IIT JEE 1981 Mathematics Question Paper with Answer and Solution

(B) In a parallelogram $ABCD$,the diagonals $AC$ and $BD$ bisect each other at the same midpoint.
The midpoint of diagonal $AC$ is given by $\frac{z_1 + z_3}{2}$.
The midpoint of diagonal $BD$ is given by $\frac{z_2 + z_4}{2}$.
Since the diagonals bisect each other,we have $\frac{z_1 + z_3}{2} = \frac{z_2 + z_4}{2}$.
Therefore,$z_1 + z_3 = z_2 + z_4$.

(C) Let $r$ be the circumradius of the equilateral triangle and $\omega$ be the cube root of unity. Let $ABC$ be the equilateral triangle with $z_1, z_2$ and $z_3$ as its vertices $A, B$ and $C$ respectively,with circumcentre $O'(z_0)$.
The vectors $O'A, O'B, O'C$ are equal in magnitude and separated by an angle of $\frac{2\pi}{3}$.
Then,we can write:
$z_1 - z_0 = r e^{i\theta}$
$z_2 - z_0 = r e^{i(\theta + \frac{2\pi}{3})} = r \omega e^{i\theta}$
$z_3 - z_0 = r e^{i(\theta + \frac{4\pi}{3})} = r \omega^2 e^{i\theta}$
Thus,$z_1 = z_0 + r e^{i\theta}$,$z_2 = z_0 + r \omega e^{i\theta}$,and $z_3 = z_0 + r \omega^2 e^{i\theta}$.
Squaring and adding these gives:
$z_1^2 + z_2^2 + z_3^2 = (z_0 + r e^{i\theta})^2 + (z_0 + r \omega e^{i\theta})^2 + (z_0 + r \omega^2 e^{i\theta})^2$
$= 3z_0^2 + 2 z_0 r e^{i\theta} (1 + \omega + \omega^2) + r^2 e^{i2\theta} (1 + \omega^2 + \omega^4)$
Since $1 + \omega + \omega^2 = 0$ and $1 + \omega^2 + \omega^4 = 1 + \omega^2 + \omega = 0$,we get:
$z_1^2 + z_2^2 + z_3^2 = 3z_0^2$.

(A) Since the coefficients $p$ and $q$ are real,the complex roots must occur in conjugate pairs.
Given one root is $x_1 = 2 + i\sqrt{3}$,the other root must be $x_2 = 2 - i\sqrt{3}$.
Using the properties of quadratic equations,the sum of the roots is $x_1 + x_2 = (2 + i\sqrt{3}) + (2 - i\sqrt{3}) = 4$.
From the equation $x^2 + px + q = 0$,the sum of the roots is $-p$. Therefore,$-p = 4$,which implies $p = -4$.
The product of the roots is $x_1 \times x_2 = (2 + i\sqrt{3})(2 - i\sqrt{3}) = 2^2 - (i\sqrt{3})^2 = 4 - (-3) = 7$.
From the equation $x^2 + px + q = 0$,the product of the roots is $q$. Therefore,$q = 7$.
Thus,$(p, q) = (-4, 7)$.

(C) Let the boxes be marked as $A, B, C$. We have to ensure that no box remains empty and all five balls are placed.
There are two possible distributions of $5$ balls into $3$ boxes such that no box is empty:
$(i)$ Two boxes contain $1$ ball each and the $3^{rd}$ box contains $3$ balls.
The number of ways to choose the balls is $^5C_1 \times ^4C_1 \times ^3C_3 = 5 \times 4 \times 1 = 20$.
Since the box containing $3$ balls can be any of the $3$ boxes,the total ways for this case is $20 \times 3 = 60$.
$(ii)$ Two boxes contain $2$ balls each and the $3^{rd}$ box contains $1$ ball.
The number of ways to choose the balls is $^5C_2 \times ^3C_2 \times ^1C_1 = 10 \times 3 \times 1 = 30$.
Since the box containing $1$ ball can be any of the $3$ boxes,the total ways for this case is $30 \times 3 = 90$.
Therefore,the total number of ways is $60 + 90 = 150$.

(B) Given $\sin \theta + \cos \theta = 1$.
Dividing both sides by $\sqrt{1^2 + 1^2} = \sqrt{2}$,we get:
$\frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta = \frac{1}{\sqrt{2}}$
$\sin \theta \cos \frac{\pi}{4} + \cos \theta \sin \frac{\pi}{4} = \sin \frac{\pi}{4}$
$\sin \left( \theta + \frac{\pi}{4} \right) = \sin \frac{\pi}{4}$
The general solution for $\sin x = \sin \alpha$ is $x = n\pi + (-1)^n \alpha$.
Therefore,$\theta + \frac{\pi}{4} = n\pi + (-1)^n \frac{\pi}{4}$
$\theta = n\pi + (-1)^n \frac{\pi}{4} - \frac{\pi}{4}$.

(D) Let the angles of the triangle be $A-d, A, A+d$. Since the sum of angles in a triangle is $180^o$,we have $(A-d) + A + (A+d) = 180^o$,which implies $3A = 180^o$,so $A = 60^o$.
Given $b:c = \sin B : \sin C = \sqrt{3} : \sqrt{2}$.
Since $B = A = 60^o$ is not necessarily true (the angles are $A-d, A, A+d$),let the angles be $B-d, B, B+d$. Then $B = 60^o$.
Using the sine rule,$\frac{\sin B}{\sin C} = \frac{b}{c} = \frac{\sqrt{3}}{\sqrt{2}}$.
Substituting $B = 60^o$,we get $\frac{\sin 60^o}{\sin C} = \frac{\sqrt{3}/2}{\sin C} = \frac{\sqrt{3}}{\sqrt{2}}$.
This simplifies to $\sin C = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$,so $C = 45^o$.
Since $A+B+C = 180^o$,we have $A + 60^o + 45^o = 180^o$,which gives $A = 75^o$.

(B) Let $ABCD$ be a rectangle with opposite vertices $A(1, 3)$ and $C(5, 1)$.
Since the diagonals of a rectangle bisect each other,the intersection point of the diagonals is the midpoint of $AC$.
Midpoint of $AC = \left(\frac{1+5}{2}, \frac{3+1}{2}\right) = (3, 2)$.
The other two vertices $B$ and $D$ lie on the line $y = 2x + c$. Since the diagonal $BD$ passes through the intersection point of the diagonals,the point $(3, 2)$ must lie on the line $y = 2x + c$.
Substituting $x = 3$ and $y = 2$ into the equation:
$2 = 2(3) + c$
$2 = 6 + c$
$c = 2 - 6 = -4$.
Thus,the value of $c$ is $-4$.

(C) The equation $|x| + |y| = 1$ represents a square with vertices at $(1, 0)$,$(0, 1)$,$(-1, 0)$,and $(0, -1)$.
The distance between consecutive vertices (e.g.,$(1, 0)$ and $(0, 1)$) is the side length $s = \sqrt{(1-0)^2 + (0-1)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.
The area of a square is given by $s^2$.
Therefore,the area $= (\sqrt{2})^2 = 2$ square units.
Alternatively,the area of the square formed by $|x| + |y| = a$ is $2a^2$. Here $a = 1$,so the area is $2(1)^2 = 2$.

(C) Let the point be $P(4, 3)$ and the circle be $x^2 + y^2 = 9$. The radius $r = 3$.
The equation of the chord of contact $AB$ is $T = 0$,which is $4x + 3y = 9$.
The distance from the origin $O(0, 0)$ to the chord $AB$ is $OQ = \frac{|4(0) + 3(0) - 9|}{\sqrt{4^2 + 3^2}} = \frac{9}{5}$.
In $\triangle OAQ$,$AQ = \sqrt{OA^2 - OQ^2} = \sqrt{3^2 - (\frac{9}{5})^2} = \sqrt{9 - \frac{81}{25}} = \sqrt{\frac{225 - 81}{25}} = \sqrt{\frac{144}{25}} = \frac{12}{5}$.
The length of the chord of contact $AB = 2 \times AQ = 2 \times \frac{12}{5} = \frac{24}{5}$.
The distance $PQ$ is the distance from $P(4, 3)$ to the line $4x + 3y - 9 = 0$,which is $PQ = \frac{|4(4) + 3(3) - 9|}{\sqrt{4^2 + 3^2}} = \frac{|16 + 9 - 9|}{5} = \frac{16}{5}$.
The area of $\triangle PAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times PQ = \frac{1}{2} \times \frac{24}{5} \times \frac{16}{5} = \frac{192}{25}$.

(A) The area of the quadrilateral $PQCR$ is the sum of the areas of two congruent right-angled triangles,$\Delta PQC$ and $\Delta PRC$.
Area $(PQCR) = 2 \times \text{Area}(\Delta PQC) = 2 \times (\frac{1}{2} \times PQ \times QC) = PQ \times r$.
Here,$r$ is the radius of the circle and $PQ$ is the length of the tangent from point $P$.
The equation of the circle is $x^2 + y^2 - 2x - 4y - 20 = 0$.
The centre $C$ is $(1, 2)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + (-2)^2 - (-20)} = \sqrt{1 + 4 + 20} = \sqrt{25} = 5$.
The length of the tangent $L = PQ = \sqrt{S_1}$,where $S_1$ is the value of the circle equation at point $P(16, 7)$.
$PQ = \sqrt{16^2 + 7^2 - 2(16) - 4(7) - 20} = \sqrt{256 + 49 - 32 - 28 - 20} = \sqrt{225} = 15$.
Therefore,the area of quadrilateral $PQCR = PQ \times r = 15 \times 5 = 75 \text{ sq. units}$.

(D) Given the equation: $\frac{x^2}{1 - r} - \frac{y^2}{1 + r} = 1$ where $r > 1$.
Since $r > 1$,let $r - 1 = p$,where $p > 0$. Then $1 - r = -p$.
Substituting this into the equation,we get: $\frac{x^2}{-p} - \frac{y^2}{1 + r} = 1$.
Multiplying by $-1$,we get: $\frac{x^2}{p} + \frac{y^2}{1 + r} = -1$.
Since the sum of two squares divided by positive constants equals a negative value $(-1)$,there are no real values of $(x, y)$ that satisfy this equation.
Therefore,the equation represents an imaginary ellipse.

(B) Given the identity $p{\lambda ^4} + q{\lambda ^3} + r{\lambda ^2} + s\lambda + t = \left| {\begin{array}{*{20}{c}}{{\lambda ^2} + 3\lambda }&{\lambda - 1}&{\lambda + 3}\\{\lambda + 1}&{2 - \lambda }&{\lambda - 4}\\{\lambda - 3}&{\lambda + 4}&{3\lambda }\end{array}} \right|$.
Since this is an identity in $\lambda$,it holds true for all values of $\lambda$.
To find the value of $t$,we substitute $\lambda = 0$ into the equation:
$t = \left| {\begin{array}{*{20}{c}}0&{ - 1}&3\\1&2&{ - 4}\\{ - 3}&4&0\end{array}} \right|$.
Expanding the determinant along the first row:
$t = 0(0 - (-16)) - (-1)(0 - 12) + 3(4 - (-6))$
$t = 0 + 1(-12) + 3(10)$
$t = -12 + 30 = 18$.
Thus,the value of $t$ is $18$.

(B) We know that $\cos^{-1}x + \sin^{-1}x = \frac{\pi}{2}$ for $x \in [-1, 1]$.
Given expression: $\cos \left[ \cos^{-1}\frac{1}{5} + \cos^{-1}\frac{1}{5} + \sin^{-1}\frac{1}{5} \right]$.
Using the property,this simplifies to: $\cos \left[ \frac{\pi}{2} + \cos^{-1}\frac{1}{5} \right]$.
Since $\cos(\frac{\pi}{2} + \theta) = -\sin\theta$,we have: $-\sin \left( \cos^{-1}\frac{1}{5} \right)$.
Let $\cos^{-1}\frac{1}{5} = \alpha$,then $\cos\alpha = \frac{1}{5}$.
Then $\sin\alpha = \sqrt{1 - \cos^2\alpha} = \sqrt{1 - (\frac{1}{5})^2} = \sqrt{1 - \frac{1}{25}} = \sqrt{\frac{24}{25}} = \frac{2\sqrt{6}}{5}$.
Thus,the expression equals $-\frac{2\sqrt{6}}{5}$.

(D) We are given the expression $a \cdot [(b + c) \times (a + b + c)]$.
Using the distributive property of the cross product,we expand the term inside the square brackets:
$(b + c) \times (a + b + c) = (b \times a) + (b \times b) + (b \times c) + (c \times a) + (c \times b) + (c \times c)$.
Since the cross product of any vector with itself is zero ($b \times b = 0$ and $c \times c = 0$),the expression simplifies to:
$(b \times a) + (b \times c) + (c \times a) + (c \times b)$.
Now,taking the dot product with $a$:
$a \cdot [(b \times a) + (b \times c) + (c \times a) + (c \times b)] = a \cdot (b \times a) + a \cdot (b \times c) + a \cdot (c \times a) + a \cdot (c \times b)$.
This can be written in terms of scalar triple products:
$[a \, b \, a] + [a \, b \, c] + [a \, c \, a] + [a \, c \, b]$.
In a scalar triple product,if any two vectors are identical,the value is $0$. Thus,$[a \, b \, a] = 0$ and $[a \, c \, a] = 0$.
Also,$[a \, c \, b] = -[a \, b \, c]$.
Substituting these values:
$0 + [a \, b \, c] + 0 - [a \, b \, c] = 0$.

(A) For $f(x)$ to be continuous at $x = 2$,we must have $\lim_{x \to 2} f(x) = f(2) = k$.
First,we evaluate the limit: $\lim_{x \to 2} \frac{x^3 + x^2 - 16x + 20}{(x - 2)^2}$.
Since substituting $x = 2$ gives the indeterminate form $\frac{0}{0}$,we factor the numerator.
By synthetic division or polynomial division,$x^3 + x^2 - 16x + 20 = (x - 2)^2(x + 5)$.
Thus,$\lim_{x \to 2} \frac{(x - 2)^2(x + 5)}{(x - 2)^2} = \lim_{x \to 2} (x + 5) = 2 + 5 = 7$.
Therefore,$k = 7$.

(C) Given the functional equation $f(x + y) = f(x)f(y)$.
Setting $y = 0$,we get $f(x + 0) = f(x)f(0)$,which implies $f(x) = f(x)f(0)$. Since $f(5) = 2$,$f(x)$ is not identically zero,so $f(0) = 1$.
By the definition of the derivative,$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$.
Substituting $f(x + h) = f(x)f(h)$,we get $f'(x) = \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h} = f(x) \lim_{h \to 0} \frac{f(h) - 1}{h}$.
Since $f(0) = 1$,this is $f'(x) = f(x) \lim_{h \to 0} \frac{f(h) - f(0)}{h} = f(x)f'(0)$.
Given $f(5) = 2$ and $f'(0) = 3$,we have $f'(5) = f(5)f'(0) = 2 \times 3 = 6$.

(D) We are given the integral $I = \int_0^1 (1 + e^{-x^2}) \,dx$.
By the linearity property of definite integrals,we can split this into two parts:
$I = \int_0^1 1 \,dx + \int_0^1 e^{-x^2} \,dx$.
The first part is $\int_0^1 1 \,dx = [x]_0^1 = 1 - 0 = 1$.
The second part is $\int_0^1 e^{-x^2} \,dx$.
The function $e^{-x^2}$ does not have an elementary antiderivative. Therefore,the integral $\int_0^1 e^{-x^2} \,dx$ cannot be expressed in terms of simple algebraic or transcendental functions.
Thus,the value of the integral is $1 + \int_0^1 e^{-x^2} \,dx$.
Comparing this with the given options,none of the options $A, B,$ or $C$ match this result.
Therefore,the correct option is $D$.

(B) To find the area bounded by the curve ${x^2} = 4y$ and the line $x = 4y - 2$,we first find their points of intersection.
From the line equation,$4y = x + 2$. Substituting this into the parabola equation ${x^2} = 4y$,we get:
${x^2} = x + 2$
${x^2} - x - 2 = 0$
$(x - 2)(x + 1) = 0$
So,$x = 2$ and $x = -1$.
When $x = 2$,$4y = 4 \implies y = 1$. Point $A(2, 1)$.
When $x = -1$,$4y = 1 \implies y = \frac{1}{4}$. Point $B(-1, \frac{1}{4})$.
The required area is the integral of the upper curve minus the lower curve from $x = -1$ to $x = 2$:
$\text{Area} = \int_{-1}^{2} \left( \frac{x+2}{4} - \frac{x^2}{4} \right) dx$
$= \frac{1}{4} \int_{-1}^{2} (x + 2 - x^2) dx$
$= \frac{1}{4} \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^{2}$
$= \frac{1}{4} \left[ \left( \frac{4}{2} + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right) \right]$
$= \frac{1}{4} \left[ \left( 6 - \frac{8}{3} \right) - \left( \frac{3 - 12 + 2}{6} \right) \right]$
$= \frac{1}{4} \left[ \frac{10}{3} - \left( -\frac{7}{6} \right) \right]$
$= \frac{1}{4} \left[ \frac{20 + 7}{6} \right] = \frac{1}{4} \times \frac{27}{6} = \frac{27}{24} = \frac{9}{8} \, \text{sq. unit}$.

(B) Let $f(x) = (ax^2 + bx + c)(1 + \cos^8 x)$.
Given $\int_0^1 f(x) \, dx = \int_0^2 f(x) \, dx$.
This implies $\int_0^2 f(x) \, dx - \int_0^1 f(x) \, dx = 0$,which simplifies to $\int_1^2 f(x) \, dx = 0$.
Since $f(x)$ is a continuous function and the integral over the interval $(1, 2)$ is zero,$f(x)$ must change sign in the interval $(1, 2)$ unless $f(x) = 0$ for all $x \in (1, 2)$.
Since $a, b, c$ are non-zero,$ax^2 + bx + c$ is not identically zero. Also,$(1 + \cos^8 x) \ge 1 > 0$ for all $x$.
Thus,$f(x)$ must take both positive and negative values in $(1, 2)$.
By the Intermediate Value Theorem,there exists at least one $x_0 \in (1, 2)$ such that $f(x_0) = 0$.
Since $(1 + \cos^8 x) \neq 0$,we must have $ax_0^2 + bx_0 + c = 0$.
Therefore,the quadratic equation $ax^2 + bx + c = 0$ has at least one root in $(1, 2)$,which is contained in $(0, 2)$.

(A) Let $E_1$ be the event that face $1$ turns up and $E_2$ be the event that face $2$ turns up.
Given probabilities are $P(E_1) = 0.1$ and $P(E_2) = 0.32$.
We are given that either face $1$ or $2$ has turned up. Let this event be $A = E_1 \cup E_2$.
Since $E_1$ and $E_2$ are mutually exclusive,$P(A) = P(E_1) + P(E_2) = 0.1 + 0.32 = 0.42$.
We need to find the conditional probability $P(E_1 | A)$,which is the probability that face $1$ has turned up given that either $1$ or $2$ has turned up.
$P(E_1 | A) = \frac{P(E_1 \cap A)}{P(A)} = \frac{P(E_1)}{P(A)} = \frac{0.1}{0.42}$.
Simplifying the fraction: $\frac{0.1}{0.42} = \frac{10}{42} = \frac{5}{21}$.

(A) Let $s^2 = \frac{a+b+c}{abc}$. Then the expression becomes $\theta = \tan^{-1}\sqrt{a^2 s^2} + \tan^{-1}\sqrt{b^2 s^2} + \tan^{-1}\sqrt{c^2 s^2}$.
Since $a, b, c$ are positive,$\theta = \tan^{-1}(as) + \tan^{-1}(bs) + \tan^{-1}(cs)$.
Using the formula $\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \tan^{-1}\left(\frac{x+y+z-xyz}{1-(xy+yz+zx)}\right)$:
$\tan \theta = \frac{as+bs+cs - (as)(bs)(cs)}{1 - (asbs + bscs + csas)} = \frac{s(a+b+c) - s^3(abc)}{1 - s^2(ab+bc+ca)}$.
Substitute $s^2 = \frac{a+b+c}{abc}$,which implies $s^2(abc) = a+b+c$:
$\tan \theta = \frac{s(a+b+c) - s(s^2 abc)}{1 - s^2(ab+bc+ca)} = \frac{s(a+b+c) - s(a+b+c)}{1 - s^2(ab+bc+ca)} = \frac{0}{1 - s^2(ab+bc+ca)} = 0$.
Alternatively,let $a=b=c=1$. Then $\theta = \tan^{-1}\sqrt{3} + \tan^{-1}\sqrt{3} + \tan^{-1}\sqrt{3} = 3 \times 60^\circ = 180^\circ = \pi$. Thus,$\tan \theta = \tan \pi = 0$.

(C) Given that $|a| = 3, |b| = 4, |c| = 5$.
Since $a$ and $(b + c)$ are perpendicular,$a \cdot (b + c) = 0 \implies a \cdot b + a \cdot c = 0$ .....$(i)$
Since $b$ and $(c + a)$ are perpendicular,$b \cdot (c + a) = 0 \implies b \cdot c + b \cdot a = 0$ .....$(ii)$
Since $c$ and $(a + b)$ are perpendicular,$c \cdot (a + b) = 0 \implies c \cdot a + c \cdot b = 0$ .....$(iii)$
Adding equations $(i), (ii),$ and $(iii)$,we get:
$2(a \cdot b + b \cdot c + c \cdot a) = 0 \implies a \cdot b + b \cdot c + c \cdot a = 0$.
Now,the modulus of $a + b + c$ is given by:
$|a + b + c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a)$
$|a + b + c|^2 = 3^2 + 4^2 + 5^2 + 2(0)$
$|a + b + c|^2 = 9 + 16 + 25 = 50$
$|a + b + c| = \sqrt{50} = 5\sqrt{2}$.

(B) Given $y = a \log |x| + b x^2 + x$.
Since $\frac{d}{dx} \log |x| = \frac{1}{x}$ for all $x \neq 0$,we differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$.
Since the function has extremum values at $x = -1$ and $x = 2$,the derivative must be zero at these points.
For $x = -1$: $\frac{a}{-1} + 2b(-1) + 1 = 0 \implies -a - 2b + 1 = 0 \implies a + 2b = 1$ (Equation $1$).
For $x = 2$: $\frac{a}{2} + 2b(2) + 1 = 0 \implies \frac{a}{2} + 4b + 1 = 0 \implies a + 8b = -2$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(a + 8b) - (a + 2b) = -2 - 1 \implies 6b = -3 \implies b = -\frac{1}{2}$.
Substituting $b = -\frac{1}{2}$ into Equation $1$: $a + 2(-\frac{1}{2}) = 1 \implies a - 1 = 1 \implies a = 2$.
Thus,$a = 2$ and $b = -\frac{1}{2}$.