IIT JEE 2003 Mathematics Question Paper with Answer and Solution

(A) Given $|z| = 1$,let $z = x + iy$,so $x^2 + y^2 = 1$.
$\omega = \frac{z - 1}{z + 1} = \frac{(x - 1) + iy}{(x + 1) + iy}$.
To find the real part,multiply the numerator and denominator by the conjugate of the denominator $(x + 1) - iy$:
$\omega = \frac{((x - 1) + iy)((x + 1) - iy)}{(x + 1)^2 + y^2} = \frac{(x^2 - 1) + i(y(x + 1) - y(x - 1)) + y^2}{(x + 1)^2 + y^2}$.
$\omega = \frac{(x^2 + y^2 - 1) + i(xy + y - xy + y)}{(x + 1)^2 + y^2} = \frac{(1 - 1) + 2iy}{(x + 1)^2 + y^2} = \frac{2iy}{(x + 1)^2 + y^2}$.
Since the real part is $0$,$\text{Re}(\omega) = 0$.

(A) Given that $a, b, c$ are in $A.P.$
$2b = a + c$ ......$(i)$
Given that $a^2, b^2, c^2$ are in $H.P.$
$b^2 = \frac{2a^2c^2}{a^2 + c^2}$
$b^2(a^2 + c^2) = 2a^2c^2$
$b^2((a+c)^2 - 2ac) = 2a^2c^2$
Substituting $a+c = 2b$ from $(i)$:
$b^2(4b^2 - 2ac) = 2a^2c^2$
$4b^4 - 2acb^2 - 2a^2c^2 = 0$
$2b^4 - acb^2 - a^2c^2 = 0$
$(2b^2 + ac)(b^2 - ac) = 0$
Since $a, b, c$ are real,$2b^2 + ac = 0$ implies $a, b, c$ are $0$ or complex,so we consider $b^2 = ac$.
If $b^2 = ac$,then $a, b, c$ are in $G.P.$
Since $a, b, c$ are in both $A.P.$ and $G.P.$,we must have $a = b = c$.

(A) The expression is $(1 + t^2)^{12}(1 + t^{12})(1 + t^{24})$.
Expanding $(1 + t^2)^{12}$ using the binomial theorem,we get $\sum_{k=0}^{12} {^{12}C_k} (t^2)^k = \sum_{k=0}^{12} {^{12}C_k} t^{2k}$.
We want the coefficient of $t^{24}$ in the product $(\sum_{k=0}^{12} {^{12}C_k} t^{2k})(1 + t^{12} + t^{24})$.
Distributing the terms:
$1$. From $1 \times (\dots)$,we need the coefficient of $t^{24}$ in $(1 + t^2)^{12}$,which is $^{12}C_{12} = 1$.
$2$. From $t^{12} \times (\dots)$,we need the coefficient of $t^{12}$ in $(1 + t^2)^{12}$,which is $^{12}C_6 = 924$.
$3$. From $t^{24} \times (\dots)$,we need the coefficient of $t^0$ in $(1 + t^2)^{12}$,which is $^{12}C_0 = 1$.
Summing these coefficients: $1 + ^{12}C_6 + 1 = ^{12}C_6 + 2$.

(C) Let the vertices be $A(0, 0)$,$B(4, 0)$,and $C(3, 4)$.
To find the orthocentre,we find the intersection of the altitudes.
$1$. Altitude from $A$ to $BC$: The slope of $BC$ is $\frac{4-0}{3-4} = -4$. The altitude is perpendicular to $BC$,so its slope is $\frac{1}{4}$. Since it passes through $A(0, 0)$,its equation is $y = \frac{1}{4}x$,or $x - 4y = 0$.
$2$. Altitude from $C$ to $AB$: The side $AB$ lies on the $x$-axis $(y=0)$,so the altitude from $C(3, 4)$ is a vertical line $x = 3$.
$3$. Intersection: Substituting $x = 3$ into $x - 4y = 0$,we get $3 - 4y = 0$,which gives $y = \frac{3}{4}$.
Thus,the orthocentre is $\left( 3, \frac{3}{4} \right)$.

(A) The given equations are $x^2 - 8x + 12 = 0$ and $y^2 - 14y + 45 = 0$.
Solving $x^2 - 8x + 12 = 0$,we get $(x - 2)(x - 6) = 0$,so $x = 2$ and $x = 6$.
Solving $y^2 - 14y + 45 = 0$,we get $(y - 5)(y - 9) = 0$,so $y = 5$ and $y = 9$.
The lines forming the square are $x = 2, x = 6, y = 5,$ and $y = 9$.
The centre of the inscribed circle is the midpoint of the square,which is the intersection of the diagonals.
The centre is given by $(\frac{2 + 6}{2}, \frac{5 + 9}{2}) = (\frac{8}{2}, \frac{14}{2}) = (4, 7)$.

(A) The parabola is $y^2 = 16x$,so $4a = 16$,which gives $a = 4$. The focus is $(a, 0) = (4, 0)$.
Any focal chord passing through $(4, 0)$ with slope $m$ has the equation $y - 0 = m(x - 4)$,or $mx - y - 4m = 0$.
This line is tangent to the circle $(x - 6)^2 + y^2 = 2$,which has center $(6, 0)$ and radius $r = \sqrt{2}$.
The perpendicular distance from the center $(6, 0)$ to the line $mx - y - 4m = 0$ must be equal to the radius $\sqrt{2}$.
Using the distance formula: $\frac{|m(6) - 0 - 4m|}{\sqrt{m^2 + (-1)^2}} = \sqrt{2}$.
$\frac{|2m|}{\sqrt{m^2 + 1}} = \sqrt{2}$.
Squaring both sides: $\frac{4m^2}{m^2 + 1} = 2$.
$4m^2 = 2m^2 + 2$.
$2m^2 = 2$,so $m^2 = 1$,which gives $m = \pm 1$.
Thus,the possible values of the slope are $\{-1, 1\}$.

(D) For the ellipse $\frac{x^2}{9} + \frac{y^2}{5} = 1$,we have $a^2 = 9$ and $b^2 = 5$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.
The coordinates of the end points of the latus rectum are $(\pm ae, \pm \frac{b^2}{a}) = (\pm 2, \pm \frac{5}{3})$.
The tangent at the point $(2, \frac{5}{3})$ is $\frac{2x}{9} + \frac{5y/3}{5} = 1$,which simplifies to $\frac{2x}{9} + \frac{y}{3} = 1$,or $\frac{x}{9/2} + \frac{y}{3} = 1$.
This tangent forms a triangle with the coordinate axes in the first quadrant with intercepts $x_0 = 9/2$ and $y_0 = 3$.
The area of this triangle is $\frac{1}{2} \times \frac{9}{2} \times 3 = \frac{27}{4}$.
Since there are four such symmetric triangles formed by the tangents at the four ends of the latus recta,the total area of the quadrilateral is $4 \times \frac{27}{4} = 27$ $sq. \text{ units}$.

(B) The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $(a \cos \theta, b \sin \theta)$ is $\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1$.
Here $a^2 = 27 \implies a = 3\sqrt{3}$ and $b^2 = 1 \implies b = 1$.
So,the tangent equation is $\frac{x \cos \theta}{3\sqrt{3}} + y \sin \theta = 1$.
The $x$-intercept is $a_x = \frac{3\sqrt{3}}{\cos \theta} = 3\sqrt{3} \sec \theta$ and the $y$-intercept is $a_y = \frac{1}{\sin \theta} = \csc \theta$.
The sum of intercepts is $f(\theta) = 3\sqrt{3} \sec \theta + \csc \theta$.
To find the minimum,set $f'(\theta) = 3\sqrt{3} \sec \theta \tan \theta - \csc \theta \cot \theta = 0$.
$3\sqrt{3} \frac{\sin \theta}{\cos^2 \theta} = \frac{\cos \theta}{\sin^2 \theta} \implies \tan^3 \theta = \frac{1}{3\sqrt{3}} = \left(\frac{1}{\sqrt{3}}\right)^3$.
Thus,$\tan \theta = \frac{1}{\sqrt{3}}$,which gives $\theta = \pi/6$.

(D) Given the limit: $\mathop {\lim }\limits_{x \to 0} \frac{[(a - n)nx - \tan x]\sin nx}{x^2} = 0$
We can rewrite the expression as: $\mathop {\lim }\limits_{x \to 0} \left( \frac{(a - n)nx - \tan x}{x} \cdot \frac{\sin nx}{x} \right) = 0$
Multiplying and dividing by $n$ in the second term: $\mathop {\lim }\limits_{x \to 0} \left( ((a - n)n - \frac{\tan x}{x}) \cdot n \cdot \frac{\sin nx}{nx} \right) = 0$
Since $\mathop {\lim }\limits_{x \to 0} \frac{\tan x}{x} = 1$ and $\mathop {\lim }\limits_{x \to 0} \frac{\sin nx}{nx} = 1$,the equation becomes:
$n((a - n)n - 1) = 0$
Since $n \neq 0$,we have $(a - n)n - 1 = 0$
$(a - n)n = 1$
$a - n = \frac{1}{n}$
$a = n + \frac{1}{n}$

(D) Using $L$'Hopital's rule,we differentiate the numerator and denominator with respect to $h$:
$\mathop {\lim }\limits_{h \to 0} \frac{{f(2h + 2 + {h^2}) - f(2)}}{{f(h - {h^2} + 1) - f(1)}} = \mathop {\lim }\limits_{h \to 0} \frac{{f'(2h + 2 + {h^2}) \cdot (2 + 2h)}}{{f'(h - {h^2} + 1) \cdot (1 - 2h)}}$
As $h \to 0$,the expression becomes:
$\frac{{f'(2) \cdot 2}}{{f'(1) \cdot 1}} = \frac{{6 \times 2}}{{4 \times 1}} = \frac{{12}}{{4}} = 3$.

(D) Given $f(x) = x^2 + 2bx + 2c^2$. Completing the square,we get $f(x) = (x + b)^2 + 2c^2 - b^2$. The minimum value of $f(x)$ is $2c^2 - b^2$.
Given $g(x) = -x^2 - 2cx + b^2$. Completing the square,we get $g(x) = b^2 + c^2 - (x + c)^2$. The maximum value of $g(x)$ is $b^2 + c^2$.
According to the condition $\min f(x) > \max g(x)$,we have $2c^2 - b^2 > b^2 + c^2$.
Simplifying the inequality,we get $c^2 > 2b^2$.
Taking the square root on both sides,we get $|c| > |b|\sqrt{2}$.

(D) The total number of ways to select $2$ numbers from $6$ numbers without replacement is $P(6, 2) = 6 \times 5 = 30$.
Let the two numbers be $x$ and $y$. We want the probability that $\min(x, y) < 4$.
It is easier to calculate the complement: the probability that $\min(x, y) \geq 4$.
This means both numbers must be chosen from the set $\{4, 5, 6\}$.
The number of ways to choose $2$ numbers from $\{4, 5, 6\}$ is $P(3, 2) = 3 \times 2 = 6$.
The probability that $\min(x, y) \geq 4$ is $\frac{6}{30} = \frac{1}{5}$.
Therefore,the probability that $\min(x, y) < 4$ is $1 - \frac{1}{5} = \frac{4}{5}$.

(B) The vertices of the triangle are $(0, 0)$,$(0, 21)$,and $(21, 0)$.
The interior points $(x, y)$ must satisfy $x > 0$,$y > 0$,and $x + y < 21$.
For a fixed $x$,$y$ can take integer values from $1$ to $21 - x - 1 = 20 - x$.
Since $y > 0$,we must have $20 - x \geq 1$,so $x$ can range from $1$ to $19$.
For each $x \in \{1, 2, \dots, 19\}$,the number of possible values for $y$ is $20 - x$.
The total number of integral points is $\sum_{x=1}^{19} (20 - x) = 19 + 18 + \dots + 1$.
Using the sum formula $\frac{n(n+1)}{2}$ with $n = 19$,we get $\frac{19 \times 20}{2} = 190$.

(A) Let the point be $(x, y)$. Since the point is equidistant from $(a_1, b_1)$ and $(a_2, b_2)$,we have:
$(x - a_1)^2 + (y - b_1)^2 = (x - a_2)^2 + (y - b_2)^2$
Expanding both sides:
$x^2 - 2a_1x + a_1^2 + y^2 - 2b_1y + b_1^2 = x^2 - 2a_2x + a_2^2 + y^2 - 2b_2y + b_2^2$
Canceling $x^2$ and $y^2$ from both sides:
$-2a_1x - 2b_1y + a_1^2 + b_1^2 = -2a_2x - 2b_2y + a_2^2 + b_2^2$
Rearranging the terms to the form $(a_1 - a_2)x + (b_1 - b_2)y + c = 0$:
$2(a_2 - a_1)x + 2(b_2 - b_1)y + (a_1^2 + b_1^2 - a_2^2 - b_2^2) = 0$
Multiplying by $-1$ to match the given form $(a_1 - a_2)x + (b_1 - b_2)y + c = 0$:
$2(a_1 - a_2)x + 2(b_1 - b_2)y + (a_2^2 + b_2^2 - a_1^2 - b_1^2) = 0$
Dividing by $2$:
$(a_1 - a_2)x + (b_1 - b_2)y + \frac{1}{2}(a_2^2 + b_2^2 - a_1^2 - b_1^2) = 0$
Comparing this with the given equation,we get $c = \frac{1}{2}(a_2^2 + b_2^2 - a_1^2 - b_1^2)$.

(B) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,where $a^2 = \cos^2 \alpha$ and $b^2 = \sin^2 \alpha$.
For a hyperbola,the foci are located at $(\pm ae, 0)$,where $e$ is the eccentricity.
The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{\sin^2 \alpha}{\cos^2 \alpha}} = \sqrt{1 + \tan^2 \alpha} = \sec \alpha$.
The abscissa of the foci is $\pm ae = \pm \sqrt{\cos^2 \alpha} \cdot \sec \alpha = \pm \cos \alpha \cdot \frac{1}{\cos \alpha} = \pm 1$.
Since the value $\pm 1$ is independent of $\alpha$,the abscissae of the foci remain constant.

(A) Let the angles of the triangle be $4x, x$,and $x$.
Since the sum of angles in a triangle is $180^{\circ}$,we have $4x + x + x = 180^{\circ}$,which implies $6x = 180^{\circ}$,so $x = 30^{\circ}$.
The angles are $120^{\circ}, 30^{\circ}$,and $30^{\circ}$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$,where $a$ is the longest side opposite to $120^{\circ}$.
Then $a = k \sin 120^{\circ}$,$b = k \sin 30^{\circ}$,and $c = k \sin 30^{\circ}$.
The ratio of the longest side to the perimeter is $\frac{a}{a+b+c} = \frac{\sin 120^{\circ}}{\sin 120^{\circ} + \sin 30^{\circ} + \sin 30^{\circ}}$.
Substituting the values: $\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2} + \frac{1}{2} + \frac{1}{2}} = \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}+2}{2}} = \frac{\sqrt{3}}{2+\sqrt{3}}$.

(A) The given system of equations is homogeneous:
$x + ay + 0z = 0$
$0x + y + az = 0$
$ax + 0y + z = 0$
For a homogeneous system to have infinite solutions (non-trivial solutions),the determinant of the coefficient matrix must be zero:
$\begin{vmatrix} 1 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1(1 \times 1 - a \times 0) - a(0 \times 1 - a \times a) + 0(0 \times 0 - 1 \times a) = 0$
$1(1) - a(-a^2) + 0 = 0$
$1 + a^3 = 0$
$a^3 = -1$
Taking the cube root on both sides,we get $a = -1$.

(D) Given $A = \begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}$.
We calculate $A^2$ as follows:
$A^2 = A \times A = \begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} \alpha^2 + 0 & 0 + 0 \\ \alpha + 1 & 0 + 1 \end{bmatrix} = \begin{bmatrix} \alpha^2 & 0 \\ \alpha + 1 & 1 \end{bmatrix}$.
We are given that $A^2 = B$,so:
$\begin{bmatrix} \alpha^2 & 0 \\ \alpha + 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}$.
Comparing the corresponding elements,we get:
$1$) $\alpha^2 = 1 \implies \alpha = \pm 1$
$2$) $\alpha + 1 = 5 \implies \alpha = 4$
Since there is no value of $\alpha$ that satisfies both equations simultaneously,there are no real values of $\alpha$ for which $A^2 = B$.

(C) The volume $V$ of the parallelepiped formed by vectors $\vec{u} = i + aj + k$,$\vec{v} = j + ak$,and $\vec{w} = ai + k$ is given by the absolute value of the scalar triple product $[\vec{u}, \vec{v}, \vec{w}]$.
$V = |\vec{u} \cdot (\vec{v} \times \vec{w})| = \left| \det \begin{bmatrix} 1 & a & 1 \\ 0 & 1 & a \\ a & 0 & 1 \end{bmatrix} \right|$
Expanding the determinant: $1(1 - 0) - a(0 - a^2) + 1(0 - a) = 1 + a^3 - a$.
Let $f(a) = a^3 - a + 1$. To find the minimum,we find the derivative $f'(a) = 3a^2 - 1$.
Setting $f'(a) = 0$,we get $3a^2 = 1$,so $a^2 = \frac{1}{3}$,which gives $a = \pm \frac{1}{\sqrt{3}}$.
Since we are looking for the volume (which must be non-negative),we evaluate the function at these points. For $a = \frac{1}{\sqrt{3}}$,$V = |(\frac{1}{\sqrt{3}})^3 - \frac{1}{\sqrt{3}} + 1| = |\frac{1}{3\sqrt{3}} - \frac{1}{\sqrt{3}} + 1| = |1 - \frac{2}{3\sqrt{3}}|$,which is positive.
Thus,the value of $a$ that minimizes the volume is $\frac{1}{\sqrt{3}}$.

(B) Given $f(x) = \frac{x}{1+x}$ for $x \in [0, \infty)$.
To check for one-one: Let $f(x_1) = f(x_2)$,then $\frac{x_1}{1+x_1} = \frac{x_2}{1+x_2}$.
$x_1(1+x_2) = x_2(1+x_1) \implies x_1 + x_1x_2 = x_2 + x_1x_2 \implies x_1 = x_2$.
Thus,$f$ is one-one.
To check for onto: The range of $f(x) = \frac{x}{1+x} = \frac{1+x-1}{1+x} = 1 - \frac{1}{1+x}$.
As $x$ ranges from $0$ to $\infty$,$1+x$ ranges from $1$ to $\infty$,so $\frac{1}{1+x}$ ranges from $1$ to $0$.
Therefore,$f(x)$ ranges from $1-1=0$ to $1-0=1$,i.e.,$[0, 1)$.
Since the codomain is $[0, \infty)$ and the range is $[0, 1)$,the range $\neq$ codomain.
Thus,$f$ is not onto.
Therefore,$f$ is one-one but not onto.

(C) Let $y = \frac{x^2 + x + 2}{x^2 + x + 1}$.
We can rewrite the function as $y = \frac{(x^2 + x + 1) + 1}{x^2 + x + 1} = 1 + \frac{1}{x^2 + x + 1}$.
Let $g(x) = x^2 + x + 1$. The minimum value of $g(x)$ occurs at $x = -1/2$,which is $g(-1/2) = (-1/2)^2 - 1/2 + 1 = 1/4 - 1/2 + 1 = 3/4$.
Since $x^2 + x + 1$ is a parabola opening upwards,its range is $[3/4, \infty)$.
Therefore,the range of $\frac{1}{x^2 + x + 1}$ is $(0, 1 / (3/4)] = (0, 4/3]$.
Adding $1$ to this range,we get the range of $f(x)$ as $(1 + 0, 1 + 4/3] = (1, 7/3]$.

(D) Lagrange's Mean Value Theorem $(LMVT)$ states that for a function $f(x)$ to be applicable on $[a, b]$,it must be continuous on $[a, b]$ and differentiable on $(a, b)$.
Checking option $(d)$: $f(x) = |x|$.
This function is continuous on $[0, 1]$,but it is not differentiable at $x = 0$,which lies within the interval $[0, 1]$.
Therefore,$LMVT$ is not applicable to $f(x) = |x|$ on $[0, 1]$.

(D) To find where $f(x)$ increases,we calculate the derivative $f'(x)$ using Leibniz's rule:
$f'(x) = \frac{d}{dx} \int_{x^2}^{x^2 + 1} e^{-t^2} dt = e^{-(x^2+1)^2} \cdot \frac{d}{dx}(x^2+1) - e^{-(x^2)^2} \cdot \frac{d}{dx}(x^2)$
$f'(x) = e^{-(x^4 + 2x^2 + 1)} \cdot (2x) - e^{-x^4} \cdot (2x)$
$f'(x) = 2x e^{-x^4} \left( e^{-(2x^2 + 1)} - 1 \right)$
Since $e^{-(2x^2 + 1)} < 1$ for all real $x$,the term $(e^{-(2x^2 + 1)} - 1)$ is always negative.
For $f'(x) > 0$,we need $2x$ to be negative,which implies $x < 0$.
Thus,$f(x)$ increases in $(-\infty, 0)$.

(A) Given curves are $y = \sqrt{x}$ (or $x = y^2$) and $2y + 3 = x$ (or $y = \frac{x-3}{2}$).
To find the intersection points,set $y^2 = 2y + 3$,which gives $y^2 - 2y - 3 = 0$.
$(y - 3)(y + 1) = 0$,so $y = 3$ or $y = -1$.
Since we are in the $1^{st}$ quadrant,$y = 3$,which implies $x = 9$.
The line $2y + 3 = x$ intersects the $x$-axis at $y = 0$,so $x = 3$.
The area is bounded by $y = \sqrt{x}$ from $x = 0$ to $x = 3$,and by the difference of the curves from $x = 3$ to $x = 9$.
Area $= \int_0^3 \sqrt{x} \, dx + \int_3^9 \left( \sqrt{x} - \frac{x-3}{2} \right) \, dx$
$= \left[ \frac{2}{3} x^{3/2} \right]_0^3 + \left[ \frac{2}{3} x^{3/2} - \frac{1}{2} \left( \frac{x^2}{2} - 3x \right) \right]_3^9$
$= \frac{2}{3}(3\sqrt{3}) + \left( \left[ \frac{2}{3}(27) - \frac{1}{2} \left( \frac{81}{2} - 27 \right) \right] - \left[ \frac{2}{3}(3\sqrt{3}) - \frac{1}{2} \left( \frac{9}{2} - 9 \right) \right] \right)$
$= 2\sqrt{3} + 18 - \frac{1}{2} \left( \frac{27}{2} \right) - 2\sqrt{3} + \frac{1}{2} \left( -\frac{9}{2} \right)$
$= 18 - \frac{27}{4} - \frac{9}{4} = 18 - \frac{36}{4} = 18 - 9 = 9 \text{ sq. units.}$

(A) We are given $I(m, n) = \int_0^1 t^m (1 + t)^n dt$.
Using integration by parts,let $u = (1 + t)^n$ and $dv = t^m dt$.
Then $du = n(1 + t)^{n-1} dt$ and $v = \frac{t^{m+1}}{m+1}$.
Applying the formula $\int u dv = uv - \int v du$:
$I(m, n) = \left[ \frac{t^{m+1}(1 + t)^n}{m+1} \right]_0^1 - \int_0^1 \frac{t^{m+1}}{m+1} \cdot n(1 + t)^{n-1} dt$.
Evaluating the boundary terms:
At $t=1$,we get $\frac{1^{m+1}(1+1)^n}{m+1} = \frac{2^n}{m+1}$.
At $t=0$,the term is $0$.
So,$I(m, n) = \frac{2^n}{m+1} - \frac{n}{m+1} \int_0^1 t^{m+1}(1 + t)^{n-1} dt$.
Since $I(m+1, n-1) = \int_0^1 t^{m+1}(1 + t)^{n-1} dt$,we have:
$I(m, n) = \frac{2^n}{m+1} - \frac{n}{m+1} I(m+1, n-1)$.

(A) The given differential equation is $(1 + t)\frac{dy}{dt} - ty = 1$.
Dividing by $(1 + t)$,we get $\frac{dy}{dt} - \frac{t}{1 + t}y = \frac{1}{1 + t}$.
This is a linear differential equation of the form $\frac{dy}{dt} + P(t)y = Q(t)$,where $P(t) = -\frac{t}{1 + t}$ and $Q(t) = \frac{1}{1 + t}$.
The integrating factor ($I$.$F$.) is $e^{\int P(t) dt} = e^{\int -\frac{t}{1 + t} dt} = e^{\int (-1 + \frac{1}{1 + t}) dt} = e^{-t + \ln(1 + t)} = (1 + t)e^{-t}$.
The general solution is $y \cdot (I.F.) = \int Q(t) \cdot (I.F.) dt + C$.
$y(1 + t)e^{-t} = \int \frac{1}{1 + t} \cdot (1 + t)e^{-t} dt + C = \int e^{-t} dt + C = -e^{-t} + C$.
Given $y(0) = -1$,we substitute $t = 0$: $-1(1 + 0)e^{0} = -e^{0} + C \Rightarrow -1 = -1 + C \Rightarrow C = 0$.
Thus,$y(1 + t)e^{-t} = -e^{-t}$,which simplifies to $y = -\frac{1}{1 + t}$.
For $t = 1$,$y(1) = -\frac{1}{1 + 1} = -\frac{1}{2}$.

(A) The set $B$ can be partitioned into four disjoint regions:
$B = (A \cap B \cap C) \cup (A \cap B \cap \bar{C}) \cup (\bar{A} \cap B \cap C) \cup (\bar{A} \cap B \cap \bar{C})$
Note that $(A \cap B \cap C) \cup (\bar{A} \cap B \cap C) = B \cap C$.
Thus,$P(B) = P(A \cap B \cap \bar{C}) + P(\bar{A} \cap B \cap \bar{C}) + P(B \cap C)$.
Substituting the given values:
$\frac{3}{4} = \frac{1}{3} + \frac{1}{3} + P(B \cap C)$
$\frac{3}{4} = \frac{2}{3} + P(B \cap C)$
$P(B \cap C) = \frac{3}{4} - \frac{2}{3} = \frac{9-8}{12} = \frac{1}{12}$.

(C) The line is given by $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$. The direction vector of the line is $\vec{b} = (1, 1, 2)$ and the normal to the plane $2x-4y+z=7$ is $\vec{n} = (2, -4, 1)$.
First,check if the line is parallel to the plane by calculating the dot product $\vec{b} \cdot \vec{n} = (1)(2) + (1)(-4) + (2)(1) = 2 - 4 + 2 = 0$. Since the dot product is $0$,the line is parallel to the plane.
For the line to lie in the plane,any point on the line must satisfy the plane equation. Let the point on the line be $(4, 2, k)$.
Substituting this point into the plane equation $2x-4y+z=7$:
$2(4) - 4(2) + k = 7$
$8 - 8 + k = 7$
$k = 7$.