Match the statements given in Column-I with the values given in Column-II. List-I: P. If $\vec{a} = \hat{j} + \sqrt{3}\hat{k}$, $\vec{b} = -\hat{j} + \sqrt{3}\hat{k}$ and $\vec{c} = 2\sqrt{3}\hat{k}$ form a triangle, then the internal angle of the triangle between $\vec{a}$ and $\vec{b}$ is Q. If $\int_a^b (f(x) - 3x) dx = a^2 - b^2$, then the value of $f(\frac{\pi}{6})$ is R. The value of $\frac{\pi^2}{\ln 3} \int_{7/6}^{5/6} \sec(\pi x) dx$ is S. The maximum value of $\text{Arg}(\frac{1}{1 - z})$ for $|z| = 1, z \neq 1$ is given by List-II: 1. $\frac{\pi}{6}$ 2. $\frac{2\pi}{3}$ 3. $\frac{\pi}{3}$ 4. $\pi$ 5. $\frac{\pi}{2}$

### Analyzing the Geometry of Vectors Imagine you are standing in the $yz$-plane. We are given three vectors: $\vec{a} = \hat{j} + \sqrt{3}\hat{k}$, $\vec{b} = -\hat{j} + \sqrt{3}\hat{k}$, and $\vec{c} = 2\sqrt{3}\hat{k}$. If you add $\vec{a}$ and $\vec{b}$, the $\hat{j}$ components cancel out, and the $\hat{k}$ components sum perfectly to $\vec{c}$. This is the definition of a closed triangle. To find the internal angle between $\vec{a}$ and $\vec{b}$, we use the dot product formula: $$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$$ Calculating this, we find $\vec{a} \cdot \vec{b} = -1 + 3 = 2$, and the magnitudes are both $2$. Thus, $\cos \theta = \frac{2}{4} = \frac{1}{2}$, which gives $\theta = \frac{\pi}{3}$. However, this $\theta$ is the angle between the vectors when their tails are joined. In a triangle, the vectors are head-to-tail, so the internal angle is the supplement: $$\pi - \frac{\pi}{3} = \frac{2\pi}{3}$$ ### The Calculus Power Move Next, we face the integral equation: $$\int_a^b (f(x) - 3x) dx = a^2 - b^2$$ When you see variable limits of integration, you should apply the **Leibniz Rule**. Differentiating both sides with respect to the upper limit $b$, the left side becomes $f(b) - 3b$. The right side, differentiating $a^2 - b^2$ with respect to $b$, gives $-2b$. Equating them, we get: $$f(b) - 3b = -2b \Rightarrow f(b) = b$$ Thus, $f(x) = x$. Finding $f(\frac{\pi}{6})$ is now trivial: it is simply **$\frac{\pi}{6}$**. ### The Precision of Integration Now, we tackle the definite integral: $$I = \int_{7/6}^{5/6} \sec(\pi x) dx$$ The standard integral of $\sec(kx)$ is $\frac{1}{k} \ln |\sec(kx) + \tan(kx)|$. Applying this, we get: $$I = \frac{1}{\pi} [\ln |\sec(\pi x) + \tan(\pi x)|]_{7/6}^{5/6}$$ At the upper limit $x = \frac{5}{6}$, we are in the second quadrant, where secant and tangent are negative. At the lower limit $x = \frac{7}{6}$, we are in the third quadrant. Substituting these values carefully, we find the integral evaluates to $\frac{1}{\pi} \ln 3$. When we multiply this by the factor $\frac{\pi^2}{\ln 3}$ provided in the question, the terms cancel out, leaving us with exactly **$\pi$**. ### The Elegance of Complex Numbers Finally, we arrive at the complex transformation $w = \frac{1}{1 - z}$ for $|z| = 1$. Let $z = \cos \theta + i \sin \theta$. The denominator $1 - z$ becomes $1 - \cos \theta - i \sin \theta$. Using half-angle identities, this is $2 \sin^2(\frac{\theta}{2}) - 2i \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2})$. When we rationalize the expression for $w$, the denominator simplifies to $1$. We are left with: $$w = \frac{1}{2} + \frac{i}{2} \cot\left(\frac{\theta}{2}\right)$$ Notice that the real part is constant. The locus of $w$ is a vertical line at $\text{Re}(w) = \frac{1}{2}$. As $\theta \to 0^+$, the imaginary part tends to infinity, meaning the maximum argument of this vector is **$\frac{\pi}{2}$**.

JEE Advanced 2011

LEVELJEE Advanced

Animated Solution for Mathematics - Complex Numbers: Match the statements given in Column-I with the values given in Column-II.

Q: Wait, why is the internal angle $\pi - \theta$ and not just $\theta$?

Great catch! The dot product formula $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$ gives the angle between the vectors when their tails are joined. In a triangle, the vectors are arranged head-to-tail. The angle between the vectors at the vertex is the supplementary angle to the angle between the vectors when tails are joined. Always visualize the vector directions!

Q: Why can we differentiate with respect to $b$ in the integral equation?

Because $b$ is the upper limit of the integral! The Leibniz Rule is specifically designed for this. It tells us that $\frac{d}{db} \int_a^b f(x) dx = f(b)$. Since $a$ is a constant, its derivative is zero. It is a powerful shortcut that saves you from solving the integral explicitly. It is the 'surgical strike' of calculus.

List-I

(P)

a = \hat{j} + 3 \hat{k}

b = - \hat{j} + 3 \hat{k}

and

c = 23 \hat{k}

form a triangle, then the internal angle of the triangle between

a

and

b

(Q)

\int_{a}^{b} (f (x) - 3 x) d x = a^{2} - b^{2}

, then the value of

f (\frac{π}{6})

(R)

The value of

\frac{π ^{2}}{l n 3} \int_{7/6}^{5/6} sec (π x) d x

(S)

The maximum value of

Arg (\frac{1}{1 - z})

for

∣ z ∣ = 1, z \neq = 1

is given by

List-II

(1)

\frac{π}{6}

(2)

\frac{2 π}{3}

(3)

\frac{π}{3}

(4)

π

(5)

\frac{π}{2}

Select Matching Pairs:

* Multiple Allowed

PMatches

QMatches

RMatches

SMatches

Visualized Solution

Match the Following: $4$ Distinct Concepts

We need to match four statements in Column-I with their values in Column-II.
The problems cover Vectors, Leibniz Rule, Definite Integration, and Complex Numbers.
Let's tackle them one by one, starting with the vector triangle.

Part A: Forming the Vector Triangle

Given vectors: $a = \hat{j} + 3 \hat{k}$ , $b = - \hat{j} + 3 \hat{k}$ , $c = 23 \hat{k}$ .
Notice that $a + b = (1 - 1) \hat{j} + (3 + 3) \hat{k} = 23 \hat{k} = c$ .
This confirms they form a closed triangle.

Part A: Angle Between $a$ and $b$

To find the angle $θ$ between $a$ and $b$ , we use the dot product.
$cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}$ .
$a \cdot b = (1) (- 1) + (3) (3) = - 1 + 3 = 2$ .
$∣ a ∣ = 1^{2} + (3)^{2} = 2$ , and $∣ b ∣ = 2$ .
$cos θ = \frac{2}{2 \times 2} = \frac{1}{2} ⟹ θ = \frac{π}{3}$ .

Part A: Internal Angle of the Triangle

The angle $θ = \frac{π}{3}$ is the angle between the vectors (tails joined).
The internal angle of the triangle at the vertex where $a$ meets $b$ is $π - θ$ .
Internal angle = $π - \frac{π}{3} = \frac{2 π}{3}$ .
So, (A) matches with (q).

Part B: Integral Equation

Given: $\int_{a}^{b} (f (x) - 3 x) d x = a^{2} - b^{2}$ .
We need to find the function $f (x)$ to evaluate $f (\frac{π}{6})$ .
We can differentiate both sides with respect to the upper limit $b$ , treating $a$ as a constant.

Part B: Applying Leibniz Rule

Differentiating w.r.t $b$ : $\frac{d}{d b} \int_{a}^{b} (f (x) - 3 x) d x = \frac{d}{d b} (a^{2} - b^{2})$ .
Using Leibniz Rule: $f (b) - 3 b = - 2 b$ .
Rearranging gives: $f (b) = b$ .
Therefore, the function is $f (x) = x$ .

Part B: Evaluating $f (\frac{π}{6})$

Since $f (x) = x$ , we substitute $x = \frac{π}{6}$ .
$f (\frac{π}{6}) = \frac{π}{6}$ .
So, (B) matches with (p).

Part C: Secant Integral Setup

Let $I = \int_{7/6}^{5/6} sec (π x) d x$ .
The standard integral formula: $\int sec (k x) d x = \frac{1}{k} ln ∣ sec (k x) + tan (k x) ∣$ .
Applying this to our integral: $I = [\frac{1}{π} ln ∣ sec (π x) + tan (π x) ∣]_{7/6}^{5/6}$ .

Part C: Substituting Limits

Upper limit $x = \frac{5}{6}$ : $sec (\frac{5 π}{6}) = - \frac{2}{3}$ , $tan (\frac{5 π}{6}) = - \frac{1}{3}$ .
Value at upper limit: $ln ∣ - \frac{3}{3} ∣ = ln 3$ .
Lower limit $x = \frac{7}{6}$ : $sec (\frac{7 π}{6}) = - \frac{2}{3}$ , $tan (\frac{7 π}{6}) = \frac{1}{3}$ .
Value at lower limit: $ln ∣ - \frac{1}{3} ∣ = ln \frac{1}{3}$ .

Part C: Final Value

$I = \frac{1}{π} (ln 3 - ln \frac{1}{3})$ .
Using log properties: $ln 3 - ln (3^{- 1/2}) = \frac{1}{2} ln 3 - (- \frac{1}{2} ln 3) = ln 3$ .
So, $I = \frac{1}{π} ln 3$ .
The expression to evaluate is $\frac{π ^{2}}{l n 3} \times I = \frac{π ^{2}}{l n 3} \times \frac{l n 3}{π} = π$ .
(C) matches with (s).

Part D: Mobius Transformation

Given $w = \frac{1}{1 - z}$ where $∣ z ∣ = 1$ and $z \neq = 1$ .
Let $z = cos θ + i sin θ$ .
$1 - z = 1 - cos θ - i sin θ = 2 sin^{2} (\frac{θ}{2}) - 2 i sin (\frac{θ}{2}) cos (\frac{θ}{2})$ .
$1 - z = 2 sin (\frac{θ}{2}) [sin (\frac{θ}{2}) - i cos (\frac{θ}{2})]$ .

Part D: Locus of $w$

$w = \frac{1}{2 s i n ( \frac{θ}{2} ) [ s i n ( \frac{θ}{2} ) - i c o s ( \frac{θ}{2} )]}$ .
Multiply numerator and denominator by $[sin (\frac{θ}{2}) + i cos (\frac{θ}{2})]$ .
The denominator becomes $sin^{2} (\frac{θ}{2}) + cos^{2} (\frac{θ}{2}) = 1$ .
$w = \frac{s i n ( \frac{θ}{2} ) + i c o s ( \frac{θ}{2} )}{2 s i n ( \frac{θ}{2} )} = \frac{1}{2} + \frac{i}{2} cot (\frac{θ}{2})$ .
The real part of $w$ is always $\frac{1}{2}$ .

Part D: Maximum Argument of $w$

The locus is the vertical line $Re (w) = \frac{1}{2}$ .
The argument of $w$ is $tan^{- 1} (\frac{Im ( w )}{Re ( w )}) = tan^{- 1} (cot (\frac{θ}{2}))$ .
As $θ \to 0^{+}$ , $cot (\frac{θ}{2}) \to \infty$ , so $Im (w) \to \infty$ .
The vector $w$ points almost vertically upwards, so $Arg (w) \to \frac{π}{2}$ .
Maximum value of $Arg (w)$ is $\frac{π}{2}$ .

Final Matching Summary

(A) Internal angle of vector triangle $\to$ (q) $\frac{2 π}{3}$
(B) Value of $f (\frac{π}{6})$ using Leibniz $\to$ (p) $\frac{π}{6}$
(C) Definite integral value $\to$ (s) $π$
(D) Maximum argument of $w$ $\to$ (t) $\frac{π}{2}$
Final Answer: A-q, B-p, C-s, D-t.

00:00 / 00:00

The Sigma Insight: Geometrical Applications of Complex Numbers

Solution Diagram

Analyzing the Geometry of Vectors

Imagine you are standing in the

y z

-plane. We are given three vectors:

a = \hat{j} + 3 \hat{k}

b = - \hat{j} + 3 \hat{k}

, and

c = 23 \hat{k}

If you add

a

and

b

, the

\hat{j}

components cancel out, and the

\hat{k}

components sum perfectly to

c

. This is the definition of a closed triangle.

To find the internal angle between

a

and

b

, we use the dot product formula:

cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}

Calculating this, we find

a \cdot b = - 1 + 3 = 2

, and the magnitudes are both

2

. Thus,

cos θ = \frac{2}{4} = \frac{1}{2}

, which gives

θ = \frac{π}{3}

However, this

θ

is the angle between the vectors when their tails are joined. In a triangle, the vectors are head-to-tail, so the internal angle is the supplement:

π - \frac{π}{3} = \frac{2 π}{3}

The Calculus Power Move

Next, we face the integral equation:

\int_{a}^{b} (f (x) - 3 x) d x = a^{2} - b^{2}

When you see variable limits of integration, you should apply the Leibniz Rule. Differentiating both sides with respect to the upper limit

b

, the left side becomes

f (b) - 3 b

The right side, differentiating

a^{2} - b^{2}

with respect to

b

, gives

- 2 b

. Equating them, we get:

f (b) - 3 b = - 2 b \Rightarrow f (b) = b

Thus,

f (x) = x

. Finding

f (\frac{π}{6})

is now trivial: it is simply $\frac{π}{6}$ .

The Precision of Integration

Now, we tackle the definite integral:

I = \int_{7/6}^{5/6} sec (π x) d x

The standard integral of

sec (k x)

\frac{1}{k} ln ∣ sec (k x) + tan (k x) ∣

. Applying this, we get:

I = \frac{1}{π} [ln ∣ sec (π x) + tan (π x) ∣]_{7/6}^{5/6}

At the upper limit

x = \frac{5}{6}

, we are in the second quadrant, where secant and tangent are negative. At the lower limit

x = \frac{7}{6}

, we are in the third quadrant.

Substituting these values carefully, we find the integral evaluates to

\frac{1}{π} ln 3

. When we multiply this by the factor

\frac{π ^{2}}{l n 3}

provided in the question, the terms cancel out, leaving us with exactly $π$ .

The Elegance of Complex Numbers

Finally, we arrive at the complex transformation

w = \frac{1}{1 - z}

for

∣ z ∣ = 1

. Let

z = cos θ + i sin θ

The denominator

1 - z

becomes

1 - cos θ - i sin θ

. Using half-angle identities, this is

2 sin^{2} (\frac{θ}{2}) - 2 i sin (\frac{θ}{2}) cos (\frac{θ}{2})

When we rationalize the expression for

w

, the denominator simplifies to

1

. We are left with:

w = \frac{1}{2} + \frac{i}{2} cot (\frac{θ}{2})

Notice that the real part is constant. The locus of

w

is a vertical line at

Re (w) = \frac{1}{2}

. As

θ \to 0^{+}

, the imaginary part tends to infinity, meaning the maximum argument of this vector is $\frac{π}{2}$ .

Common Traps & FAQs

Wait, why is the internal angle

π - θ

and not just

θ

Great catch! The dot product formula

cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}

gives the angle between the vectors when their tails are joined. In a triangle, the vectors are arranged head-to-tail. The angle between the vectors at the vertex is the supplementary angle to the angle between the vectors when tails are joined. Always visualize the vector directions!

Why can we differentiate with respect to

b

in the integral equation?

Because

b

is the upper limit of the integral! The Leibniz Rule is specifically designed for this. It tells us that

\frac{d}{d b} \int_{a}^{b} f (x) d x = f (b)

. Since

a

is a constant, its derivative is zero. It is a powerful shortcut that saves you from solving the integral explicitly. It is the 'surgical strike' of calculus.

How did the real part of

w

become a constant in the complex problem?

That is the beauty of the Mobius transformation! By substituting

z = cos θ + i sin θ

and simplifying, the imaginary components cancel out in a way that leaves the real part independent of

θ

. It is a geometric property—the unit circle maps to a straight line under this specific transformation. It is not a coincidence; it is mathematical elegance.

Animated Solution for Mathematics - Complex Numbers: Match the statements given in Column-I with the values given in Column-II.

List-I

List-II

Select Matching Pairs:

Visualized Solution

Match the Following: $4$ Distinct Concepts

Part A: Forming the Vector Triangle

Part A: Angle Between $a$ and $b$

Part A: Internal Angle of the Triangle

Part B: Integral Equation

Part B: Applying Leibniz Rule

Part B: Evaluating $f (\frac{π}{6})$

Part C: Secant Integral Setup

Part C: Substituting Limits

Part C: Final Value

Part D: Mobius Transformation

Part D: Locus of $w$

Part D: Maximum Argument of $w$

Final Matching Summary

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Geometry of Vectors

The Calculus Power Move

The Precision of Integration

The Elegance of Complex Numbers

Similar Questions

If for the complex numbers $z$ satisfying $∣ z - 2 - 2 i ∣ \leq 1$ , the maximum value of $∣3 i z + 6∣$ is attained at $a + ib$ , then $a + b$ is equal to .

Comprehension Passage

Area of $S =$

$min_{z \in S} ∣1 - 3 i - z ∣ =$

Let a complex number be $w = 1 - 3 i$ . Let another complex number $z$ be such that $∣ z w ∣ = 1$ and $ar g (z) - ar g (w) = \frac{π}{2}$ . Then the area of the triangle with vertices origin, $z$ and $w$ is equal to:

Let O be the origin, the point A be $z_{1} = 3 + 22 i,$ the point $B (z_{2})$ be such that $3 ∣ z_{2} ∣ = ∣ z_{1} ∣$ and $a r g (z_{2}) = a r g (z_{1}) + \frac{π}{6}$ . Then

For any integer $k$ , let $α_{k} = cos (\frac{k π}{7}) + i sin (\frac{k π}{7})$ , where $i = - 1$ . The value of the expression $\frac{\sum _{k = 1}^{12} ∣ α _{k + 1} - α _{k} ∣}{\sum _{k = 1}^{3} ∣ α _{4 k - 1} - α _{4 k - 2} ∣}$ is .........

For a non-zero complex number $z$ , let $ar g (z)$ denote the principal argument with $- π < ar g (z) \leq π$ . Then, which of the following statement(s) is (are) FALSE ?

Let the curve $z (1 + i) + \overline{z} (1 - i) = 4, z \in C$ , divide the region $∣ z - 3∣ \leq 1$ into two parts of areas $α$ and $β$ . Then $∣ α - β ∣$ equals:

If $z_{1}, z_{2}$ are complex numbers such that $Re (z_{1}) = ∣ z_{1} - 1∣$ , $Re (z_{2}) = ∣ z_{2} - 1∣$ , and $arg (z_{1} - z_{2}) = \frac{π}{6}$ , then $Im (z_{1} + z_{2})$ is equal to

Animated Solution for Mathematics - Complex Numbers: Match the statements given in Column-I with the values given in Column-II.

List-I

List-II

Select Matching Pairs:

Visualized Solution

Match the Following: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4 Distinct Concepts

Part A: Forming the Vector Triangle

Part A: Angle Between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Part A: Internal Angle of the Triangle

Part B: Integral Equation

Part B: Applying Leibniz Rule

Part B: Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(6π​)

Part C: Secant Integral Setup

Part C: Substituting Limits

Part C: Final Value

Part D: Mobius Transformation

Part D: Locus of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } w

Part D: Maximum Argument of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } w

Final Matching Summary

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Geometry of Vectors

The Calculus Power Move

The Precision of Integration

The Elegance of Complex Numbers

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If for the complex numbers z satisfying ∣z−2−2i∣≤1, the maximum value of ∣3iz+6∣ is attained at a+ib, then a+b is equal to .

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Area of S=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } minz∈S​∣1−3i−z∣=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let O be the origin, the point A be z1​=3​+22​i, the point B(z2​) be such that 3​∣z2​∣=∣z1​∣ and arg(z2​)=arg(z1​)+6π​. Then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For a non-zero complex number z, let arg(z) denote the principal argument with −π<arg(z)≤π. Then, which of the following statement(s) is (are) FALSE ?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the curve z(1+i)+z(1−i)=4,z∈C, divide the region ∣z−3∣≤1 into two parts of areas α and β. Then ∣α−β∣ equals:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z1​,z2​ are complex numbers such that Re(z1​)=∣z1​−1∣, Re(z2​)=∣z2​−1∣, and arg(z1​−z2​)=6π​, then Im(z1​+z2​) is equal to

Match the Following: $4$ Distinct Concepts

Part A: Angle Between $a$ and $b$

Part B: Evaluating $f (\frac{π}{6})$

Part D: Locus of $w$

Part D: Maximum Argument of $w$

If for the complex numbers $z$ satisfying $∣ z - 2 - 2 i ∣ \leq 1$ , the maximum value of $∣3 i z + 6∣$ is attained at $a + ib$ , then $a + b$ is equal to .

Area of $S =$

$min_{z \in S} ∣1 - 3 i - z ∣ =$

Let O be the origin, the point A be $z_{1} = 3 + 22 i,$ the point $B (z_{2})$ be such that $3 ∣ z_{2} ∣ = ∣ z_{1} ∣$ and $a r g (z_{2}) = a r g (z_{1}) + \frac{π}{6}$ . Then

For a non-zero complex number $z$ , let $ar g (z)$ denote the principal argument with $- π < ar g (z) \leq π$ . Then, which of the following statement(s) is (are) FALSE ?

Let the curve $z (1 + i) + \overline{z} (1 - i) = 4, z \in C$ , divide the region $∣ z - 3∣ \leq 1$ into two parts of areas $α$ and $β$ . Then $∣ α - β ∣$ equals:

If $z_{1}, z_{2}$ are complex numbers such that $Re (z_{1}) = ∣ z_{1} - 1∣$ , $Re (z_{2}) = ∣ z_{2} - 1∣$ , and $arg (z_{1} - z_{2}) = \frac{π}{6}$ , then $Im (z_{1} + z_{2})$ is equal to