IIT JEE 1982 Mathematics Question Paper with Answer and Solution

(B) First,simplify the expression inside the parenthesis:
$\frac{1 + i}{1 - i} = \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i} = \frac{(1 + i)^2}{1^2 - i^2} = \frac{1 + 2i + i^2}{1 - (-1)} = \frac{1 + 2i - 1}{2} = \frac{2i}{2} = i$
Given that ${\left( \frac{1 + i}{1 - i} \right)^m} = 1$,we substitute the simplified form:
$i^m = 1$
We know that the powers of $i$ follow the cycle $i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1$.
Therefore,the least positive integral value of $m$ for which $i^m = 1$ is $m = 4$.

(D) Given inequality: $|z - 4| < |z - 2|$
Squaring both sides: $|z - 4|^2 < |z - 2|^2$
Let $z = x + iy$. Then $|(x - 4) + iy|^2 < |(x - 2) + iy|^2$
$(x - 4)^2 + y^2 < (x - 2)^2 + y^2$
$x^2 - 8x + 16 + y^2 < x^2 - 4x + 4 + y^2$
$-8x + 16 < -4x + 4$
$12 < 4x$
$x > 3$
Since $x = \text{Re}(z)$,the region is $\text{Re}(z) > 3$.
Thus,the correct option is $D$.

(A) Given the equation $\left| \frac{z - 5i}{z + 5i} \right| = 1$.
Substituting $z = x + iy$,we get $\left| \frac{x + i(y - 5)}{x + i(y + 5)} \right| = 1$.
Using the property $\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}$,we have $|x + i(y - 5)| = |x + i(y + 5)|$.
Squaring both sides,we get $x^2 + (y - 5)^2 = x^2 + (y + 5)^2$.
Expanding the squares: $x^2 + y^2 - 10y + 25 = x^2 + y^2 + 10y + 25$.
Simplifying the equation: $-10y = 10y$,which implies $20y = 0$,so $y = 0$.
The equation $y = 0$ represents the real axis.

(A) Given that $a_1, a_2, a_3, ......., a_n$ are in $A.P.$ with common difference $d = a_{i+1} - a_i$.
We know that $a_n = a_1 + (n - 1)d$,which implies $d = \frac{a_n - a_1}{n - 1}$.
Rationalizing each term of the sum:
$\frac{1}{\sqrt{a_i} + \sqrt{a_{i+1}}} = \frac{\sqrt{a_{i+1}} - \sqrt{a_i}}{a_{i+1} - a_i} = \frac{\sqrt{a_{i+1}} - \sqrt{a_i}}{d}$.
Summing these terms from $i = 1$ to $n-1$:
$\sum_{i=1}^{n-1} \frac{1}{\sqrt{a_i} + \sqrt{a_{i+1}}} = \frac{1}{d} ((\sqrt{a_2} - \sqrt{a_1}) + (\sqrt{a_3} - \sqrt{a_2}) + ....... + (\sqrt{a_n} - \sqrt{a_{n-1}}))$.
This is a telescoping sum,which simplifies to:
$\frac{1}{d} (\sqrt{a_n} - \sqrt{a_1})$.
Substituting $d = \frac{a_n - a_1}{n - 1}$:
$= \frac{n - 1}{a_n - a_1} (\sqrt{a_n} - \sqrt{a_1}) = \frac{n - 1}{(\sqrt{a_n} - \sqrt{a_1})(\sqrt{a_n} + \sqrt{a_1})} (\sqrt{a_n} - \sqrt{a_1}) = \frac{n - 1}{\sqrt{a_n} + \sqrt{a_1}}$.

(C) Let the first term of the $G.P.$ be $a$ and the common ratio be $r$.
Given that the third term $T_3 = ar^2 = 4$.
We need to find the product of the first $5$ terms:
$P = a \times (ar) \times (ar^2) \times (ar^3) \times (ar^4)$
$P = a^5 \times r^{(1+2+3+4)} = a^5 \times r^{10}$
$P = (ar^2)^5$
Substituting the value $ar^2 = 4$:
$P = 4^5$.

(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) Given equation: $e^{\sin x} - e^{-\sin x} - 4 = 0$
Let $y = e^{\sin x}$. Since $e^{-\sin x} = \frac{1}{y}$,the equation becomes:
$y - \frac{1}{y} - 4 = 0$
Multiplying by $y$ $(y \neq 0)$:
$y^2 - 4y - 1 = 0$
Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$y = \frac{4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5}$
Since $y = e^{\sin x} > 0$,we must have $y = 2 + \sqrt{5}$ (as $2 - \sqrt{5} < 0$).
Taking the natural logarithm on both sides:
$\sin x = \ln(2 + \sqrt{5})$
We know that $\sqrt{4} < \sqrt{5} < \sqrt{9}$,so $2 < \sqrt{5} < 3$. Thus,$4 < 2 + \sqrt{5} < 5$.
Since $e^1 \approx 2.718$,$\ln(2 + \sqrt{5}) > \ln(e) = 1$.
Since $\sin x \leq 1$ for all real $x$,the equation $\sin x = \ln(2 + \sqrt{5})$ has no real solution.
Therefore,the number of real roots is $0$ (None).

(C) Let $\alpha$ be the common root of the equations $ax^2 + bx + c = 0$ and $bx^2 + cx + a = 0$.
Then $a\alpha^2 + b\alpha + c = 0$ and $b\alpha^2 + c\alpha + a = 0$.
Using the method of cross-multiplication:
$\frac{\alpha^2}{b(a) - c(c)} = \frac{\alpha}{c(b) - a(a)} = \frac{1}{a(c) - b(b)}$
$\frac{\alpha^2}{ab - c^2} = \frac{\alpha}{bc - a^2} = \frac{1}{ac - b^2}$
From the first two ratios,$\alpha = \frac{bc - a^2}{ab - c^2}$.
From the last two ratios,$\alpha = \frac{ac - b^2}{bc - a^2}$.
Equating the two expressions for $\alpha$:
$(bc - a^2)^2 = (ab - c^2)(ac - b^2)$
$b^2c^2 + a^4 - 2a^2bc = a^2bc - ab^3 - c^3a + b^2c^2$
$a^4 + ab^3 + ac^3 = 3a^2bc$
Since $a \neq 0$,dividing by $a$ gives $a^3 + b^3 + c^3 = 3abc$.
Therefore,$\frac{a^3 + b^3 + c^3}{abc} = 3$.

(A) There are $10$ digits in total: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.
$A$ $9$ digit number cannot have $0$ at the first place.
Total ways to arrange $9$ distinct digits out of $10$ is $^{10}P_9 = \frac{10!}{1!} = 10!$.
However,this includes cases where $0$ is at the first place. If $0$ is fixed at the first place,we need to arrange $8$ distinct digits from the remaining $9$ digits in the remaining $8$ positions,which is $^9P_8 = \frac{9!}{1!} = 9!$.
Therefore,the number of $9$ digit numbers with all digits different is $^{10}P_9 - ^9P_8 = 10! - 9! = (10 - 1) \times 9! = 9 \times 9!$.

(A) Let $f(n) = 7^{2n} + 2^{3n-3} \cdot 3^{n-1}$.
For $n = 1$,$f(1) = 7^{2(1)} + 2^{3(1)-3} \cdot 3^{1-1} = 7^2 + 2^0 \cdot 3^0 = 49 + 1 = 50$.
For $n = 2$,$f(2) = 7^{2(2)} + 2^{3(2)-3} \cdot 3^{2-1} = 7^4 + 2^3 \cdot 3^1 = 2401 + 8 \cdot 3 = 2401 + 24 = 2425$.
Since $50$ and $2425$ are both divisible by $25$,the expression is always divisible by $25$.

(C) To find the sum of the coefficients in the expansion of a polynomial $P(x)$,we substitute $x = 1$.
Given the expression $(1 + x - 3x^2)^{2163}$,we set $x = 1$.
Sum of coefficients $= (1 + 1 - 3(1)^2)^{2163}$
$= (1 + 1 - 3)^{2163}$
$= (-1)^{2163}$
Since $2163$ is an odd number,$(-1)^{2163} = -1$.

(C) We know that $\sin 54^\circ = \cos 36^\circ = \frac{\sqrt{5}+1}{4}$.
Consider the expression $E = \sin 12^\circ \sin 48^\circ \sin 54^\circ$.
Using the formula $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$:
$E = \frac{1}{2} [\cos(48^\circ - 12^\circ) - \cos(48^\circ + 12^\circ)] \sin 54^\circ$
$E = \frac{1}{2} [\cos 36^\circ - \cos 60^\circ] \sin 54^\circ$
Since $\sin 54^\circ = \cos 36^\circ$ and $\cos 60^\circ = 1/2$:
$E = \frac{1}{2} [\cos 36^\circ - 1/2] \cos 36^\circ$
$E = \frac{1}{2} [\cos^2 36^\circ - \frac{1}{2} \cos 36^\circ]$
Substituting $\cos 36^\circ = \frac{\sqrt{5}+1}{4}$:
$E = \frac{1}{2} [(\frac{\sqrt{5}+1}{4})^2 - \frac{1}{2} (\frac{\sqrt{5}+1}{4})]$
$E = \frac{1}{2} [\frac{5+1+2\sqrt{5}}{16} - \frac{\sqrt{5}+1}{8}]$
$E = \frac{1}{2} [\frac{6+2\sqrt{5}}{16} - \frac{2\sqrt{5}+2}{16}]$
$E = \frac{1}{2} [\frac{4}{16}] = \frac{1}{8}$.

(B) The line passes through $(a \cos \alpha, a \sin \alpha)$ and $(a \cos \beta, a \sin \beta)$.
Its equation is $x \cos \frac{\alpha + \beta}{2} + y \sin \frac{\alpha + \beta}{2} = a \cos \frac{\alpha - \beta}{2}$.
The foot of the perpendicular from the origin $(0,0)$ to the line $x \cos \theta + y \sin \theta = p$ is $(p \cos \theta, p \sin \theta)$.
Here,$p = a \cos \frac{\alpha - \beta}{2}$ and $\theta = \frac{\alpha + \beta}{2}$.
Thus,the foot of the perpendicular is $\left( a \cos \frac{\alpha - \beta}{2} \cos \frac{\alpha + \beta}{2}, a \cos \frac{\alpha - \beta}{2} \sin \frac{\alpha + \beta}{2} \right)$.
Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$ and $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,we get:
$x = \frac{a}{2} [\cos \alpha + \cos \beta]$ and $y = \frac{a}{2} [\sin \alpha + \sin \beta]$.
Therefore,the correct option is $B$.

(B) Let the two fixed coplanar points be $A(0, 0)$ and $B(a, 0)$.
Let the moving point $P$ be $(x, y)$.
According to the given condition,the ratio of distances is constant,say $\lambda$:
$\frac{PA}{PB} = \lambda \implies \frac{\sqrt{x^2 + y^2}}{\sqrt{(x-a)^2 + y^2}} = \lambda$
Squaring both sides:
$x^2 + y^2 = \lambda^2 ((x-a)^2 + y^2)$
$x^2 + y^2 = \lambda^2 (x^2 - 2ax + a^2 + y^2)$
$(1 - \lambda^2)x^2 + (1 - \lambda^2)y^2 + 2a\lambda^2 x - a^2\lambda^2 = 0$
Since $\lambda \ne 1$,we can divide by $(1 - \lambda^2)$:
$x^2 + y^2 + \frac{2a\lambda^2}{1 - \lambda^2}x - \frac{a^2\lambda^2}{1 - \lambda^2} = 0$
This is the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$. Thus,the locus is a circle.

(B) Let $A = (a, 0)$,$B = (-a, 0)$,and $P = (x, y)$.
Given the condition $\frac{PA}{PB} = k$,we have $\frac{PA^2}{PB^2} = k^2$.
Substituting the coordinates,we get $\frac{(x - a)^2 + y^2}{(x + a)^2 + y^2} = k^2$.
$(x - a)^2 + y^2 = k^2((x + a)^2 + y^2)$.
$(x^2 - 2ax + a^2 + y^2) = k^2(x^2 + 2ax + a^2 + y^2)$.
$(x^2 + y^2)(1 - k^2) - 2ax(1 + k^2) + a^2(1 - k^2) = 0$.
If $k = 1$,the equation becomes $-4ax = 0$,which represents a straight line (the perpendicular bisector of $AB$),not a circle.
Therefore,for the locus to be a circle,$k$ must not be equal to $1$.

(D) Step $1$: The two women choose their chairs from the chairs marked $1$ to $4$. Since the order of selection matters (as the chairs are numbered),the number of ways is $^4{P_2} = 4 \times 3 = 12$.
Step $2$: After the women have occupied $2$ chairs,there are $8 - 2 = 6$ chairs remaining.
Step $3$: The three men choose their chairs from the remaining $6$ chairs. The number of ways is $^6{P_3} = 6 \times 5 \times 4 = 120$.
Step $4$: The total number of arrangements is $^4{P_2} \times ^6{P_3} = 12 \times 120 = 1440$.
Since $1440$ is not among the given options,the correct choice is $(d)$.

(B) Let $N_i$ be the number of students who gave exactly $i$ wrong answers.
Since no student gave more than $k$ wrong answers,the total number of wrong answers is $\sum_{i=1}^{k} i \cdot N_i$.
We are given that $a_i$ is the number of students who gave at least $i$ wrong answers.
Thus,$a_i = N_i + N_{i+1} + \dots + N_k$.
This implies $N_i = a_i - a_{i+1}$ for $i < k$ and $N_k = a_k$.
The total number of wrong answers is $1 \cdot N_1 + 2 \cdot N_2 + \dots + k \cdot N_k$.
Substituting the values of $N_i$:
$= 1(a_1 - a_2) + 2(a_2 - a_3) + \dots + (k-1)(a_{k-1} - a_k) + k(a_k)$
$= a_1 + (2-1)a_2 + (3-2)a_3 + \dots + (k-(k-1))a_k$
$= a_1 + a_2 + a_3 + \dots + a_k$.

(D) Given the equation of the line is $ax + by + c = 0$.
We are given the condition $3a + 2b + 4c = 0$.
Dividing the condition by $4$,we get $\frac{3}{4}a + \frac{2}{4}b + c = 0$,which simplifies to $\frac{3}{4}a + \frac{1}{2}b + c = 0$.
Comparing this with the equation $ax + by + c = 0$,we can see that the lines pass through the fixed point $(x, y) = (3/4, 1/2)$ for all values of $a, b, c$.
Thus,the lines are concurrent at the point $(3/4, 1/2)$.

(B) To find the value of $A$,we can use the method of substitution since the given equation holds for all values of $x$.
Let $x = 1$.
Substituting $x = 1$ into the determinant:
$\left| {\begin{array}{*{20}{c}}{{1^2} + 1}&{1 + 1}&{1 - 2}\\ {2(1)^2 + 3(1) - 1}&{3(1)}&{3(1) - 3}\\ {{1^2} + 2(1) + 3}&{2(1) - 1}&{2(1) - 1}\end{array}} \right| = A(1) - 12$
$\left| {\begin{array}{*{20}{c}}2&2&{ - 1}\\ 4&3&0\\ 6&1&1\end{array}} \right| = A - 12$
Now,evaluate the determinant:
$2(3 \times 1 - 0 \times 1) - 2(4 \times 1 - 0 \times 6) + (-1)(4 \times 1 - 3 \times 6) = A - 12$
$2(3) - 2(4) - 1(4 - 18) = A - 12$
$6 - 8 + 14 = A - 12$
$12 = A - 12$
$A = 24$.

(A) Let $\Delta = \left| \begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array} \right|$.
Expanding the determinant,we get:
$\Delta = a(cb - a^2) - b(b^2 - ac) + c(ba - c^2)$
$= abc - a^3 - b^3 + abc + abc - c^3$
$= 3abc - (a^3 + b^3 + c^3)$
$= -(a^3 + b^3 + c^3 - 3abc)$.
Using the algebraic identity $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$,we have:
$\Delta = -(a + b + c) \times \frac{1}{2} [(a - b)^2 + (b - c)^2 + (c - a)^2]$.
Since $a, b, c$ are positive,$(a + b + c) > 0$. Since $a, b, c$ are not all equal,at least one of $(a - b)^2, (b - c)^2, (c - a)^2$ must be positive,so the sum of squares is positive.
Therefore,$\Delta$ is negative.

(D) Given the vector equation: $(x + 3y - 4z)i + (x - 3y + 5z)j + (3x + y)k = \lambda xi + \lambda yj + \lambda zk$.
Comparing the coefficients of $i, j,$ and $k$ on both sides, we get the following system of linear equations:
$(1 - \lambda)x + 3y - 4z = 0$ ... $(i)$
$x - (3 + \lambda)y + 5z = 0$ ... $(ii)$
$3x + y - \lambda z = 0$ ... $(iii)$
Since $(x, y, z) \ne (0, 0, 0)$, the system has a non-trivial solution, which implies the determinant of the coefficient matrix must be zero:
$\begin{vmatrix} 1 - \lambda & 3 & -4 \\ 1 & -(3 + \lambda) & 5 \\ 3 & 1 & -\lambda \end{vmatrix} = 0$.
Expanding the determinant:
$(1 - \lambda) [\lambda(3 + \lambda) - 5] - 3 [-\lambda - 15] - 4 [1 + 3(3 + \lambda)] = 0$
$(1 - \lambda) [\lambda^2 + 3\lambda - 5] + 3\lambda + 45 - 4 [1 + 9 + 3\lambda] = 0$
$(\lambda^2 + 3\lambda - 5 - \lambda^3 - 3\lambda^2 + 5\lambda) + 3\lambda + 45 - 40 - 12\lambda = 0$
$-\lambda^3 - 2\lambda^2 - \lambda = 0$
$-\lambda(\lambda^2 + 2\lambda + 1) = 0$
$-\lambda(\lambda + 1)^2 = 0$.
Wait, re-evaluating the determinant expansion:
$(1 - \lambda)(\lambda^2 + 3\lambda - 5) - 3(-\lambda - 15) - 4(1 + 9 + 3\lambda) = 0$
$(\lambda^2 + 3\lambda - 5 - \lambda^3 - 3\lambda^2 + 5\lambda) + 3\lambda + 45 - 40 - 12\lambda = 0$
$-\lambda^3 - 2\lambda^2 - \lambda = 0 \Rightarrow \lambda(\lambda^2 + 2\lambda + 1) = 0 \Rightarrow \lambda(\lambda + 1)^2 = 0$.
Checking the original equations again:
$x + 3y - 4z = \lambda x \Rightarrow (1-\lambda)x + 3y - 4z = 0$
$x - 3y + 5z = \lambda y \Rightarrow x - (3+\lambda)y + 5z = 0$
$3x + y = \lambda z \Rightarrow 3x + y - \lambda z = 0$
Determinant: $(1-\lambda)(\lambda^2+3\lambda-5) - 3(-\lambda-15) - 4(1+9+3\lambda) = 0$
$-\lambda^3 - 2\lambda^2 - \lambda + 0 = 0 \Rightarrow \lambda = 0, -1$. Thus, the values are $0, -1$.

(D) The scalar triple product is defined as $|(a \times b) \cdot c| = |a| |b| |c| |\sin \theta| |\cos \alpha|$,where $\theta$ is the angle between $a$ and $b$,and $\alpha$ is the angle between $(a \times b)$ and $c$.
Given $|(a \times b) \cdot c| = |a| |b| |c|$,we have $|\sin \theta| |\cos \alpha| = 1$.
Since the maximum value of $|\sin \theta|$ and $|\cos \alpha|$ is $1$,this equality holds only if $|\sin \theta| = 1$ and $|\cos \alpha| = 1$.
$|\sin \theta| = 1 \Rightarrow \theta = \frac{\pi}{2}$,which means $a \perp b$ $(a \cdot b = 0)$.
$|\cos \alpha| = 1 \Rightarrow \alpha = 0$ or $\pi$,which means $c$ is parallel to the vector $(a \times b)$.
Since $(a \times b)$ is perpendicular to both $a$ and $b$,$c$ must also be perpendicular to both $a$ and $b$.
Therefore,$c \perp a$ $(c \cdot a = 0)$ and $c \perp b$ $(c \cdot b = 0)$.
Thus,$a, b, c$ are mutually perpendicular,implying $a \cdot b = b \cdot c = c \cdot a = 0$.

(D) Given $y = f\left( \frac{2x - 1}{x^2 + 1} \right).$ Let $t = \frac{2x - 1}{x^2 + 1}.$
By the chain rule,$\frac{dy}{dx} = f'(t) \cdot \frac{dt}{dx}.$
Since $f'(x) = \sin(x^2),$ we have $f'(t) = \sin(t^2) = \sin \left( \frac{2x - 1}{x^2 + 1} \right)^2.$
Now,calculate $\frac{dt}{dx}$ using the quotient rule:
$\frac{dt}{dx} = \frac{(x^2 + 1)(2) - (2x - 1)(2x)}{(x^2 + 1)^2} = \frac{2x^2 + 2 - 4x^2 + 2x}{(x^2 + 1)^2} = \frac{-2x^2 + 2x + 2}{(x^2 + 1)^2}.$
Therefore,$\frac{dy}{dx} = \frac{-2x^2 + 2x + 2}{(x^2 + 1)^2} \sin \left( \frac{2x - 1}{x^2 + 1} \right)^2.$

(B) Let $I = \int_0^\pi x f(\sin x) dx$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:
$I = \int_0^\pi (\pi - x) f(\sin(\pi - x)) dx$.
Since $\sin(\pi - x) = \sin x$,this becomes:
$I = \int_0^\pi (\pi - x) f(\sin x) dx = \pi \int_0^\pi f(\sin x) dx - \int_0^\pi x f(\sin x) dx$.
$I = \pi \int_0^\pi f(\sin x) dx - I$.
$2I = \pi \int_0^\pi f(\sin x) dx$.
$I = \frac{\pi}{2} \int_0^\pi f(\sin x) dx$.

(A) determinant of order $2$ is given by $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$.
Each element can be either $0$ or $1$,so there are $2^4 = 16$ possible determinants.
The value of the determinant is $ad - bc$.
For the value to be positive,$ad - bc > 0$,which implies $ad > bc$.
Since $a, b, c, d \in \{0, 1\}$,the possible values for $ad$ and $bc$ are $0$ or $1$.
$ad - bc = 1$ occurs when $ad = 1$ and $bc = 0$.
$ad = 1$ implies $a=1$ and $d=1$.
$bc = 0$ implies $(b, c) \in \{(0, 0), (0, 1), (1, 0)\}$.
These correspond to the matrices:
$\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1$,
$\begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} = 1$,
$\begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} = 1$.
There are $3$ such cases.
Thus,the probability is $\frac{3}{16}$.

(C) By the definition of conditional probability,we have:
$P\left( \frac{\overline{A}}{\overline{B}} \right) = \frac{P(\overline{A} \cap \overline{B})}{P(\overline{B})}$
Using De Morgan's Law,we know that $\overline{A} \cap \overline{B} = \overline{A \cup B}$.
Therefore,$P(\overline{A} \cap \overline{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B)$.
Substituting this into the expression,we get:
$P\left( \frac{\overline{A}}{\overline{B}} \right) = \frac{1 - P(A \cup B)}{P(\overline{B})}$