Animated Solution for Mathematics - Definite Integration: Match the statement in Column-I with the values in Column-II

A Journey Through Four Dimensions

Welcome, future engineers! Today, we are not just solving a problem; we are embarking on a grand tour of the JEE Advanced syllabus.

We have a multi-concept challenge that spans the vast landscapes of 3D Geometry, Inverse Trigonometry, Vector Algebra, and Calculus. Take a deep breath, as this is about seeing the connections between these domains.

The Geometry of Laser Beams

Imagine two laser beams crossing through space, defined by lines $L_1$ and $L_2$. We need to find a line from the origin that intersects both.

We define a point $P$ on the first line using parameter $\lambda$ as $(2+\lambda ,1-2\lambda ,-1+\lambda )$. Similarly, we define point $Q$ on the second line using $\mu$ as:

The origin $O$, $P$, and $Q$ are collinear, meaning the direction ratios of $\overrightarrow{OP}$ and $\overrightarrow{OQ}$ must be proportional:

Solving this system yields $\lambda =3$ and $\mu =\frac{1}{3}$. Substituting these back gives the fixed points $P(5,-5,2)$ and $Q(\frac{10}{3},-\frac{10}{3},\frac{4}{3})$. The distance squared, $d^2$, between these two points is 6.

The Triangle of Truth

We face the inverse trigonometry equation: ${\tan }^{-1}(x+3)-{\tan }^{-1}(x-3)={\sin }^{-1}(\frac{3}{5})$.

Visualize a right-angled triangle where the opposite side is $3$ and the hypotenuse is $5$. By the Pythagorean theorem, the adjacent side is $4$, so ${\sin }^{-1}(\frac{3}{5})={\tan }^{-1}(\frac{3}{4})$.

Using the identity ${\tan }^{-1}A-{\tan }^{-1}B={\tan }^{-1}(\frac{A-B}{1+AB})$, we simplify the left side:

Equating the arguments, we get $\frac{6}{x^2-8}=\frac{3}{4}$, which simplifies to $x^2=16$. Thus, the solutions are $x=\pm 4$.

The Vector Waltz

Given $\vec{a}=\mu \vec{b}+4\vec{c}$ and the condition $\vec{a}\cdot \vec{b}=0$, we substitute $\vec{a}$ into the dot product:

We then use the second condition, $2|\vec{b}+\vec{c}|=|\vec{b}-\vec{a}|$. Squaring both sides and expanding, we substitute our dot product relationship.

The vector terms vanish, leaving us with the quadratic equation ${\mu }^2-5\mu =0$. This yields the final values $\mu =0$ or $\mu =5$.

The Calculus Symphony

Finally, we tackle the integral $I=\frac{2}{\pi }{\int }_{-\pi }^{\pi }\frac{\sin (\frac{9x}{2})}{\sin (\frac{x}{2})}dx$. This expression is a hidden trigonometric series:

Integrating this term by term from $-\pi$ to $\pi$, we know that the integral of $\cos (nx)$ over a full period is zero. We are left only with the integral of $1$:

Multiplying by the factor $\frac{2}{\pi }$ outside the integral, we arrive at the final result of 4.