IIT JEE 1980 Mathematics Question Paper with Answer and Solution

(B) Given equation: $\frac{(1 + i)x - 2i}{3 + i} + \frac{(2 - 3i)y + i}{3 - i} = i$
Multiply by the common denominator $(3+i)(3-i) = 9 - i^2 = 10$:
$((1 + i)x - 2i)(3 - i) + ((2 - 3i)y + i)(3 + i) = 10i$
$(3x - ix + 3ix - i^2x - 6i + 2i^2) + (6y + 2iy - 9iy - 3i^2y + 3i + i^2) = 10i$
$(3x - ix + 3ix + x - 6i - 2) + (6y + 2iy - 9iy + 3y + 3i - 1) = 10i$
$(4x + 2ix - 6i - 2) + (9y - 7iy + 3i - 1) = 10i$
$(4x + 9y - 3) + i(2x - 7y - 3) = 10i$
Equating real and imaginary parts:
Real part: $4x + 9y - 3 = 0 \implies 4x + 9y = 3$
Imaginary part: $2x - 7y - 3 = 10 \implies 2x - 7y = 13$
Solving the system:
From $2x - 7y = 13$,$x = \frac{13 + 7y}{2}$.
Substitute into $4x + 9y = 3$: $2(13 + 7y) + 9y = 3 \implies 26 + 14y + 9y = 3 \implies 23y = -23 \implies y = -1$.
Then $x = \frac{13 + 7(-1)}{2} = \frac{6}{2} = 3$.
Thus,$x = 3$ and $y = -1$.

(C) Let the number of sides of the polygon be $n$.
The sum of the interior angles of a polygon with $n$ sides is $(n - 2) \times 180^o$.
Since the angles are in $A.P.$ with first term $a = 120^o$ and common difference $d = 5^o$,the sum of the angles is given by $\frac{n}{2}[2a + (n - 1)d]$.
Equating the two expressions:
$\frac{n}{2}[2(120) + (n - 1)5] = (n - 2)180$
$n[240 + 5n - 5] = 360(n - 2)$
$5n^2 + 235n = 360n - 720$
$5n^2 - 125n + 720 = 0$
Dividing by $5$:
$n^2 - 25n + 144 = 0$
$(n - 9)(n - 16) = 0$
So,$n = 9$ or $n = 16$.
If $n = 16$,the largest angle is $T_{16} = a + 15d = 120^o + 15(5^o) = 120^o + 75^o = 195^o$.
Since an interior angle of a convex polygon must be less than $180^o$,$n = 16$ is rejected.
Therefore,the number of sides is $n = 9$.

(C) The given equation is $(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0$.
Expanding the terms,we get:
$(x^2 - (a+b)x + ab) + (x^2 - (b+c)x + bc) + (x^2 - (c+a)x + ca) = 0$.
Combining like terms,we obtain the quadratic equation:
$3x^2 - 2(a + b + c)x + (ab + bc + ca) = 0$.
For a quadratic equation $Ax^2 + Bx + C = 0$,the discriminant is $D = B^2 - 4AC$.
Here,$A = 3$,$B = -2(a + b + c)$,and $C = (ab + bc + ca)$.
$D = [-2(a + b + c)]^2 - 4(3)(ab + bc + ca)$
$D = 4(a^2 + b^2 + c^2 + 2ab + 2bc + 2ca) - 12(ab + bc + ca)$
$D = 4(a^2 + b^2 + c^2 - ab - bc - ca)$
$D = 2[2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca]$
$D = 2[(a-b)^2 + (b-c)^2 + (c-a)^2]$.
Since the sum of squares of real numbers is always non-negative,$D \ge 0$.
Therefore,the roots are always real.

(C) Given that ${x^2} + px + 1$ is a factor of $a{x^3} + bx + c$,we can write:
$a{x^3} + bx + c = (x^2 + px + 1)(ax + k)$
where $k$ is a constant.
Expanding the right side:
$a{x^3} + bx + c = ax^3 + (ap + k)x^2 + (p k + a)x + k$
Comparing the coefficients of like powers of $x$ on both sides:
Coefficient of $x^2$: $ap + k = 0 \Rightarrow k = -ap$
Coefficient of $x$: $pk + a = b$
Constant term: $k = c$
Substituting $k = c$ into $k = -ap$,we get $c = -ap \Rightarrow p = -c/a$.
Substituting $p = -c/a$ and $k = c$ into $pk + a = b$:
$(-c/a)(c) + a = b$
$-c^2/a + a = b$
$a^2 - c^2 = ab$.

(C) We use the Pascal's identity: $^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r}$.
The given expression is $S = ^{47}C_4 + \sum_{r=1}^5 {}^{52-r}C_3$.
Expanding the summation:
$S = ^{47}C_4 + (^{51}C_3 + ^{50}C_3 + ^{49}C_3 + ^{48}C_3 + ^{47}C_3)$.
Rearranging terms:
$S = (^{47}C_4 + ^{47}C_3) + ^{48}C_3 + ^{49}C_3 + ^{50}C_3 + ^{51}C_3$.
Using $^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r}$:
$^{47}C_4 + ^{47}C_3 = ^{48}C_4$.
Now,$S = (^{48}C_4 + ^{48}C_3) + ^{49}C_3 + ^{50}C_3 + ^{51}C_3$.
Using the identity again:
$^{48}C_4 + ^{48}C_3 = ^{49}C_4$.
Continuing this process:
$S = (^{49}C_4 + ^{49}C_3) + ^{50}C_3 + ^{51}C_3 = ^{50}C_4 + ^{50}C_3 + ^{51}C_3$.
$S = (^{50}C_4 + ^{50}C_3) + ^{51}C_3 = ^{51}C_4 + ^{51}C_3$.
$S = ^{52}C_4$.
Thus,the correct option is $C$.

(A) The total number of words of $5$ letters that can be formed using $10$ different letters (with repetition allowed) is $10^5 = 100000$.
The number of words in which no letter is repeated is given by the permutation formula $^{10}P_5 = 10 \times 9 \times 8 \times 7 \times 6 = 30240$.
The number of words with at least one letter repeated is equal to the total number of words minus the number of words with no letter repeated.
Required number of words $= 100000 - 30240 = 69760$.

(C) Using the Binomial Theorem,we expand $101^{50}$ and $99^{50}$ around $100$:
$101^{50} = (100 + 1)^{50} = 100^{50} + 50 \times 100^{49} + \frac{50 \times 49}{2} \times 100^{48} + \dots$ $(i)$
$99^{50} = (100 - 1)^{50} = 100^{50} - 50 \times 100^{49} + \frac{50 \times 49}{2} \times 100^{48} - \dots$ $(ii)$
Subtracting $(ii)$ from $(i)$:
$101^{50} - 99^{50} = 2 \times (50 \times 100^{49} + \frac{50 \times 49 \times 48}{6} \times 100^{47} + \dots)$
Since the right side is clearly greater than $100^{50}$,we have:
$101^{50} - 99^{50} > 100^{50}$
Therefore,$101^{50} > 100^{50} + 99^{50}$.

(B) Given $A = \sin^2 \theta + \cos^4 \theta$.
Since $\cos^2 \theta \le 1$,we have $\cos^4 \theta \le \cos^2 \theta$.
Therefore,$A = \sin^2 \theta + \cos^4 \theta \le \sin^2 \theta + \cos^2 \theta = 1$.
Thus,$A \le 1$.
Now,express $A$ in terms of $\cos^2 \theta$:
$A = (1 - \cos^2 \theta) + \cos^4 \theta = \cos^4 \theta - \cos^2 \theta + 1$.
Let $x = \cos^2 \theta$,where $0 \le x \le 1$.
Then $A = x^2 - x + 1 = (x - \frac{1}{2})^2 + \frac{3}{4}$.
Since $(x - \frac{1}{2})^2 \ge 0$,the minimum value is $\frac{3}{4}$ when $x = \frac{1}{2}$.
Thus,$\frac{3}{4} \le A \le 1$.

(A) Given $\alpha + \beta - \gamma = \pi ,$ so $\gamma = \alpha + \beta - \pi .$
Consider the expression $E = \sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma .$
Using the identity $\sin^2 A - \sin^2 B = \sin(A - B)\sin(A + B),$
$E = \sin^2 \alpha + \sin(\beta - \gamma)\sin(\beta + \gamma).$
Since $\beta - \gamma = \pi - \alpha,$ we have $\sin(\beta - \gamma) = \sin(\pi - \alpha) = \sin \alpha.$
Also,$\beta + \gamma = \beta + (\alpha + \beta - \pi) = \alpha + 2\beta - \pi.$
Alternatively,using $\gamma = \alpha + \beta - \pi,$
$E = \sin^2 \alpha + \sin^2 \beta - \sin^2(\alpha + \beta - \pi) = \sin^2 \alpha + \sin^2 \beta - \sin^2(\alpha + \beta).$
Using $\sin^2 A - \sin^2 B = \sin(A-B)\sin(A+B),$
$E = \sin^2 \alpha - \sin(\alpha + \beta - \beta)\sin(\alpha + \beta + \beta) = \sin^2 \alpha - \sin \alpha \sin(\alpha + 2\beta).$
$E = \sin \alpha [\sin \alpha - \sin(\alpha + 2\beta)].$
Using $\sin C - \sin D = 2\cos(\frac{C+D}{2})\sin(\frac{C-D}{2}),$
$E = \sin \alpha [2\cos(\alpha + \beta)\sin(-\beta)] = -2\sin \alpha \sin \beta \cos(\alpha + \beta).$
Since $\alpha + \beta = \pi + \gamma,$ $\cos(\alpha + \beta) = \cos(\pi + \gamma) = -\cos \gamma.$
Therefore,$E = -2\sin \alpha \sin \beta (-\cos \gamma) = 2\sin \alpha \sin \beta \cos \gamma.$

(D) Let $AB = h$. Since $C$ is the midpoint of $AB$,$AC = \frac{h}{2}$ and $CB = \frac{h}{2}$.
Given $AP = n \cdot AB = nh$.
In $\triangle PAC$,$\tan(\angle APC) = \frac{AC}{AP} = \frac{h/2}{nh} = \frac{1}{2n}$.
In $\triangle PAB$,$\tan(\angle APB) = \frac{AB}{AP} = \frac{h}{nh} = \frac{1}{n}$.
We know $\alpha = \angle APB - \angle APC$.
Therefore,$\tan \alpha = \tan(\angle APB - \angle APC) = \frac{\tan(\angle APB) - \tan(\angle APC)}{1 + \tan(\angle APB) \cdot \tan(\angle APC)}$.
Substituting the values: $\tan \alpha = \frac{\frac{1}{n} - \frac{1}{2n}}{1 + (\frac{1}{n})(\frac{1}{2n})} = \frac{\frac{1}{2n}}{1 + \frac{1}{2n^2}} = \frac{\frac{1}{2n}}{\frac{2n^2 + 1}{2n^2}} = \frac{1}{2n} \cdot \frac{2n^2}{2n^2 + 1} = \frac{n}{2n^2 + 1}$.
Thus,$n = (2n^2 + 1)\tan \alpha$.

(B) The given line is $5x - y = 1$.
Any line perpendicular to $5x - y = 1$ is of the form $x + 5y = k$.
Writing this in intercept form: $\frac{x}{k} + \frac{y}{k/5} = 1$.
The intercepts on the coordinate axes are $a = k$ and $b = k/5$.
The area of the triangle formed by the line and the coordinate axes is given by $\frac{1}{2} |ab| = 5$.
Substituting the values: $\frac{1}{2} |k \cdot \frac{k}{5}| = 5$.
$|k^2| = 50$,which implies $k = \pm \sqrt{50} = \pm 5\sqrt{2}$.
Therefore,the equation of the line $L$ is $x + 5y = \pm 5\sqrt{2}$.

(D) The equations of the lines are:
$L_1: x + 2y - 3 = 0$
$L_2: 3x + 4y - 7 = 0$
$L_3: 2x + 3y - 4 = 0$
$L_4: 4x + 5y - 6 = 0$
First,check for parallel lines by comparing slopes $(m = -A/B)$:
$m_1 = -1/2, m_2 = -3/4, m_3 = -2/3, m_4 = -4/5$.
Since no slopes are equal,no two lines are parallel. Thus,they cannot form a rectangle.
Next,check for concurrency for any three lines. For $L_1, L_2, L_3$:
$\Delta = \begin{vmatrix} 1 & 2 & -3 \\ 3 & 4 & -7 \\ 2 & 3 & -4 \end{vmatrix} = 1(-16 + 21) - 2(-12 + 14) - 3(9 - 8) = 5 - 4 - 3 = -2 \neq 0$.
Since the determinant is not zero,the lines are not concurrent.
Therefore,the correct option is $D$.

(D) The equation of the circle is $x^2 + y^2 - 2x + 4y + 3 = 0$.
Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1$ and $f = 2$.
The center of the circle is $(-g, -f) = (1, -2)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + 2^2 - 3} = \sqrt{1 + 4 - 3} = \sqrt{2}$.
Since the sides of the square are parallel to the coordinate axes,the vertices of the square are at a distance $r$ from the center along the diagonals.
The vertices are $(1 \pm r \cos(45^\circ), -2 \pm r \sin(45^\circ)) = (1 \pm \sqrt{2} \cdot \frac{1}{\sqrt{2}}, -2 \pm \sqrt{2} \cdot \frac{1}{\sqrt{2}}) = (1 \pm 1, -2 \pm 1)$.
The vertices are $(2, -1), (0, -1), (0, -3), (2, -3)$.
Comparing these with the given options,none of the options match the calculated vertices.
Therefore,the correct option is $(d)$.

(C) The radius $r$ of the circle is the distance between the centre $(1, 2)$ and the point $(4, 6)$ on the circle.
Using the distance formula,$r = \sqrt{(4 - 1)^2 + (6 - 2)^2}$.
$r = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
The area of the circle is given by the formula $A = \pi r^2$.
Substituting $r = 5$,we get $A = \pi (5)^2 = 25\pi \text{ sq. units}$.

(B) Given limit: $L = \mathop {\lim }\limits_{x \to 0} \frac{{x({2^x} - 1)}}{{1 - \cos x}}$
Using the standard limits $\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} - 1}}{x} = \log 2$ and $1 - \cos x = 2\sin^2(\frac{x}{2})$:
$L = \mathop {\lim }\limits_{x \to 0} \frac{{x({2^x} - 1)}}{{2\sin^2(\frac{x}{2})}}$
$L = \mathop {\lim }\limits_{x \to 0} \left( \frac{{{2^x} - 1}}{x} \right) \cdot \left( \frac{x^2}{2\sin^2(\frac{x}{2})} \right)$
$L = (\log 2) \cdot \mathop {\lim }\limits_{x \to 0} \frac{1}{2 \left( \frac{\sin(x/2)}{x} \right)^2}$
Since $\mathop {\lim }\limits_{x \to 0} \frac{\sin(x/2)}{x/2} = 1$,we have $\mathop {\lim }\limits_{x \to 0} \frac{\sin(x/2)}{x} = \frac{1}{2}$:
$L = (\log 2) \cdot \frac{1}{2 \cdot (1/2)^2} = (\log 2) \cdot \frac{1}{2 \cdot (1/4)} = (\log 2) \cdot 2 = 2\log 2 = \log 4$.

(A) Given: $P(A) = 0.25$,$P(B) = 0.50$,and $P(A \cap B) = 0.14$.
We need to find the probability that neither $A$ nor $B$ occurs,which is $P(A^c \cap B^c)$.
By De Morgan's Law,$P(A^c \cap B^c) = P((A \cup B)^c) = 1 - P(A \cup B)$.
Using the Addition Theorem of Probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = 0.25 + 0.50 - 0.14 = 0.61$.
Therefore,$P(A^c \cap B^c) = 1 - 0.61 = 0.39$.

(B) Given that $P(A) = 0.5$ and $P(B) = 0.3$.
Since $A$ and $B$ are mutually exclusive events,$P(A \cap B) = 0$.
The probability of happening neither $A$ nor $B$ is $P(\overline{A} \cap \overline{B})$.
By De Morgan's Law,$P(\overline{A} \cap \overline{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B)$.
For mutually exclusive events,$P(A \cup B) = P(A) + P(B) = 0.5 + 0.3 = 0.8$.
Therefore,the required probability is $1 - 0.8 = 0.2$.

(C) Given $A = \sin^2 x + \cos^4 x$.
Let $t = \sin^2 x$,then $0 \le t \le 1$. Since $\cos^2 x = 1 - \sin^2 x = 1 - t$,we have $\cos^4 x = (1 - t)^2$.
Substituting these into the expression for $A$:
$A = t + (1 - t)^2 = t + 1 - 2t + t^2 = t^2 - t + 1$.
This is a quadratic expression in $t$ where $t \in [0, 1]$.
The vertex of the parabola $f(t) = t^2 - t + 1$ is at $t = -(-1)/(2 \times 1) = 1/2$.
Since $1/2$ is within the interval $[0, 1]$,the minimum value is $f(1/2) = (1/2)^2 - (1/2) + 1 = 1/4 - 1/2 + 1 = 3/4$.
The maximum value occurs at the boundaries $t=0$ or $t=1$:
$f(0) = 0^2 - 0 + 1 = 1$.
$f(1) = 1^2 - 1 + 1 = 1$.
Thus,the range of $A$ is $\frac{3}{4} \le A \le 1$.

(B) The equation of a circle passing through the intersection of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 + \lambda S_2 = 0$.
Here,$S_1 = x^2 + y^2 - 6x + 8 = 0$ and $S_2 = x^2 + y^2 - 6 = 0$.
The family of circles is $(x^2 + y^2 - 6x + 8) + \lambda(x^2 + y^2 - 6) = 0$.
Since the circle passes through the point $(1, 1)$,we substitute $x = 1$ and $y = 1$ into the equation:
$(1^2 + 1^2 - 6(1) + 8) + \lambda(1^2 + 1^2 - 6) = 0$
$(1 + 1 - 6 + 8) + \lambda(1 + 1 - 6) = 0$
$4 + \lambda(-4) = 0$
$4 - 4\lambda = 0 \implies \lambda = 1$.
Substituting $\lambda = 1$ back into the family equation:
$(x^2 + y^2 - 6x + 8) + 1(x^2 + y^2 - 6) = 0$
$2x^2 + 2y^2 - 6x + 2 = 0$
Dividing by $2$,we get $x^2 + y^2 - 3x + 1 = 0$.

(C) Let $\Delta = \left| {\begin{array}{*{20}{c}}{{b^2} + {c^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{c^2} + {a^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{a^2} + {b^2}}\end{array}} \right|$.
Apply the operation ${R_1} \to {R_1} - ({R_2} + {R_3})$:
$\Delta = \left| {\begin{array}{*{20}{c}}{-2{c^2}}&{-2{c^2}}&{0}\\{{b^2}}&{{c^2} + {a^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{a^2} + {b^2}}\end{array}} \right|$.
Alternatively,taking common factors from columns or rows:
$\Delta = 2{a^2}{b^2}{c^2} \left| {\begin{array}{*{20}{c}}{\frac{{{b^2} + {c^2}}}{{{a^2}}}}&1&1\\1&{\frac{{{c^2} + {a^2}}}{{{b^2}}}}&1\\1&1&{\frac{{{a^2} + {b^2}}}{{{c^2}}}}\end{array}} \right|$.
By simplifying the determinant,we get:
$\Delta = 4{a^2}{b^2}{c^2}$.
Trick: Put $a=1, b=1, c=1$ is not ideal as options might overlap. Let $a=1, b=2, c=3$:
$\Delta = \left| {\begin{array}{*{20}{c}}{13}&1&1\\4&10&4\\9&9&5\end{array}} \right| = 13(50-36) - 1(20-36) + 1(36-90) = 13(14) + 16 - 54 = 182 + 16 - 54 = 144$.
Checking options: $4{a^2}{b^2}{c^2} = 4(1)^2(2)^2(3)^2 = 4 \times 1 \times 4 \times 9 = 144$. Hence,option $C$ is correct.

(C) We know that $1 = \sin^2 \frac{x}{4} + \cos^2 \frac{x}{4}$ and $\sin \frac{x}{2} = 2 \sin \frac{x}{4} \cos \frac{x}{4}$.
Substituting these into the integral,we get:
$\int \sqrt{\sin^2 \frac{x}{4} + \cos^2 \frac{x}{4} + 2 \sin \frac{x}{4} \cos \frac{x}{4}} \, dx$
$= \int \sqrt{(\sin \frac{x}{4} + \cos \frac{x}{4})^2} \, dx$
$= \int (\sin \frac{x}{4} + \cos \frac{x}{4}) \, dx$
$= -4 \cos \frac{x}{4} + 4 \sin \frac{x}{4} + c$
$= 4(\sin \frac{x}{4} - \cos \frac{x}{4}) + c$.

(B) Given,the probability of event $A$ occurring in one trial is $P(A) = 0.4$.
Therefore,the probability of event $A$ not occurring in one trial is $P(\bar{A}) = 1 - 0.4 = 0.6$.
For $n = 3$ independent trials,the probability that event $A$ does not happen at all is $P(\text{none}) = (P(\bar{A}))^3 = (0.6)^3 = 0.216$.
The probability that event $A$ happens at least once is given by $P(\text{at least once}) = 1 - P(\text{none})$.
$P(\text{at least once}) = 1 - 0.216 = 0.784$.

(A) For $y$ to be a real value,the expression under the square root must be non-negative:
$\frac{(x + 1)(x - 3)}{(x - 2)} \ge 0$
We use the sign scheme (Wavy Curve Method) for the expression $f(x) = \frac{(x + 1)(x - 3)}{(x - 2)}$.
The critical points are $x = -1, 2, 3$.
Testing the intervals:
$1$. For $x \in (-1, 2)$,let $x = 0$: $\frac{(1)(-3)}{(-2)} = \frac{3}{2} > 0$ (Valid).
$2$. For $x \in [3, \infty)$,let $x = 4$: $\frac{(5)(1)}{(2)} = \frac{5}{2} > 0$ (Valid).
$3$. At $x = -1$,$y = 0$ (Valid).
$4$. At $x = 3$,$y = 0$ (Valid).
$5$. At $x = 2$,the expression is undefined.
Combining these,the solution is $x \in [-1, 2) \cup [3, \infty)$.

(A) Given $y = 5x(1 - x)^{-2/3} + \cos^2(2x + 1)$.
Applying the product rule and chain rule:
$\frac{dy}{dx} = 5 \left[ (1 - x)^{-2/3} + x \cdot \left( -\frac{2}{3} \right)(1 - x)^{-5/3}(-1) \right] + 2\cos(2x + 1)(-\sin(2x + 1))(2)$.
$\frac{dy}{dx} = 5(1 - x)^{-2/3} + \frac{10x}{3(1 - x)^{5/3}} - 4\cos(2x + 1)\sin(2x + 1)$.
Using $2\sin\theta\cos\theta = \sin(2\theta)$,we have $4\cos(2x + 1)\sin(2x + 1) = 2\sin(4x + 2)$.
$\frac{dy}{dx} = \frac{5(1 - x) + \frac{10x}{3}}{(1 - x)^{5/3}} - 2\sin(4x + 2) = \frac{5 - 5x + \frac{10x}{3}}{(1 - x)^{5/3}} - 2\sin(4x + 2)$.
$\frac{dy}{dx} = \frac{5(3 - 3x + 2x)}{3(1 - x)^{5/3}} - 2\sin(4x + 2) = \frac{5(3 - x)}{3(1 - x)^{5/3}} - 2\sin(4x + 2)$.