IIT JEE 1989 Mathematics Question Paper with Answer and Solution

(B) Since the triangle with vertices $z_1 = a + i$,$z_2 = 1 + bi$,and $z_3 = 0$ is equilateral,we have the condition $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$.
Substituting the values,we get $(a + i)^2 + (1 + bi)^2 + 0 = (a + i)(1 + bi)$.
Expanding both sides: $(a^2 - 1 + 2ai) + (1 - b^2 + 2bi) = a + abi + i - b$.
Simplifying: $(a^2 - b^2) + 2i(a + b) = (a - b) + i(1 + ab)$.
Equating real and imaginary parts:
$a^2 - b^2 = a - b$ ... $(i)$
$2(a + b) = 1 + ab$ ... $(ii)$
From $(i)$,$(a - b)(a + b) = (a - b)$,which implies $(a - b)(a + b - 1) = 0$.
So,either $a = b$ or $a + b = 1$.
Case $1$: If $a = b$,then from $(ii)$,$2(2a) = 1 + a^2$,so $a^2 - 4a + 1 = 0$.
Solving for $a$,$a = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}$.
Since $0 < a < 1$,we must have $a = b = 2 - \sqrt{3}$.
Case $2$: If $a + b = 1$,then $b = 1 - a$. Substituting into $(ii)$,$2(1) = 1 + a(1 - a)$,which gives $a^2 - a + 1 = 0$. This equation has no real roots.
Thus,the only solution is $a = b = 2 - \sqrt{3}$.

(D) Let $p = -q$ where $q > 0$. The cube roots of $p$ are $\alpha = -q^{1/3}$,$\beta = -q^{1/3}\omega$,and $\gamma = -q^{1/3}\omega^2$,where $\omega = \frac{-1 + i\sqrt{3}}{2}$ is the complex cube root of unity.
Substituting these into the expression:
$\frac{x(-q^{1/3}) + y(-q^{1/3}\omega) + z(-q^{1/3}\omega^2)}{x(-q^{1/3}\omega) + y(-q^{1/3}\omega^2) + z(-q^{1/3}\omega^3)} = \frac{-(x + y\omega + z\omega^2)}{-\omega(x + y\omega + z\omega^2)} = \frac{1}{\omega} = \omega^2$.
Since $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$,none of the given options match.

(D) Given the equation $|x|^2 - 3|x| + 2 = 0$.
Let $|x| = t$,where $t \ge 0$.
The equation becomes $t^2 - 3t + 2 = 0$.
Factoring the quadratic equation: $(t - 1)(t - 2) = 0$.
This gives $t = 1$ or $t = 2$.
Since $t = |x|$,we have $|x| = 1$ or $|x| = 2$.
For $|x| = 1$,the solutions are $x = 1$ and $x = -1$.
For $|x| = 2$,the solutions are $x = 2$ and $x = -2$.
Thus,the real solutions are $x \in \{1, -1, 2, -2\}$.
The total number of real solutions is $4$.

(D) For the given equation to be meaningful,we must have $x > 0$. Taking the logarithm base $2$ on both sides:
$(\frac{3}{4}(\log_2 x)^2 + \log_2 x - \frac{5}{4}) \log_2 x = \log_2 \sqrt{2} = \frac{1}{2}$.
Let $t = \log_2 x$. Then the equation becomes:
$(\frac{3}{4}t^2 + t - \frac{5}{4}) t = \frac{1}{2}$.
Multiplying by $4$:
$(3t^2 + 4t - 5) t = 2 \Rightarrow 3t^3 + 4t^2 - 5t - 2 = 0$.
By testing roots,we find $(t - 1)$ is a factor:
$(t - 1)(3t^2 + 7t + 2) = 0 \Rightarrow (t - 1)(3t + 1)(t + 2) = 0$.
Thus,$t = 1, -2, -1/3$.
Since $t = \log_2 x$,we have $x = 2^1 = 2$,$x = 2^{-2} = 1/4$,and $x = 2^{-1/3} = 1/\sqrt[3]{2}$.
All three solutions are real,and $1/\sqrt[3]{2}$ is irrational. Therefore,all statements $(A)$,$(B)$,and $(C)$ are correct.

(D) Given that $\alpha$ is a root of $a^2x^2 + bx + c = 0$,we have $a^2\alpha^2 + b\alpha + c = 0$,which implies $b\alpha + c = -a^2\alpha^2$.
Given that $\beta$ is a root of $a^2x^2 - bx - c = 0$,we have $a^2\beta^2 - b\beta - c = 0$,which implies $b\beta + c = a^2\beta^2$.
Let $f(x) = a^2x^2 + 2bx + 2c$.
Evaluating $f(x)$ at $\alpha$: $f(\alpha) = a^2\alpha^2 + 2(b\alpha + c) = a^2\alpha^2 + 2(-a^2\alpha^2) = -a^2\alpha^2$. Since $a \ne 0$ and $\alpha > 0$,$f(\alpha) < 0$.
Evaluating $f(x)$ at $\beta$: $f(\beta) = a^2\beta^2 + 2(b\beta + c) = a^2\beta^2 + 2(a^2\beta^2) = 3a^2\beta^2$. Since $a \ne 0$ and $\beta > 0$,$f(\beta) > 0$.
Since $f(\alpha) < 0$ and $f(\beta) > 0$,by the Intermediate Value Theorem,there exists a root $\gamma$ of $f(x) = 0$ such that $\alpha < \gamma < \beta$.

(A) number is divisible by $3$ if the sum of its digits is divisible by $3$. The sum of all given digits ${0, 1, 2, 3, 4, 5}$ is $15$. To form a $5$-digit number,we must exclude one digit such that the sum of the remaining $5$ digits is divisible by $3$.
Case $1$: Exclude $0$. The remaining digits are ${1, 2, 3, 4, 5}$. The sum is $15$,which is divisible by $3$. The number of $5$-digit numbers is $5! = 120$.
Case $2$: Exclude $3$. The remaining digits are ${0, 1, 2, 4, 5}$. The sum is $12$,which is divisible by $3$. The number of $5$-digit numbers is $5! - 4! = 120 - 24 = 96$ (subtracting cases where $0$ is in the first position).
Total ways = $120 + 96 = 216$.

(B) Given equation: $\sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos 3x$
Rearranging terms: $(\sin 3x + \sin x) - 3\sin 2x = (\cos 3x + \cos x) - 3\cos 2x$
Using sum-to-product formulas: $2\sin 2x \cos x - 3\sin 2x = 2\cos 2x \cos x - 3\cos 2x$
$\sin 2x(2\cos x - 3) = \cos 2x(2\cos x - 3)$
$(\sin 2x - \cos 2x)(2\cos x - 3) = 0$
Since $2\cos x - 3 \neq 0$ (as $\cos x$ cannot be $1.5$),we have $\sin 2x = \cos 2x$
$\tan 2x = 1$
$2x = n\pi + \frac{\pi}{4}$
$x = \frac{n\pi}{2} + \frac{\pi}{8}$

(C) We know that if the origin is shifted to $(h, k)$,then the new coordinates $(x', y')$ are given by $(x - h, y - k)$.
Given the original point $(x, y) = (4, 5)$ and the new origin $(h, k) = (1, -2)$.
Substituting these values,we get:
$x' = 4 - 1 = 3$
$y' = 5 - (-2) = 5 + 2 = 7$
Therefore,the new coordinates are $(3, 7)$.

(B) The definition of an ellipse is the locus of a point such that the sum of its distances from two fixed points (foci) is a constant,provided that the constant is greater than the distance between the two fixed points.
Given $PA + PB = 4$,where $4$ is the constant sum.
If the distance $AB < 4$,the locus of $P$ is an ellipse.
If $AB = 4$,the locus is the line segment $AB$.
If $AB > 4$,the locus is an empty set.
Assuming $AB < 4$,the locus is an ellipse.

(B) The center of the circle is the point of intersection of the diameters $2x - 3y = 5$ and $3x - 4y = 7$.
Solving these equations: $2x - 3y = 5$ (multiplied by $3$) $\Rightarrow 6x - 9y = 15$ and $3x - 4y = 7$ (multiplied by $2$) $\Rightarrow 6x - 8y = 14$.
Subtracting the first from the second: $(6x - 8y) - (6x - 9y) = 14 - 15 \Rightarrow y = -1$.
Substituting $y = -1$ into $2x - 3y = 5$: $2x - 3(-1) = 5$ $\Rightarrow 2x + 3 = 5$ $\Rightarrow 2x = 2$ $\Rightarrow x = 1$.
So,the center $(h, k) = (1, -1)$.
Given area $= 154$,we have $\pi r^2 = 154$ $\Rightarrow \frac{22}{7} r^2 = 154$ $\Rightarrow r^2 = 154 \times \frac{7}{22} = 49$ $\Rightarrow r = 7$.
The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$.
$(x - 1)^2 + (y + 1)^2 = 7^2$.
$x^2 - 2x + 1 + y^2 + 2y + 1 = 49$.
$x^2 + y^2 - 2x + 2y + 2 = 49$.
$x^2 + y^2 - 2x + 2y = 47$.

(A) The given equation of the circle is $x^2 + y^2 - 6x + 2y = 0$.
Comparing this with the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g = -6 \implies g = -3$ and $2f = 2 \implies f = 1$.
The center of the circle is $(-g, -f) = (3, -1)$.
$A$ diameter of the circle always passes through its center.
We need the equation of a line passing through the origin $(0, 0)$ and the center $(3, -1)$.
The slope $m$ of the line is $\frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 0}{3 - 0} = -\frac{1}{3}$.
The equation of the line is $y - 0 = -\frac{1}{3}(x - 0)$,which simplifies to $3y = -x$ or $x + 3y = 0$.

(A) The equation of the circle is $x^2 + y^2 = 4$. The point of contact is $P(1, \sqrt{3})$.
The equation of the tangent at $(x_1, y_1)$ is $xx_1 + yy_1 = r^2$. Substituting the values,we get $x(1) + y(\sqrt{3}) = 4$,which simplifies to $x + \sqrt{3}y = 4$.
The tangent intersects the $x$-axis at $y = 0$,so $x = 4$. The point is $A(4, 0)$.
The normal at $(1, \sqrt{3})$ passes through the center $(0, 0)$ and the point $(1, \sqrt{3})$. Its equation is $y = \sqrt{3}x$.
The normal intersects the $x$-axis at the origin $O(0, 0)$.
The triangle is formed by the vertices $O(0, 0)$,$P(1, \sqrt{3})$,and $A(4, 0)$.
The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Area $= \frac{1}{2} |0(\sqrt{3} - 0) + 1(0 - 0) + 4(0 - \sqrt{3})| = \frac{1}{2} |-4\sqrt{3}| = 2\sqrt{3}$ square units.

(A) The first circle is $(x - 1)^2 + (y - 3)^2 = r^2$,with center $C_1 = (1, 3)$ and radius $r_1 = r$.
The second circle is $x^2 + y^2 - 8x + 2y + 8 = 0$. Rewriting in standard form: $(x^2 - 8x + 16) + (y^2 + 2y + 1) = -8 + 16 + 1$,which is $(x - 4)^2 + (y + 1)^2 = 9 = 3^2$. Thus,center $C_2 = (4, -1)$ and radius $r_2 = 3$.
The distance between the centers $d = \sqrt{(4 - 1)^2 + (-1 - 3)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5$.
For two circles to intersect at two distinct points,the condition is $|r_1 - r_2| < d < r_1 + r_2$.
$1$) $r_1 + r_2 > d$ $\Rightarrow r + 3 > 5$ $\Rightarrow r > 2$.
$2$) $|r_1 - r_2| < d$ $\Rightarrow |r - 3| < 5$ $\Rightarrow -5 < r - 3 < 5$ $\Rightarrow -2 < r < 8$. Since $r$ must be positive,$0 < r < 8$.
Combining $r > 2$ and $r < 8$,we get $2 < r < 8$.

(C) The given limit is of the form $\mathop {\lim }\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$,which is the definition of the derivative $f'(a)$ for the function $f(x) = x^2 \sin x$.
First,find the derivative of $f(x) = x^2 \sin x$ with respect to $x$ using the product rule:
$f'(x) = \frac{d}{dx}(x^2) \sin x + x^2 \frac{d}{dx}(\sin x)$
$f'(x) = 2x \sin x + x^2 \cos x$
Evaluating at $x = a$ gives:
$f'(a) = 2a \sin a + a^2 \cos a$
Alternatively,applying $L$-Hospital's rule with respect to $h$:
$\mathop {\lim }\limits_{h \to 0} \frac{\frac{d}{dh}((a+h)^2 \sin(a+h) - a^2 \sin a)}{1}$
$= \mathop {\lim }\limits_{h \to 0} (2(a+h) \sin(a+h) + (a+h)^2 \cos(a+h))$
$= 2a \sin a + a^2 \cos a$

(B) Given,$P(X) = 0.3$ and $P(Y) = 0.2$.
Since the events are independent,the probability that both $X$ and $Y$ fail is $P(X \cap Y) = P(X) \times P(Y) = 0.3 \times 0.2 = 0.06$.
The probability that either $X$ or $Y$ fails is given by the addition theorem of probability:
$P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$
Substituting the values:
$P(X \cup Y) = 0.3 + 0.2 - 0.06 = 0.44$.

(C) Given the roots of $x^2 + px + q = 0$ are $\alpha$ and $\beta$,we have $\alpha + \beta = -p$ and $\alpha\beta = q$.
Given the roots of $x^2 - xr + s = 0$ are $\alpha^4$ and $\beta^4$,we have $\alpha^4 + \beta^4 = r$ and $\alpha^4\beta^4 = s$.
Consider the equation $x^2 - 4qx + 2q^2 - r = 0$. The discriminant $D$ is given by:
$D = (-4q)^2 - 4(1)(2q^2 - r) = 16q^2 - 8q^2 + 4r = 8q^2 + 4r$.
Substituting $q = \alpha\beta$ and $r = \alpha^4 + \beta^4$:
$D = 8(\alpha\beta)^2 + 4(\alpha^4 + \beta^4) = 4(2\alpha^2\beta^2 + \alpha^4 + \beta^4) = 4(\alpha^2 + \beta^2)^2$.
Since $(\alpha^2 + \beta^2)^2 \ge 0$,it follows that $D \ge 0$.
Therefore,the roots of the equation $x^2 - 4qx + 2q^2 - r = 0$ are always real.

(C) The figure consists of $8$ squares arranged in three rows: top row ($2$ squares),middle row ($4$ squares),and bottom row ($2$ squares).
We need to place $6$ '$X$'s in these $8$ squares.
The total number of ways to place $6$ '$X$'s in $8$ squares is $^8C_6 = \frac{8 \times 7}{2 \times 1} = 28$.
We must ensure each row contains at least one '$X$'.
Let $R_1, R_2, R_3$ be the top,middle,and bottom rows respectively.
Total squares = $2 + 4 + 2 = 8$.
If the top row $(R_1)$ has no '$X$',then all $6$ '$X$'s must be placed in the remaining $8 - 2 = 6$ squares. This can be done in $^6C_6 = 1$ way.
If the bottom row $(R_3)$ has no '$X$',then all $6$ '$X$'s must be placed in the remaining $8 - 2 = 6$ squares. This can be done in $^6C_6 = 1$ way.
Note that it is impossible for both the top and bottom rows to have no '$X$' simultaneously,as that would require placing $6$ '$X$'s in only $4$ squares,which is impossible.
Thus,the number of invalid ways is $1 + 1 = 2$.
The required number of ways is $28 - 2 = 26$.

(B) Since $a, b,$ and $c$ are coplanar,there exist scalars $x, y, z$ (not all zero) such that $xa + yb + zc = 0$ $(i)$.
Taking the dot product of $(i)$ with $a$,we get $x(a \cdot a) + y(a \cdot b) + z(a \cdot c) = 0$ $(ii)$.
Taking the dot product of $(i)$ with $b$,we get $x(b \cdot a) + y(b \cdot b) + z(b \cdot c) = 0$ $(iii)$.
Since $x, y, z$ are not all zero,the system of equations $(i), (ii),$ and $(iii)$ has a non-trivial solution.
Therefore,the determinant of the coefficients must be zero:
$\left| \begin{array}{ccc} a & b & c \\ a \cdot a & a \cdot b & a \cdot c \\ b \cdot a & b \cdot b & b \cdot c \end{array} \right| = 0$.

(B) Given that $|a + b| = |a - b|$.
Squaring both sides,we get $|a + b|^2 = |a - b|^2$.
Using the property $|x|^2 = x \cdot x$,we have $(a + b) \cdot (a + b) = (a - b) \cdot (a - b)$.
Expanding the dot product,we get $a \cdot a + 2(a \cdot b) + b \cdot b = a \cdot a - 2(a \cdot b) + b \cdot b$.
Subtracting $a \cdot a + b \cdot b$ from both sides,we get $2(a \cdot b) = -2(a \cdot b)$.
This implies $4(a \cdot b) = 0$,which means $a \cdot b = 0$.
Since the dot product of two non-zero vectors is zero,the vectors $a$ and $b$ are perpendicular to each other.

(B) The component of vector $a$ along $b$ is given by the formula: $\text{proj}_{b} a = \frac{(a \cdot b)b}{|b|^2}$.
Given $a = 4i + 6j$ and $b = 3j + 4k$.
First,calculate the dot product $a \cdot b = (4i + 6j) \cdot (0i + 3j + 4k) = (4 \times 0) + (6 \times 3) + (0 \times 4) = 0 + 18 + 0 = 18$.
Next,calculate the square of the magnitude of $b$: $|b|^2 = 3^2 + 4^2 = 9 + 16 = 25$.
Substituting these values into the formula: $\text{proj}_{b} a = \frac{18}{25}(3j + 4k)$.

(C) Given that $a \cdot b = 0$. This implies that either $a = 0$,$b = 0$,or $a \perp b$.
Also,given that $a \times b = 0$. This implies that either $a = 0$,$b = 0$,or $a \parallel b$.
For both conditions to be satisfied simultaneously,$a$ and $b$ cannot be both perpendicular and parallel unless at least one of them is a null vector (zero vector).
Therefore,either $a = 0$ or $b = 0$.

(C) Let $\vec{a} = 6i + 2j + 3k$ and $\vec{b} = 3i - 6j - 2k$.
The vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by $\vec{a} \times \vec{b}$.
$\vec{a} \times \vec{b} = \begin{vmatrix} i & j & k \\ 6 & 2 & 3 \\ 3 & -6 & -2 \end{vmatrix} = i(-4 - (-18)) - j(-12 - 9) + k(-36 - 6) = 14i + 21j - 42k$.
The magnitude of this vector is $|\vec{a} \times \vec{b}| = \sqrt{14^2 + 21^2 + (-42)^2} = \sqrt{196 + 441 + 1764} = \sqrt{2401} = 49$.
The unit vector perpendicular to both is $\pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \pm \frac{14i + 21j - 42k}{49} = \pm \frac{2i + 3j - 6k}{7}$.
Comparing with the given options,the correct unit vector is $\frac{2i + 3j - 6k}{7}$.

(C) The volume of a parallelepiped with edges represented by vectors $\vec{a}$,$\vec{b}$,and $\vec{c}$ is given by the absolute value of the scalar triple product $|\vec{a} \cdot (\vec{b} \times \vec{c})|$,which is equal to the determinant of the matrix formed by the components of the vectors.
Given vectors are $\vec{a} = -12i + 0j + \alpha k$,$\vec{b} = 0i + 3j - 1k$,and $\vec{c} = 2i + 1j - 15k$.
The volume is given by:
$|\det(\vec{a}, \vec{b}, \vec{c})| = 546$
$\left| \begin{matrix} -12 & 0 & \alpha \\ 0 & 3 & -1 \\ 2 & 1 & -15 \end{matrix} \right| = \pm 546$
Expanding the determinant along the first row:
$-12(3(-15) - (-1)(1)) - 0(...) + \alpha(0(1) - 3(2)) = \pm 546$
$-12(-45 + 1) + \alpha(-6) = \pm 546$
$-12(-44) - 6\alpha = \pm 546$
$528 - 6\alpha = \pm 546$
Case $1$: $528 - 6\alpha = 546 \Rightarrow -6\alpha = 18 \Rightarrow \alpha = -3$
Case $2$: $528 - 6\alpha = -546 \Rightarrow -6\alpha = -1074 \Rightarrow \alpha = 179$
Comparing with the given options,the correct value is $\alpha = -3$.

(A) Given $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$.
First,find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = \sec \theta \tan \theta + \sin \theta = \tan \theta (\sec \theta + \cos \theta)$.
$\frac{dy}{d\theta} = n \sec^{n-1} \theta (\sec \theta \tan \theta) + n \cos^{n-1} \theta \sin \theta = n \tan \theta (\sec^n \theta + \cos^n \theta)$.
Now,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{n \tan \theta (\sec^n \theta + \cos^n \theta)}{\tan \theta (\sec \theta + \cos \theta)} = \frac{n(\sec^n \theta + \cos^n \theta)}{\sec \theta + \cos \theta}$.
Squaring both sides:
$\left( \frac{dy}{dx} \right)^2 = \frac{n^2(\sec^n \theta + \cos^n \theta)^2}{(\sec \theta + \cos \theta)^2}$.
Using the identity $(a+b)^2 = (a-b)^2 + 4ab$:
$(\sec^n \theta + \cos^n \theta)^2 = (\sec^n \theta - \cos^n \theta)^2 + 4 \sec^n \theta \cos^n \theta = y^2 + 4$.
Similarly,$(\sec \theta + \cos \theta)^2 = (\sec \theta - \cos \theta)^2 + 4 \sec \theta \cos \theta = x^2 + 4$.
Thus,$\left( \frac{dy}{dx} \right)^2 = \frac{n^2(y^2 + 4)}{x^2 + 4}$,which implies $(x^2 + 4) \left( \frac{dy}{dx} \right)^2 = n^2(y^2 + 4)$.

(B) The function $f(x) = |1 - x^2|$ changes its sign at $x = -1$ and $x = 1$.
We can split the integral as:
$\int_{-2}^{2} |1 - x^2| \, dx = \int_{-2}^{-1} |1 - x^2| \, dx + \int_{-1}^{1} |1 - x^2| \, dx + \int_{1}^{2} |1 - x^2| \, dx$
In the interval $[-2, -1]$,$1 - x^2 \le 0$,so $|1 - x^2| = -(1 - x^2) = x^2 - 1$.
In the interval $[-1, 1]$,$1 - x^2 \ge 0$,so $|1 - x^2| = 1 - x^2$.
In the interval $[1, 2]$,$1 - x^2 \le 0$,so $|1 - x^2| = -(1 - x^2) = x^2 - 1$.
Thus,the integral becomes:
$\int_{-2}^{-1} (x^2 - 1) \, dx + \int_{-1}^{1} (1 - x^2) \, dx + \int_{1}^{2} (x^2 - 1) \, dx$
$= [\frac{x^3}{3} - x]_{-2}^{-1} + [x - \frac{x^3}{3}]_{-1}^{1} + [\frac{x^3}{3} - x]_{1}^{2}$
$= ((\frac{-1}{3} + 1) - (\frac{-8}{3} + 2)) + ((1 - \frac{1}{3}) - (-1 + \frac{1}{3})) + ((\frac{8}{3} - 2) - (\frac{1}{3} - 1))$
$= (\frac{2}{3} - (-\frac{2}{3})) + (\frac{2}{3} - (-\frac{2}{3})) + (\frac{2}{3} - (-\frac{2}{3}))$
$= \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = \frac{12}{3} = 4.$

(D) Given that $E$ and $F$ are independent events,we have $P(E \cap F) = P(E) \cdot P(F)$.
$(a)$ $P(E \cap F^c) = P(E) - P(E \cap F) = P(E) - P(E)P(F) = P(E)(1 - P(F)) = P(E)P(F^c)$. Thus,$E$ and $F^c$ are independent.
$(b)$ $P(E^c \cap F^c) = 1 - P(E \cup F) = 1 - [P(E) + P(F) - P(E \cap F)] = 1 - P(E) - P(F) + P(E)P(F) = (1 - P(E))(1 - P(F)) = P(E^c)P(F^c)$. Thus,$E^c$ and $F^c$ are independent.
$(c)$ Since $E$ and $F$ are independent,$P(E/F) = P(E)$. Similarly,$P(E^c/F^c) = P(E^c)$. Therefore,$P(E/F) + P(E^c/F^c) = P(E) + P(E^c) = 1$.
Since all statements are correct,the answer is $(d)$.

(A) Let $I = \int_0^a f(x)g(x) dx$.
Using the property $\int_0^a h(x) dx = \int_0^a h(a - x) dx$,we have:
$I = \int_0^a f(a - x)g(a - x) dx$.
Given $f(x) = f(a - x)$ and $g(a - x) = 2 - g(x)$,we substitute these into the integral:
$I = \int_0^a f(x)[2 - g(x)] dx$.
$I = 2\int_0^a f(x) dx - \int_0^a f(x)g(x) dx$.
$I = 2\int_0^a f(x) dx - I$.
$2I = 2\int_0^a f(x) dx$.
$I = \int_0^a f(x) dx$.

(B) Let $A$ be the event that a sum of $5$ occurs,$B$ be the event that a sum of $7$ occurs,and $C$ be the event that neither a sum of $5$ nor a sum of $7$ occurs.
For a pair of dice,the total number of outcomes is $36$.
The outcomes for sum $5$ are $(1,4), (2,3), (3,2), (4,1)$,so $P(A) = \frac{4}{36} = \frac{1}{9}$.
The outcomes for sum $7$ are $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$,so $P(B) = \frac{6}{36} = \frac{1}{6}$.
The probability of neither $5$ nor $7$ is $P(C) = 1 - (P(A) + P(B)) = 1 - (\frac{1}{9} + \frac{1}{6}) = 1 - (\frac{2+3}{18}) = 1 - \frac{5}{18} = \frac{13}{18}$.
The probability that $A$ occurs before $B$ is given by the sum of the infinite geometric series:
$P = P(A) + P(C)P(A) + P(C)^2 P(A) + \dots = \frac{P(A)}{1 - P(C)}$.
Substituting the values: $P = \frac{\frac{1}{9}}{1 - \frac{13}{18}} = \frac{\frac{1}{9}}{\frac{5}{18}} = \frac{1}{9} \times \frac{18}{5} = \frac{2}{5}$.
Thus,the correct option is $B$.