3)

Now, we can solve equations (1), (2), and (3) for u, v, and w:

$24 u + d = - 144$ implies $24 u = - 144 - d$
$12 v + d = - 36$ implies $12 v = - 36 - d$
$8 w + d = - 16$ implies $8 w = - 16 - d$

Now, we can substitute these expressions for 2u, 2v, and 2w back into the equation of the circle:

x^{2} + y^{2} + z^{2} - (12 - d / 12) x - (6 - d / 6) y - (4 - d / 4) z + d = 0

Simplify further:

x^{2} + y^{2} + z^{2} - (12 - d / 12) x - (6 - d / 6) y - (4 - d / 4) z + d = 0

Now, the equation of the circle is in the form:

x^{2} + y^{2} + z^{2} + A x + B y + C z + D = 0

Where:

A = - (12 - d / 12)

B = - (6 - d / 6)

C = - (4 - d / 4)

D = d

The values of A, B, C, and D can vary as d takes any real value. So, the equation of the circle circumscribing the triangle ABC is:

x^{2} + y^{2} + z^{2} - (12 - d / 12) x - (6 - d / 6) y - (4 - d / 4) z + d = 0

Where d can be any real number.

(ii) Prove that the plane

z = 0

cuts the enveloping cone of the sphere

x^{2} + y^{2} + z^{2} = 11

which has the vertex at

(2, 4, 1)

in a rectangular hyperbola.

Answer:

Given the equation of the sphere

S = x^{2} + y^{2} + z^{2} - 11

and the coordinates of its vertex

(α, β, γ) = (2, 4, 1)

Calculate $S_{1}$ and the equation of the plane $T$ :
$\begin{aligned} S_{1} & = α^{2} + β^{2} + γ^{2} - 11 = 2^{2} + 4^{2} + 1^{2} - 11 = 10 \\ T & = 2 x + 4 y + z - 11 \end{aligned}$
Write the equation of the enveloping cone of the sphere:
The equation of the enveloping cone can be written as $T^{2} = S S_{1}$ , where $T$ is the equation of the plane and $S$ is the equation of the sphere.
$(2 x + 4 y + z - 11)^{2} = (x^{2} + y^{2} + z^{2} - 11) (10)$
Substitute $z = 0$ into the equation:
To check if the plane $z = 0$ cuts the cone, put $z = 0$ in the above equation:
$10 (x^{2} + y^{2} - 11) = (2 x + 4 y - 11)^{2}$
Simplify the equation:
Further simplify the equation:
$6 x^{2} + 6 y^{2} - 16 xy + 44 x - 88 y - 231 = 0$

The resulting equation is a rectangular hyperbola, which proves that the plane

z = 0

cuts the enveloping cone of the sphere in a rectangular hyperbola.

(a) फलन $f (x) = 2 x^{3} - 9 x^{2} + 12 x + 6$ का अंतराल $[2, 3]$ पर अधिकतम और न्यूनतम मान ज्ञात कीजिए।

Find the maximum and the minimum value of the function

f (x) = 2 x^{3} - 9 x^{2} + 12 x + 6

on the interval

[2, 3]

Answer:

Introduction

The problem asks us to find the maximum and minimum values of the function

f (x) = 2 x^{3} - 9 x^{2} + 12 x + 6

on the interval

[2, 3]

. To find these extremum points, we’ll use calculus methods, specifically by finding the derivative of the function and setting it equal to zero to find critical points. We’ll then evaluate the function at these critical points and the endpoints of the interval to determine the maximum and minimum values.

Work/Calculations

Step 1: Find the Derivative of $f (x)$

The first derivative of

f (x)

will give us the rate of change of the function at any given point

x

f^{'} (x) = \frac{d}{d x} (2 x^{3} - 9 x^{2} + 12 x + 6)

After calculating, we find that the derivative of

f (x)

is:

f^{'} (x) = 6 x^{2} - 18 x + 12

Step 2: Find the Critical Points

To find the critical points, we set

f^{'} (x) = 0

and solve for

x

6 x^{2} - 18 x + 12 = 0

After solving, we find that the critical points are

x = 1

and

x = 2

Step 3: Evaluate $f (x)$ at Critical Points and Endpoints

To find the maximum and minimum values of

f (x)

on the interval

[2, 3]

, we need to evaluate

f (x)

at the critical points and the endpoints of the interval. The critical points within the interval are

x = 2

and the endpoints are

x = 2

and

x = 3

Let’s substitute the values into

f (x) = 2 x^{3} - 9 x^{2} + 12 x + 6

and calculate:

$f (2)$
$f (3)$

Let’s start by calculating

f (2)

After substituting the values, we get:

f (2) = 2 \times 2^{3} - 9 \times 2^{2} + 12 \times 2 + 6 = 10

Next, let’s calculate

f (3)

After substituting the values, we get:

f (3) = 2 \times 3^{3} - 9 \times 3^{2} + 12 \times 3 + 6 = 15

Conclusion

We evaluated

f (x)

at the critical point

x = 2

and the endpoint

x = 3

within the interval

[2, 3]

$f (2) = 10$
$f (3) = 15$

Therefore, the maximum value of

f (x)

on the interval

[2, 3]

15

x = 3

, and the minimum value is

10

x = 2

(b) सिद्ध कीजिए कि साधारणतः किसी एक बिंदु से परवलयज

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

पर तीन अभिलंब बनाए जा सकते हैं, लेकिन अगर बिंदु सतह

27 a (x^{2} + y^{2}) + 8 (a - z)^{3} = 0

पर स्थित है, तो इन तीन अभिलंबों में से दो अभिलंब एक ही हैं।

Prove that, in general, three normals can be drawn from a given point to the paraboloid

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

, but if the point lies on the surface

27 a (x^{2} + y^{2}) + 8 (a - z)^{3} = 0

then two of the three normals coincide.

Answer:

Introduction:
We are tasked with proving that, in general, three normals can be drawn from a given point to the paraboloid

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

. However, if the point lies on the surface

27 a (x^{2} + y^{2}) + 8 (a - z)^{3} = 0

, then two of the three normals coincide.

Work/Calculations:

The equation of the normal at $(x_{1}, y_{1}, z_{1})$ to the paraboloid is given by:

\frac{x - x_{1}}{x_{1}} = \frac{y - y_{1}}{y_{1}} = \frac{z - z_{1}}{- a}

This passes through a given point $(α, β, γ)$ if:

\frac{α - x_{1}}{x_{1}} = \frac{β - y_{1}}{y_{1}} = \frac{γ - z_{1}}{- a} = λ

Solving for $x_{1}$ , $y_{1}$ , and $z_{1}$ in terms of $α$ , $β$ , $γ$ , and $λ$ using the above equations:

x_{1} = \frac{α}{1 + λ}; y_{1} = \frac{β}{1 + λ}; z_{1} = γ + a λ (from (1))

Also, $(x_{1}, y_{1}, z_{1})$ lies on the given paraboloid, so:

x_{1}^{2} + y_{1}^{2} = 2 a z_{1} (from the equation of the paraboloid)

Substituting the expressions for $x_{1}$ , $y_{1}$ , and $z_{1}$ from step 3 into the equation above:

{(\frac{α}{1 + λ})}^{2} + {(\frac{β}{1 + λ})}^{2} = 2 a (γ + a λ) (from (1))

This equation can be simplified to:

α^{2} + β^{2} = 2 a (γ + a λ) (1 + λ)^{2} (from (2))

Equation (2) is a cubic in $λ$ and has three values of $λ$ that satisfy it, leading to three points on the paraboloid normal at which pass through $(α, β, γ)$ .
Rewriting equation (2) as a function $f (λ)$ :

f (λ) + 2 a (1 + λ)^{2} (γ + a λ) - (α^{2} - β^{2}) = 0 (from (3))

To find the condition that equation (3) has two equal roots, we need to solve for $λ$ where both $f (λ) = 0$ and $f^{'} (λ) = 0$ .
Differentiating $f (λ)$ with respect to $λ$ to find $f^{'} (λ)$ :

f^{'} (λ) = 2 a (1 + λ)^{2} (a) + 4 a (1 + λ) (γ + a λ) = 0

Solving for $λ$ in $f^{'} (λ) = 0$ :

a (1 + λ) + 2 (γ + a λ) = 0 (since 1 + λ \neq 0)

Simplifying and solving for $λ$ :

(a + 2 γ) + λ (3 a) = 0 \Rightarrow λ = - \frac{a + 2 γ}{3 a}

Substituting this value of $λ$ into equation (3):

2 a {[1 - \frac{a + 2 γ}{3 a}]}^{2} [γ - \frac{a (a + 2 γ)}{3 a}] = α^{2} + β^{2}

Further simplifying this equation:

2 a [2 (a - γ)]^{2} [a (γ - a)] = 27 a^{3} (α^{2} + β^{2})

Finally, the equation simplifies to:

27 a (α^{2} + β^{2}) + 8 (a - γ)^{3} = 0

Conclusion:
We have successfully shown that the locus of the point

(α, β, γ)

27 a (x^{2} + y^{2}) + 8 (a - z)^{3} = 0

. This proves the given statement.

A = (\begin{array}{rrrr} 5 & 7 & 2 & 1 \\ 1 & 1 & - 8 & 1 \\ 2 & 3 & 5 & 0 \\ 3 & 4 & - 3 & 1 \end{array})

(i) आव्यूह

A

की कोटि ज्ञात कीजिए।
(ii) उपसमष्टि

V = {(x_{1}, x_{2}, x_{3}, x_{4}) \in R^{4} ∣ A (\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}) = 0}

की विमा ज्ञात कीजिए।

Let

A = (\begin{array}{rrrr} 5 & 7 & 2 & 1 \\ 1 & 1 & - 8 & 1 \\ 2 & 3 & 5 & 0 \\ 3 & 4 & - 3 & 1 \end{array})

(i) Find the rank of matrix

A

(ii) Find the dimension of the subspace

V = {(x_{1}, x_{2}, x_{3}, x_{4}) \in R^{4} ∣ A (\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}) = 0}

Answer:

Rank [\begin{array}{cccc} 5 & 7 & 2 & 1 \\ 1 & 1 & - 8 & 1 \\ 2 & 3 & 5 & 0 \\ 3 & 4 & - 3 & 1 \end{array}]

Now, reduce this matrix

\begin{aligned} R_{1} \leftarrow R_{1} \div 5 \\ = [\begin{array}{cccc} 1 & \frac{7}{5} & \frac{2}{5} & \frac{1}{5} \\ 1 & 1 & - 8 & 1 \\ 2 & 3 & 5 & 0 \\ 3 & 4 & - 3 & 1 \end{array}] \end{aligned}

\begin{aligned} \begin{aligned} R_{2} \leftarrow R_{2} - R_{1} \\ = [\begin{array}{cccc} 1 & \frac{7}{5} & \frac{2}{5} & \frac{1}{5} \\ 0 & - \frac{2}{5} & - \frac{42}{5} & \frac{4}{5} \\ 2 & 3 & 5 & 0 \\ 3 & 4 & - 3 & 1 \end{array}] \end{aligned} \\ \begin{aligned} R_{3} \leftarrow R_{3} - 2 \times R_{1} \\ = [\begin{array}{cccc} 1 & \frac{7}{5} & \frac{2}{5} & \frac{1}{5} \\ 0 & - \frac{2}{5} & - \frac{42}{5} & \frac{4}{5} \\ 0 & \frac{1}{5} & \frac{21}{5} & - \frac{2}{5} \\ 3 & 4 & - 3 & 1 \end{array}] \end{aligned} \end{aligned}

\begin{aligned} R_{4} \leftarrow R_{4} - 3 \times R_{1} \\ = [\begin{array}{cccc} 1 & \frac{7}{5} & \frac{2}{5} & \frac{1}{5} \\ 0 & - \frac{2}{5} & - \frac{42}{5} & \frac{4}{5} \\ 0 & \frac{1}{5} & \frac{21}{5} & - \frac{2}{5} \\ 0 & - \frac{1}{5} & - \frac{21}{5} & \frac{2}{5} \end{array}] \end{aligned}

\begin{aligned} R_{2} \leftarrow R_{2} \times - \frac{5}{2} \\ = [\begin{array}{cccc} 1 & \frac{7}{5} & \frac{2}{5} & \frac{1}{5} \\ 0 & 1 & 21 & - 2 \\ 0 & \frac{1}{5} & \frac{21}{5} & - \frac{2}{5} \\ 0 & - \frac{1}{5} & - \frac{21}{5} & \frac{2}{5} \end{array}] \\ R_{3} \leftarrow R_{3} - \frac{1}{5} \times R_{2} \\ = [\begin{array}{cccc} 1 & \frac{7}{5} & \frac{2}{5} & \frac{1}{5} \\ 0 & 1 & 21 & - 2 \\ 0 & 0 & 0 & 0 \\ 0 & - \frac{1}{5} & - \frac{21}{5} & \frac{2}{5} \end{array}] \end{aligned}

\begin{matrix} R_{4} \leftarrow R_{4} + \frac{1}{5} \times R_{2} \\ = [\begin{array}{cccc} 1 & \frac{7}{5} & \frac{2}{5} & \frac{1}{5} \\ 0 & 1 & 21 & - 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}] \end{matrix}

The rank of a matrix is the number of non all-zeros rows

∴ Rank = 2

(ii) Find the dimension of the subspace

V

The dimension of the subspace

V

is related to the rank of

A

by the formula:

Dimension of

V = n - Rank (A)

Here,

n

is the number of columns in

A

, which is 4 .
Dimension of

V = 4 - 2 = 2

(a) कैले-हैमिल्टन प्रमेय का कथन लिखिए। इस प्रमेय का उपयोग करके $A^{100}$ का मान ज्ञात कीजिए, जहाँ

A = [\begin{array}{lll} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}]

State the Cayley-Hamilton theorem. Use this theorem to find

A^{100}

, where

A = [\begin{array}{lll} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}]

Answer:

Cayley-Hamilton Theorem:
The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation over a commutative ring, such as the real or complex field. In other words, if

λ I - A

is the characteristic equation of a matrix

A

, where

I

is the identity matrix and

λ

is a scalar, then

A

satisfies the equation

| λ I - A | = 0

First, let’s find the characteristic equation for matrix $A$ :

\begin{aligned} | λ I - A | & = | \begin{array}{ccc} λ - 1 & 0 & 0 \\ - 1 & λ & - 1 \\ 0 & - 1 & λ \end{array} | \\ = (λ - 1) | \begin{array}{cc} λ & - 1 \\ - 1 & λ \end{array} | \\ = (λ - 1) (λ^{2} - 1) \\ = (λ - 1) (λ + 1) (λ - 1) \end{aligned}

Now, according to the Cayley-Hamilton theorem, $A$ satisfies its characteristic equation, so $A$ satisfies:

A^{3} - A^{2} - A + I = 0

Rearranging the equation, we find:

A^{3} = A^{2} + A - I

We can now calculate higher powers of $A$ based on this result:

\begin{aligned} A^{4} & = A^{3} + A^{2} - A \\ = (A^{2} + A - I) + A^{2} - A \\ = 2 A^{2} - I \end{aligned}

Similarly, we can find $A^{5}$ as:

\begin{aligned} A^{5} & = 2 A^{3} - A \\ = 2 (A^{2} + A - I) - A \\ = 2 A^{2} + 2 A - 2 I - A \\ = 2 A^{2} + A - 2 I \end{aligned}

Continuing this pattern, we find:

A^{6} = 3 A^{2} - 2 I

Use the General Form $A^{2 n} = n A^{2} - (n - 1) I$

The general form

A^{2 n} = n A^{2} - (n - 1) I

can be used to find

A^{100}

For

n = 50

, the expression becomes:

A^{100} = 50 A^{2} - 49 I

Substitute the Values

Let’s substitute the values of

A^{2}

and

I

into the formula:

A^{2} = [\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}]

I = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]

After substituting, we get:

A^{100} = 50 [\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}] - 49 [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]

Calculate $A^{100}$

After calculating, we find that

A^{100}

is:

A^{100} = [\begin{array}{lll} 1 & 0 & 0 \\ 50 & 1 & 0 \\ 50 & 0 & 1 \end{array}]

Conclusion

Using the general form

A^{2 n} = n A^{2} - (n - 1) I

, we found that

A^{100}

for the given matrix

A

is:

A^{100} = [\begin{array}{lll} 1 & 0 & 0 \\ 50 & 1 & 0 \\ 50 & 0 & 1 \end{array}]

(b) बिंदु

P

से गुजरने वाली दीर्घवृत्तज

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

की अभिलंब जीवा की लंबाई ज्ञात कीजिए और सिद्ध कीजिए कि अगर यह

4 P G_{3}

के समान है, जहाँ

G_{3}

वह बिंदु है जहाँ

P

से गुजरने वाली अभिलंब जीवा

x y

-तल पर मिलती है, तो

P

शंकु

\frac{x^{2}}{a^{6}} (2 c^{2} - a^{2}) + \frac{y^{2}}{b^{6}} (2 c^{2} - b^{2}) + \frac{z^{2}}{c^{4}} = 0

पर स्थित है।

Find the length of the normal chord through a point

P

of the ellipsoid

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

and prove that if it is equal to

4 P G_{3}

, where

G_{3}

is the point where the normal chord through

P

meets the

x y

-plane, then

P

lies on the cone

\frac{x^{2}}{a^{6}} (2 c^{2} - a^{2}) + \frac{y^{2}}{b^{6}} (2 c^{2} - b^{2}) + \frac{z^{2}}{c^{4}} = 0

Answer:

Equation of the Normal:
Let

P

(α, β, γ)

, then the equation of the normal to the given ellipsoid at

P (α, β, γ)

are

\frac{x - α}{\frac{p α}{a^{2}}} = \frac{y - β}{\frac{P β}{b^{2}}} = \frac{z - r}{\frac{P γ}{c^{2}}} = r \to (1) where \frac{1}{P^{2}} = \frac{α^{2}}{a^{4}} + \frac{β^{2}}{b^{4}} + \frac{γ^{2}}{c^{4}} \to (2)

Therefore, the coordinates of any point

Q

on the normal (1) are

(α + \frac{P α}{a^{2}} r, β + \frac{P β}{b^{2}} r, γ + \frac{p γ}{c^{2}} r)

, where

r

is the distance of

Q

from

P

If $Q$ lies on the given ellipsoid i.e., $PQ$ is the normal chord, then

\begin{aligned} \frac{1}{a^{2}} {(α + \frac{p α}{a^{2}} r)}^{2} + \frac{1}{b^{2}} {(β + \frac{p β}{b^{2}} r)}^{2} + \frac{1}{c^{2}} {(γ + \frac{p γ}{c^{2}} r)}^{2} = 1 \\ \Rightarrow r^{2} P^{2} (\frac{α^{2}}{a^{6}} + \frac{β^{2}}{b^{6}} + \frac{γ^{2}}{c^{6}}) + 2 r P (\frac{α^{2}}{a^{4}} + \frac{β^{2}}{b^{4}} + \frac{γ^{2}}{c^{4}}) + (\frac{α^{2}}{a^{2}} + \frac{β^{2}}{b^{2}} + \frac{γ^{2}}{c^{2}}) = 1 \end{aligned}

\Rightarrow r^{2} p^{2} (\frac{α^{2}}{a^{4}} + \frac{β^{2}}{b^{6}} + \frac{γ^{2}}{c^{6}}) + 2 r p (\frac{1}{p^{2}}) = 0 [

from (2) and

Σ (\frac{α^{2}}{a^{2}}) = 1

p (α, β, γ)

lies on the given conicoid ]

\Rightarrow r = - \frac{2}{p^{3} (\frac{α^{2}}{a^{6}} + \frac{β^{2}}{b^{6}} + \frac{γ^{2}}{c^{6}})} = length of normal chord P Q \to (3)

Also,
Let the normal at

P (α, β, γ)

meets the coordinate plane viz.

y z, z x

and

x y

planes at

G_{1}, G_{2}, G_{3}

Then putting

x = 0, y = 0

and

z = 0

in succession in the equation (1), we have respectively,

P G_{1} = - \frac{a^{2}}{P}, P G_{2} = - \frac{b^{2}}{p} and P G_{3} = - \frac{c^{2}}{p} \to (4)

Given, $P Q = 4 P G_{3}$

\begin{aligned} \Rightarrow p Q = 4 (- \frac{c^{2}}{p}) \\ \Rightarrow r = - \frac{2}{p^{3} (\frac{α^{2}}{a^{6}} + \frac{β^{2}}{b^{6}} + \frac{γ^{2}}{c^{6}})} = 4 (- \frac{c^{2}}{p}) \\ \Rightarrow 2 c^{2} (\frac{α^{2}}{a^{6}} + \frac{β^{2}}{b^{6}} + \frac{γ^{2}}{c^{6}}) = \frac{1}{p^{2}} \\ = (\frac{α^{2}}{a^{4}} + \frac{β^{2}}{b^{4}} + \frac{γ^{2}}{c^{4}}) \to from (2) \\ \Rightarrow \frac{α^{2}}{a^{6}} (2 c^{2} - a^{2}) + \frac{β^{2}}{b^{2}} (2 c^{2} - b^{2}) + \frac{γ^{2}}{c^{6}} (2 c^{2} - c^{2}) = 0 \\ \Rightarrow \frac{α^{2}}{a^{6}} (2 c^{2} - a^{2}) + \frac{β^{2}}{b^{2}} (2 c^{2} - b^{2}) + \frac{γ^{2}}{c^{6}} (c^{2}) = 0 \\ \Rightarrow \frac{α^{2}}{a^{6}} (2 c^{2} - a^{2}) + \frac{β^{2}}{b^{2}} (2 c^{2} - b^{2}) + \frac{γ^{2}}{c^{4}} = 0 \end{aligned}

Conclusion:
We have found the length of the normal chord through a point

P

on the given ellipsoid and proved that if this length is equal to

4 P G_{3}

, then

P

lies on the cone defined by the equation

\frac{α^{2}}{a^{6}} (2 c^{2} - a^{2}) + \frac{β^{2}}{b^{2}} (2 c^{2} - b^{2}) + \frac{γ^{2}}{c^{4}} = 0

Hence proved.

u = \sin^{- 1} \sqrt{\frac{x^{1 / 3} + y^{1 / 3}}{x^{1 / 2} + y^{1 / 2}}}

है, तो दर्शाइए कि

\sin^{2} u, x

और

y

का

- \frac{1}{6}

घातविशिष्ट समांगी फलन है। अतएव दर्शाइए कि

x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}} = \frac{\tan u}{12} (\frac{13}{12} + \frac{\tan^{2} u}{12})

(ii) जैकोबियन विधि का व्यवहार करते हुए दर्शाइए कि अगर

f^{'} (x) = \frac{1}{1 + x^{2}}

और

f (0) = 0

है, तो

f (x) + f (y) = f (\frac{x + y}{1 - x y})

(i) If

u = \sin^{- 1} \sqrt{\frac{x^{1 / 3} + y^{1 / 3}}{x^{1 / 2} + y^{1 / 2}}}

then show that

\sin^{2} u

is a homogeneous function of

x

and

y

of degree

- \frac{1}{6}

.
Hence show that

x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}} = \frac{\tan u}{12} (\frac{13}{12} + \frac{\tan^{2} u}{12})

Answer:

Introduction:
We are given the equation

u = \sin^{- 1} \sqrt{\frac{x^{1 / 3} + y^{1 / 3}}{x^{1 / 2} + y^{1 / 2}}}

and asked to show that

\sin^{2} u

is a homogeneous function of

x

and

y

of degree

- \frac{1}{6}

. Additionally, we need to demonstrate that

x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}} = \frac{\tan u}{12} (\frac{13}{12} + \frac{\tan^{2} u}{12})

Homogeneity Analysis:

Step 1: Defining $f$ and Finding Degree of Homogeneity):

Let’s start by defining

f = \sin u

and then find the degree of homogeneity.

f = \sin u = \sqrt{\frac{x^{1 / 3} + y^{1 / 3}}{x^{1 / 2} + y^{1 / 2}}}

The degree of homogeneity is found as:

Degree = \frac{1}{3} - \frac{1}{2} = - \frac{1}{6}

Step 2: Expressing $f$ in terms of $x$ and $y$ :

Now, we express

f

in terms of

x

and

y

f = x^{- \frac{1}{12}} g (\frac{y}{x})

Step 3: Determining Homogeneous Function of $u$ :

Next, we find

\sin^{2} u

as a homogeneous function of

x

and

y

\sin^{2} u = x^{(- \frac{1}{6}) g (\frac{y}{x})}

The degree is confirmed to be

- \frac{1}{6}

Derivatives and Further Analysis:

Step 4: Partial Derivatives:

Now, let’s calculate the partial derivatives

\frac{\partial f}{\partial x}

and

\frac{\partial f}{\partial y}

using the homogeneous function properties:

x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n \cdot f where n = - \frac{1}{6}

Step 5: Derivative of $\sin u$ and Simplification:

Since

f = \sin u

, we differentiate with respect to

x

x \cos u \frac{\partial u}{\partial x} + y \cos u \frac{\partial u}{\partial y} = - \frac{1}{12} \sin u

After some simplification and using trigonometric identities, we get:

x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = - \frac{1}{12} \tan u

Step 6: Second Derivative and Further Simplification:

Differentiate equation 1 partially with respect to

x

and

y

and add the resulting equations. Then simplify as follows:

\begin{aligned} x \frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial u}{\partial x} + y \frac{\partial^{2} u}{\partial x \partial y} = - \frac{1}{12} \sec^{2} u \frac{\partial u}{\partial x} \\ y^{2} \frac{\partial u}{\partial y} + y \frac{\partial u}{\partial y} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} = (x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}) (1 - \frac{1}{12} \sec^{2} u) \\ [1 + \frac{1}{12} (\tan^{2} α + 1)] = - \frac{1}{12} \tan u (1 + \frac{1}{12} \sec^{2} u) \\ = \frac{\tan u}{12} (\frac{13}{12} + \frac{\tan^{2} u}{12}) \end{aligned}

Conclusion:

We have shown that

\sin^{2} u

is indeed a homogeneous function of

x

and

y

with a degree of

- \frac{1}{6}

. Additionally, we have derived and simplified the expression

x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}}

as given in the problem statement.

(ii) Using the Jacobian method, show that if

f^{'} (x) = \frac{1}{1 + x^{2}}

and

f (0) = 0

, then

f (x) + f (y) = f (\frac{x + y}{1 - x y})

Answer:

Work/Calculations

Step 1: Integrate $f^{'} (x)$ to find $f (x)$

Given

f^{'} (x) = \frac{1}{1 + x^{2}}

, we integrate to find

f (x)

\int f^{'} (x) d x = \int \frac{1}{1 + x^{2}} d x

f (x) = \tan^{- 1} (x) + c

Since

f (0) = 0

, we find that

c = 0

. Therefore,

f (x) = \tan^{- 1} (x)

Step 2: Define $α$ and $β$

α = f (x) + f (y) = \tan^{- 1} (x) + \tan^{- 1} (y)

β = f (\frac{x + y}{1 - x y}) = \tan^{- 1} (\frac{x + y}{1 - x y})

Step 3: Calculate the Partial Derivatives of $α$ and $β$

For

α

, we have:

\frac{\partial α}{\partial x} = \frac{1}{1 + x^{2}}

\frac{\partial α}{\partial y} = \frac{1}{1 + y^{2}}

For

β

, we differentiate

β

with respect to

x

\frac{\partial β}{\partial x} = \frac{\partial}{\partial x} [\tan^{- 1} (\frac{x + y}{1 - x y})]

After calculating, we find:

\frac{\partial β}{\partial x} = \frac{1}{1 + x^{2}}

Similarly, to find

\frac{\partial β}{\partial y}

, we differentiate

β

with respect to

y

\frac{\partial β}{\partial y} = \frac{\partial}{\partial y} [\tan^{- 1} (\frac{x + y}{1 - x y})]

After calculating, we find:

\frac{\partial β}{\partial y} = \frac{1}{1 + y^{2}}

Step 4: Check if the Partial Derivatives are Equal

To prove that

f (x) + f (y) = f (\frac{x + y}{1 - x y})

, it’s sufficient to show that

\frac{\partial α}{\partial x} = \frac{\partial β}{\partial x}

and

\frac{\partial α}{\partial y} = \frac{\partial β}{\partial y}

Since

\frac{\partial α}{\partial x} = \frac{\partial β}{\partial x}

and

\frac{\partial α}{\partial y} = \frac{\partial β}{\partial y}

, we can conclude that

f (x) + f (y) = f (\frac{x + y}{1 - x y})

Conclusion with Jacobian Method

In the Jacobian method, we aim to show that two functions

α

and

β

are equal by proving that their partial derivatives with respect to the same variables are equal. In this case,

α = f (x) + f (y)

and

β = f (\frac{x + y}{1 - x y})

We calculated the partial derivatives

\frac{\partial α}{\partial x}

\frac{\partial α}{\partial y}

\frac{\partial β}{\partial x}

, and

\frac{\partial β}{\partial y}

and found:

\frac{\partial α}{\partial x} = \frac{1}{1 + x^{2}} = \frac{\partial β}{\partial x}

\frac{\partial α}{\partial y} = \frac{1}{1 + y^{2}} = \frac{\partial β}{\partial y}

Since both sets of partial derivatives are equal, the Jacobian method confirms that

α

and

β

are indeed the same function under the transformation of variables. Therefore, we can conclude that:

f (x) + f (y) = f (\frac{x + y}{1 - x y})

This successfully proves the given equation using the Jacobian method.

खण्ड-B / SECTION-B

(a) अवकल समीकरण

(2 y \sin x + 3 y^{4} \sin x \cos x) d x - (4 y^{3} \cos^{2} x + \cos x) d y = 0

को हल कीजिए।

Solve the differential equation

(2 y \sin x + 3 y^{4} \sin x \cos x) d x - (4 y^{3} \cos^{2} x + \cos x) d y = 0

Answer:

Introduction:
We are given the differential equation:

(2 y \sin x + 3 y^{4} \sin x \cos x) d x - (4 y^{3} \cos^{2} x + \cos x) d y = 0

Work/Calculations:

Step 1: Identify $M$ and $N$ :
In the given equation, we have

M = 2 y \sin x + 3 y^{4} \sin x \cos x

and

N = - (4 y^{3} \cos^{2} x + \cos x)

Step 2: Calculate Partial Derivatives:
Calculate the partial derivatives of

M

and

N

\frac{\partial M}{\partial y} = 2 \sin x + 12 y^{3} \sin x \cos x

\frac{\partial N}{\partial x} = 8 y^{3} \sin x \cos x + \sin x

Step 3: Check for Exactness:
Now, check if the equation is exact by comparing

\frac{\partial M}{\partial y}

and

\frac{\partial N}{\partial x}

\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} = \frac{2 \sin x + 12 y^{3} \sin x \cos x - (8 y^{3} \sin x \cos x + \sin x)}{- (4 y^{3} \cos^{2} x + \cos x)} = - \tan x

Step 4: Find the Integration Factor:
The integration factor is given by

e^{- \int \tan x d x} = \cos x

Step 5: Multiply and Simplify:
Multiply the entire equation by the integration factor

\cos x

\cos x (2 y \sin x + 3 y^{4} \sin x \cos x) d x - \cos x (4 y^{3} \cos^{2} x + \cos x) d y = 0

This makes the equation exact.

Step 6: Integrate:
Now, integrate both sides with respect to

x

to find the solution:

\int (y \sin 2 x + \frac{3 y^{2}}{2} \sin 2 x \cos x) d x = C

Further integrate and simplify to find the solution:

- \frac{y \cos 2 x}{2} - \frac{3 y^{2}}{4} \cdot \frac{\cos 3 x}{3} - \frac{3 y^{2}}{4} \cos x = C

Conclusion:
The solution to the given differential equation is:

- \frac{y \cos 2 x}{2} - \frac{3 y^{2}}{4} \cdot \frac{\cos 3 x}{3} - \frac{3 y^{2}}{4} \cos x = C

(b) अवकल समीकरण

\frac{d^{2} y}{d x^{2}} - 4 \frac{d y}{d x} + 4 y = 3 x^{2} e^{2 x} \sin 2 x

का पूर्ण हल ज्ञात कीजिए।

Determine the complete solution of the differential equation

\frac{d^{2} y}{d x^{2}} - 4 \frac{d y}{d x} + 4 y = 3 x^{2} e^{2 x} \sin 2 x

Answer:

Auxiliary Equation:

To solve the given differential equation, we start by finding the auxiliary equation, which is given by

D^{2} - 4 D + 4 = 0

where

D

represents the derivative operator. Solving this equation yields two identical roots,

D = 2

. Thus, the complementary function (C.F.) is expressed as:

C . F . = (c_{1} + c_{2} x) e^{2 x}

where

c_{1}

and

c_{2}

are arbitrary constants.

Particular Integral (P.I.):

\begin{aligned} P.I. = \frac{1}{(D - 2)^{2}} 3 x^{2} e^{2 x} \sin 2 x = 3 e^{2 x} \frac{1}{(D + 2 - 2)^{2}} x^{2} \sin 2 x = 3 e^{2 x} \frac{1}{D^{2}} x^{2} \sin 2 x \\ = & 3 e^{2 x} \frac{1}{D} \int x^{2} \sin 2 x d x = 8 e^{2 x} \frac{1}{D} [x^{2} (- \frac{\cos 2 x}{2}) - \int (2 x) (- \frac{\cos 2 x}{2}) d x], integrating by parts \\ = & 3 e^{2 x} \frac{1}{D} [- \frac{1}{2} x^{2} \cos 2 x + \int x \cos 2 x d x] = 8 e^{2 x} \frac{1}{D} [- \frac{1}{2} x^{2} \cos 2 x + x (\frac{\sin 2 x}{2}) - \int 1 \cdot \frac{\sin 2 x}{2} d x] \\ = & 3 e^{2 x} \frac{1}{D} [- \frac{1}{2} x^{2} \cos 2 x + \frac{1}{2} x \sin 2 x + \frac{1}{4} \cos 2 x] = 8 e^{2 x} \int (- \frac{1}{2} x^{2} \cos 2 x + \frac{1}{2} x \sin 2 x + \frac{1}{4} \cos 2 x) d x \\ = & 3 e^{2 x} [- \frac{1}{2} \int x^{2} \cos 2 x d x + \frac{1}{2} \int x \sin 2 x d x + \frac{1}{4} \int \cos 2 x d x] \\ = & 3 e^{2 x} [- \frac{1}{2} {x^{2} (\frac{1}{2} \sin 2 x) - \int 2 x (\frac{1}{2} \sin 2 x) d x} + \frac{1}{2} \int x \sin 2 x d x + \frac{1}{8} \sin 2 x] \\ = & 3 e^{2 x} [- \frac{1}{4} x^{2} \sin 2 x d x + \frac{1}{2} \int x \sin 2 x d x + \frac{1}{2} \int x \sin 2 x d x + (1 / 8) \times \sin 2 x] \end{aligned}

Free UPSC Mathematics Optional Paper-1 2019 Solutions: View Online | UPSC Maths Solution | IAS Maths Solution

Introduction

Work/Calculations

Step 1: Define the Function and the Limit

Step 2: Simplify the Function

Step 3: Apply L’Hôpital’s Rule

Conclusion

Introduction

Work/Calculations

Step 1: Definition of Continuity

Step 2: Prove Continuity for f ( x , b ) f ( x , b ) f(x,b)f(x, b)f(x,b) at x = a x = a x=ax = ax=a

Step 3: Prove Continuity for f ( a , y ) f ( a , y ) f(a,y)f(a, y)f(a,y) at y = b y = b y=by = by=b

Conclusion

Conclusion

Introduction

Work/Calculations

Step 1: Calculate A B A B ABABAB

Step 2: Solve the System of Equations

Introduction

Work/Calculations

Step 1: Breaking It Down

Step 2: Checking the Derivative

Conclusion

Introduction

Work/Calculations

Step 1: Find the Derivative of f ( x ) f ( x ) f(x)f(x)f(x)

Step 2: Find the Critical Points

Step 3: Evaluate f ( x ) f ( x ) f(x)f(x)f(x) at Critical Points and Endpoints

Conclusion

Use the General Form A 2 n = n A 2 − ( n − 1 ) I A 2 n = n A 2 − ( n − 1 ) I A^(2n)=nA^(2)-(n-1)IA^{2n} = n A^2 – (n – 1) IA2n=nA2−(n−1)I

Substitute the Values

Calculate A 100 A 100 A^(100)A^{100}A100

Conclusion

Work/Calculations

Step 1: Integrate f ′ ( x ) f ′ ( x ) f^(‘)(x)f'(x)f′(x) to find f ( x ) f ( x ) f(x)f(x)f(x)

Step 2: Define α α alpha\alphaα and β β beta\betaβ

Step 3: Calculate the Partial Derivatives of α α alpha\alphaα and β β beta\betaβ

Step 4: Check if the Partial Derivatives are Equal

Conclusion with Jacobian Method

Auxiliary Equation:

Particular Integral (P.I.):

Step 2: Prove Continuity for $f (x, b)$ at $x = a$

Step 3: Prove Continuity for $f (a, y)$ at $y = b$

Step 1: Calculate $A B$

Step 1: Find the Derivative of $f (x)$

Step 3: Evaluate $f (x)$ at Critical Points and Endpoints

Use the General Form $A^{2 n} = n A^{2} - (n - 1) I$

Calculate $A^{100}$

Step 1: Integrate $f^{'} (x)$ to find $f (x)$

Step 2: Define $α$ and $β$

Step 3: Calculate the Partial Derivatives of $α$ and $β$