Match the statements given in Column-I...

MathematicsGeometrical Applications of Complex NumbersJEE Advanced 2011Difficult

Match the statements given in Column-I with the values given in Column-II.

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}

Answer:P → 2, Q → 1, R → 4, S → 5

Visualized Solution (English)

Introduction to the Matching Problem

Objective: Match Column-I statements with Column-II values.
Key Concepts involved:
1. Vector Algebra: Internal angles of a triangle.
2. Calculus: Leibniz Rule for differentiating integrals.
3. Integration: Definite integral of $sec (π x)$ .
4. Complex Numbers: Locus and Argument of Mobius transformations.

Internal Angle of a Vector Triangle

Given: $a = (0, 1, 3)$ , $b = (0, - 1, 3)$ , $c = (0, 0, 23)$
Observation: $a + b = c$
Angle between vectors $a$ and $b$ :
$cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣} = \frac{0 - 1 + 3}{2 \cdot 2} = \frac{1}{2} \Rightarrow θ = \frac{π}{3}$
Internal angle of triangle = $π - θ = π - \frac{π}{3} = \frac{2 π}{3}$

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f (x) using Leibniz Rule

Equation: $\int_{a}^{b} (f (x) - 3 x) d x = a^{2} - b^{2}$
$\int_{a}^{b} f (x) d x - [\frac{3}{2} x^{2}]_{a}^{b} = a^{2} - b^{2}$
$\int_{a}^{b} f (x) d x = \frac{1}{2} b^{2} - \frac{1}{2} a^{2}$
Differentiating w.r.t. $b$ using Leibniz Rule:
$f (b) = b \Rightarrow f (x) = x$
Therefore, $f (\frac{π}{6}) = \frac{π}{6}$

Evaluating the Secant Integral

Integral: $I = \int_{7/6}^{5/6} sec (π x) d x$
$I = [\frac{1}{π} ln ∣ sec π x + tan π x ∣]_{7/6}^{5/6}$
At $x = \frac{5}{6}$ : $ln ∣ - \frac{2}{3} - \frac{1}{3} ∣ = ln 3$
At $x = \frac{7}{6}$ : $ln ∣ - \frac{2}{3} + \frac{1}{3} ∣ = ln \frac{1}{3}$
$I = \frac{1}{π} (ln 3 - ln \frac{1}{3}) = \frac{1}{π} ln 3$
Final Value: $\frac{π ^{2}}{l n 3} \cdot \frac{1}{π} ln 3 = π$

Maximum Argument in Complex Plane

Let $z = e^{i θ} = cos θ + i sin θ$
$w = \frac{1}{1 - z} = \frac{1}{1 - c o s θ - i s i n θ}$
Rationalizing gives: $w = \frac{1}{2} + \frac{i}{2} cot (\frac{θ}{2})$
Locus: Vertical line $Re (w) = \frac{1}{2}$
As $θ \to 0^{+}$ , $Im (w) \to \infty \Rightarrow Arg (w) \to \frac{π}{2}$
Maximum value of $Arg (w) = \frac{π}{2}$

Final Matching Summary

Final Match Results:
(A) $\to$ (q): $\frac{2 π}{3}$
(B) $\to$ (p): $\frac{π}{6}$
(C) $\to$ (s): $π$
(D) $\to$ (t): $\frac{π}{2}$
Key Takeaway: Always look for geometric shortcuts in complex numbers and vector problems.

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Question 1:

$\int_{0}^{π /2} sin x d x =$

(A)

\frac{π}{8} (1 + 2)

(B)

\frac{π}{4} (1 + 2)

(C)

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(D)

\frac{π}{4 2}

Answer:A

Question 2:

If $lim_{x \to a} \frac{\int _{a}^{x} f ( x ) d x - ( \frac{x - a}{2} ) ( f ( x ) + f ( a ))}{( x - a ) ^{3}} = 0$ , then $f (x)$ is of maximum degree

(A)

(B)

(C)

(D)

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Question 3:

If $f^{''} (x) < 0\forall x \in (a, b)$ and $c$ is a point such that $a < c < b$ , and $(c, f (c))$ is the point lying on the curve for which $F (c)$ is maximum, then $f^{'} (c)$ is equal to

(A)

\frac{f ( b ) - f ( a )}{b - a}

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