IIT JEE 1999 Mathematics Question Paper with Answer and Solution

(C) Let $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ be the complex cube root of unity,where $\omega^3 = 1$.
Given expression is $4 + 5\omega^{334} + 3\omega^{365}$.
Since $\omega^3 = 1$,we have $\omega^{334} = \omega^{333} \cdot \omega = (\omega^3)^{111} \cdot \omega = 1^{111} \cdot \omega = \omega$.
Similarly,$\omega^{365} = \omega^{363} \cdot \omega^2 = (\omega^3)^{121} \cdot \omega^2 = 1^{121} \cdot \omega^2 = \omega^2$.
Substituting these values into the expression:
$4 + 5\omega + 3\omega^2$.
Using the property $1 + \omega + \omega^2 = 0$,we have $\omega^2 = -1 - \omega$.
Substituting this:
$4 + 5\omega + 3(-1 - \omega) = 4 + 5\omega - 3 - 3\omega = 1 + 2\omega$.
Substituting $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$:
$1 + 2\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 1 - 1 + i\sqrt{3} = i\sqrt{3}$.

(D) Let $x_n$ be the side length of square $S_n$. Given that the side of $S_n$ equals the diagonal of $S_{n+1}$,we have $x_n = x_{n+1} \sqrt{2}$.
This implies $x_{n+1} = \frac{x_n}{\sqrt{2}}$.
Thus,$x_n = x_1 \left(\frac{1}{\sqrt{2}}\right)^{n-1}$.
The area of $S_n$ is $A_n = x_n^2 = x_1^2 \left(\frac{1}{2}\right)^{n-1} = \frac{100}{2^{n-1}}$.
We want $A_n < 1$,so $\frac{100}{2^{n-1}} < 1$,which means $2^{n-1} > 100$.
Since $2^6 = 64$ and $2^7 = 128$,we require $n-1 \ge 7$,which implies $n \ge 8$.
Therefore,for $n = 8, 9, 10, \dots$,the area is less than $1 \ cm^2$.

(B) Given equation is $(5 + \sqrt{2})x^2 - (4 + \sqrt{5})x + 8 + 2\sqrt{5} = 0$.
Let the roots be $x_1$ and $x_2$.
Sum of roots: $x_1 + x_2 = \frac{4 + \sqrt{5}}{5 + \sqrt{2}}$.
Product of roots: $x_1x_2 = \frac{8 + 2\sqrt{5}}{5 + \sqrt{2}} = \frac{2(4 + \sqrt{5})}{5 + \sqrt{2}} = 2(x_1 + x_2)$.
Harmonic mean $H = \frac{2x_1x_2}{x_1 + x_2}$.
Substituting $x_1x_2 = 2(x_1 + x_2)$ into the formula:
$H = \frac{2 \times 2(x_1 + x_2)}{x_1 + x_2} = 4$.

(A) Given equation is $x^2 - 2ax + a^2 + a - 3 = 0$.
For real roots,the discriminant $D \ge 0$:
$D = (-2a)^2 - 4(1)(a^2 + a - 3) \ge 0$
$4a^2 - 4a^2 - 4a + 12 \ge 0$
$-4a + 12 \ge 0 \Rightarrow a \le 3$.
Let $f(x) = x^2 - 2ax + a^2 + a - 3$. Since the roots are less than $3$,the vertex of the parabola $x = -b/(2a) = a$ must be less than $3$,and $f(3) > 0$:
$1$) $a < 3$
$2$) $f(3) = 3^2 - 2a(3) + a^2 + a - 3 > 0$
$9 - 6a + a^2 + a - 3 > 0$
$a^2 - 5a + 6 > 0$
$(a - 2)(a - 3) > 0$
This implies $a < 2$ or $a > 3$.
Combining $a \le 3$,$a < 3$,and ($a < 2$ or $a > 3$),we get $a < 2$.

(C) The expansion is given by $(1 + x)^m(1 - x)^n = (1 + mx + \frac{m(m - 1)}{2}x^2 + \dots)(1 - nx + \frac{n(n - 1)}{2}x^2 - \dots)$.
Multiplying the terms,we get $1 + (m - n)x + [\frac{n(n - 1)}{2} - mn + \frac{m(m - 1)}{2}]x^2 + \dots$.
Given the coefficient of $x$ is $3$,we have $m - n = 3$,so $n = m - 3$.
Given the coefficient of $x^2$ is $-6$,we have $\frac{n^2 - n}{2} - mn + \frac{m^2 - m}{2} = -6$.
Substituting $n = m - 3$ into the equation:
$\frac{(m - 3)(m - 4)}{2} - m(m - 3) + \frac{m^2 - m}{2} = -6$.
Multiplying by $2$: $(m^2 - 7m + 12) - 2(m^2 - 3m) + (m^2 - m) = -12$.
$m^2 - 7m + 12 - 2m^2 + 6m + m^2 - m = -12$.
$-2m + 12 = -12$.
$-2m = -24$,which gives $m = 12$.

(D) We analyze the sum $a(n) = \sum_{k=1}^{2^n-1} \frac{1}{k}$.
By grouping terms,we can observe that $a(n) = 1 + (\frac{1}{2} + \frac{1}{3}) + (\frac{1}{4} + \dots + \frac{1}{7}) + \dots + (\frac{1}{2^{n-1}} + \dots + \frac{1}{2^n-1})$.
Each group of the form $\sum_{k=2^{m-1}}^{2^m-1} \frac{1}{k}$ is greater than $\frac{1}{2}$.
Since there are $n$ such groups,$a(n) > \frac{n}{2}$.
For $n=200$,$a(200) > \frac{200}{2} = 100$.
Also,it is a known property that $a(n) \le n$ for these sums.
Thus,$a(100) \le 100$ and $a(200) > 100$ are both true.

(A) In $\triangle PQR$,$\angle R = \frac{\pi}{2}$,so $P + Q = \frac{\pi}{2}$,which implies $\frac{P}{2} + \frac{Q}{2} = \frac{\pi}{4}$.
Given that $\tan(\frac{P}{2})$ and $\tan(\frac{Q}{2})$ are roots of $ax^2 + bx + c = 0$,we have:
Sum of roots: $\tan(\frac{P}{2}) + \tan(\frac{Q}{2}) = -\frac{b}{a}$
Product of roots: $\tan(\frac{P}{2}) \tan(\frac{Q}{2}) = \frac{c}{a}$
Using the identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$:
$\tan(\frac{P}{2} + \frac{Q}{2}) = \tan(\frac{\pi}{4}) = 1$
$\frac{\tan(\frac{P}{2}) + \tan(\frac{Q}{2})}{1 - \tan(\frac{P}{2}) \tan(\frac{Q}{2})} = 1$
Substituting the sum and product of roots:
$\frac{-b/a}{1 - c/a} = 1$
$\frac{-b/a}{(a-c)/a} = 1$
$-b = a - c$
$c = a + b$.

(D) Let the equation of the line $L_1$ passing through the origin be $y = mx$.
Substituting $y = mx$ into the circle equation $x^2 + y^2 - x + 3y = 0$:
$x^2 + m^2x^2 - x + 3mx = 0$
$x[x(1 + m^2) - (1 - 3m)] = 0$
The points of intersection are $(0, 0)$ and $(\frac{1 - 3m}{1 + m^2}, \frac{m(1 - 3m)}{1 + m^2})$.
The length of the intercept $I_1$ is $\sqrt{(\frac{1 - 3m}{1 + m^2})^2 + (\frac{m(1 - 3m)}{1 + m^2})^2} = |\frac{1 - 3m}{1 + m^2}| \sqrt{1 + m^2} = \frac{|1 - 3m|}{\sqrt{1 + m^2}}$.
For the line $L_2: x + y = 1$,substitute $y = 1 - x$ into the circle equation:
$x^2 + (1 - x)^2 - x + 3(1 - x) = 0$
$x^2 + x^2 - 2x + 1 - x + 3 - 3x = 0$
$2x^2 - 6x + 4 = 0 \Rightarrow x^2 - 3x + 2 = 0$.
The roots are $x = 1, 2$. Corresponding $y$ values are $y = 0, -1$.
The length of the intercept $I_2$ is $\sqrt{(1 - 2)^2 + (0 - (-1))^2} = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$.
Equating $I_1^2 = I_2^2$: $\frac{(1 - 3m)^2}{1 + m^2} = 2$.
$(1 - 3m)^2 = 2(1 + m^2) \Rightarrow 1 - 6m + 9m^2 = 2 + 2m^2$.
$7m^2 - 6m - 1 = 0 \Rightarrow (7m + 1)(m - 1) = 0$.
So,$m = 1$ or $m = -1/7$.
The lines are $y = x$ $(x - y = 0)$ and $y = -\frac{1}{7}x$ $(x + 7y = 0)$.

(D) The equation of the circle is ${x^2} + {y^2} - px - qy = 0$.
Let the midpoint of a chord be $(h, 0)$ on the $x$-axis.
The equation of the chord with midpoint $(x_1, y_1)$ is $T = S_1$,where $T = xx_1 + yy_1 - \frac{p}{2}(x + x_1) - \frac{q}{2}(y + y_1)$ and $S_1 = x_1^2 + y_1^2 - px_1 - qy_1$.
Substituting $(x_1, y_1) = (h, 0)$,we get:
$xh - \frac{p}{2}(x + h) - \frac{q}{2}y = h^2 - ph$.
Since the chord passes through $(p, q)$,we substitute $x = p$ and $y = q$:
$ph - \frac{p}{2}(p + h) - \frac{q^2}{2} = h^2 - ph$.
Multiplying by $2$:
$2ph - p^2 - ph - q^2 = 2h^2 - 2ph$.
Rearranging terms:
$2h^2 - 3ph + p^2 + q^2 = 0$.
For two distinct chords to exist,the quadratic equation in $h$ must have two distinct real roots.
Thus,the discriminant $D > 0$:
$D = (-3p)^2 - 4(2)(p^2 + q^2) > 0$.
$9p^2 - 8p^2 - 8q^2 > 0$.
$p^2 - 8q^2 > 0$.
Therefore,$p^2 > 8q^2$.

(D) The given ellipse is $4x^2 + 9y^2 = 1$,which can be written as $\frac{x^2}{1/4} + \frac{y^2}{1/9} = 1$.
The slope of the line $8x = 9y$ is $m = \frac{8}{9}$.
Let the point of tangency be $(x_1, y_1)$. The equation of the tangent at $(x_1, y_1)$ is $4xx_1 + 9yy_1 = 1$.
The slope of this tangent is $-\frac{4x_1}{9y_1}$.
Since the tangent is parallel to the line,we have $-\frac{4x_1}{9y_1} = \frac{8}{9}$,which simplifies to $x_1 = -2y_1$.
Substituting $x_1 = -2y_1$ into the ellipse equation $4x_1^2 + 9y_1^2 = 1$:
$4(-2y_1)^2 + 9y_1^2 = 1$
$16y_1^2 + 9y_1^2 = 1$
$25y_1^2 = 1 \implies y_1 = \pm \frac{1}{5}$.
If $y_1 = \frac{1}{5}$,then $x_1 = -2(\frac{1}{5}) = -\frac{2}{5}$.
If $y_1 = -\frac{1}{5}$,then $x_1 = -2(-\frac{1}{5}) = \frac{2}{5}$.
Thus,the points are $\left( -\frac{2}{5}, \frac{1}{5} \right)$ and $\left( \frac{2}{5}, -\frac{1}{5} \right)$.

(B) The equation of the chord of contact for a point $(h, k)$ with respect to the hyperbola $S: x^2 - y^2 - 9 = 0$ is given by $T = 0$,which is $xh - yk - 9 = 0$.
Comparing this with the given chord $x = 9$ (or $x - 9 = 0$),we get $h = 1$ and $k = 0$.
The equation of the pair of tangents from a point $(h, k)$ to the hyperbola is given by $SS_1 = T^2$.
Here,$S = x^2 - y^2 - 9$,$S_1 = (1)^2 - (0)^2 - 9 = 1 - 9 = -8$,and $T = x(1) - y(0) - 9 = x - 9$.
Substituting these into the formula: $(x^2 - y^2 - 9)(-8) = (x - 9)^2$.
$-8x^2 + 8y^2 + 72 = x^2 - 18x + 81$.
Rearranging the terms: $9x^2 - 8y^2 - 18x + 9 = 0$.

(B) Let $M, P$,and $C$ be the events of passing in Mathematics,Physics,and Chemistry respectively.
Given:
$P(M \cup P \cup C) = \frac{75}{100} = \frac{3}{4}$
$P(\text{at least two}) = P(M \cap P) + P(P \cap C) + P(M \cap C) - 2P(M \cap P \cap C) = \frac{50}{100} = \frac{1}{2}$
$P(\text{exactly two}) = P(M \cap P) + P(P \cap C) + P(M \cap C) - 3P(M \cap P \cap C) = \frac{40}{100} = \frac{2}{5}$
Subtracting the two equations:
$P(M \cap P \cap C) = \frac{1}{2} - \frac{2}{5} = \frac{1}{10}$
Using the inclusion-exclusion principle:
$P(M \cup P \cup C) = (m + p + c) - (mp + pc + mc) + mpc = \frac{3}{4}$
From the 'at least two' condition:
$(mp + pc + mc) - 2mpc = \frac{1}{2} \Rightarrow (mp + pc + mc) = \frac{1}{2} + 2(\frac{1}{10}) = \frac{7}{10}$
Substituting these into the inclusion-exclusion equation:
$(m + p + c) - \frac{7}{10} + \frac{1}{10} = \frac{3}{4}$
$m + p + c = \frac{3}{4} + \frac{6}{10} = \frac{15 + 12}{20} = \frac{27}{20}$.

(A) Given that ${x_1}, {x_2}, {x_3}$ and ${y_1}, {y_2}, {y_3}$ are in $G$.$P$. with the same common ratio $r$,we have:
${x_2} = r{x_1}, {x_3} = {r^2}{x_1}$
${y_2} = r{y_1}, {y_3} = {r^2}{y_1}$
To check if the points are collinear,we calculate the area of the triangle formed by these points:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Substituting the values:
$\text{Area} = \frac{1}{2} |x_1(ry_1 - r^2y_1) + rx_1(r^2y_1 - y_1) + r^2x_1(y_1 - ry_1)|$
$\text{Area} = \frac{1}{2} |x_1y_1(r - r^2) + x_1y_1(r^3 - r) + x_1y_1(r^2 - r^3)|$
$\text{Area} = \frac{1}{2} |x_1y_1(r - r^2 + r^3 - r + r^2 - r^3)| = 0$
Since the area is $0$,the points lie on a straight line.

(D) Given ${a_1 = h_1 = 2}$ and ${a_{10} = h_{10} = 3}$.
For the $A.P.$,${a_{10} = a_1 + 9d = 3}$,so ${2 + 9d = 3}$,which gives ${d = \frac{1}{9}}$.
Thus,${a_4 = a_1 + 3d = 2 + 3(\frac{1}{9}) = 2 + \frac{1}{3} = \frac{7}{3}}$.
For the $H.P.$,the reciprocals $\frac{1}{h_n}$ are in $A.P.$ Let ${H_n = \frac{1}{h_n}}$.
Then ${H_1 = \frac{1}{2}}$ and ${H_{10} = \frac{1}{3}}$.
${H_{10} = H_1 + 9D = \frac{1}{3}}$,so ${9D = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6}}$,which gives ${D = -\frac{1}{54}}$.
Then ${H_7 = H_1 + 6D = \frac{1}{2} + 6(-\frac{1}{54}) = \frac{1}{2} - \frac{1}{9} = \frac{9-2}{18} = \frac{7}{18}}$.
Since ${h_7 = \frac{1}{H_7}}$,we have ${h_7 = \frac{18}{7}}$.
Therefore,${a_4 h_7 = \frac{7}{3} \times \frac{18}{7} = 6}$.

(D) We know that $1 + \sec \theta = 1 + \frac{1}{\cos \theta} = \frac{\cos \theta + 1}{\cos \theta} = \frac{2\cos^2(\theta/2)}{\cos \theta}$.
Given ${f_n}(\theta ) = \tan(\theta/2) \cdot (1 + \sec \theta) \cdot (1 + \sec 2\theta) \dots (1 + \sec 2^n \theta)$.
Substituting the identity: ${f_n}(\theta ) = \frac{\sin(\theta/2)}{\cos(\theta/2)} \cdot \frac{2\cos^2(\theta/2)}{\cos \theta} \cdot \frac{2\cos^2 \theta}{\cos 2\theta} \dots \frac{2\cos^2(2^{n-1}\theta)}{\cos(2^n \theta)}$.
Simplifying step by step:
${f_n}(\theta ) = \frac{2\sin(\theta/2)\cos(\theta/2)}{\cos \theta} \cdot \frac{2\cos^2 \theta}{\cos 2\theta} \dots = \frac{\sin \theta}{\cos \theta} \cdot \frac{2\cos^2 \theta}{\cos 2\theta} \dots = \tan \theta \cdot \frac{2\cos^2 \theta}{\cos 2\theta} \dots = \tan 2\theta \cdot \frac{2\cos^2 2\theta}{\cos 4\theta} \dots$
Continuing this process,we get ${f_n}(\theta ) = \tan(2^n \theta)$.
Now checking the options:
$1$) ${f_2}(\pi/16) = \tan(2^2 \cdot \pi/16) = \tan(\pi/4) = 1$.
$2$) ${f_3}(\pi/32) = \tan(2^3 \cdot \pi/32) = \tan(\pi/4) = 1$.
$3$) ${f_4}(\pi/64) = \tan(2^4 \cdot \pi/64) = \tan(\pi/4) = 1$.
Thus,all the above are correct.

(C) Given $x = t^2 + t + 1$ and $y = t^2 - t + 1$.
Adding the two equations: $x + y = 2t^2 + 2 = 2(t^2 + 1)$.
Subtracting the two equations: $x - y = 2t$,which implies $t = \frac{x - y}{2}$.
Substituting $t$ into the expression for $x + y$:
$x + y = 2\left(\left(\frac{x - y}{2}\right)^2 + 1\right) = 2\left(\frac{(x - y)^2}{4} + 1\right) = \frac{(x - y)^2}{2} + 2$.
Multiplying by $2$: $2(x + y) = (x - y)^2 + 4$.
Expanding: $2x + 2y = x^2 - 2xy + y^2 + 4$.
Rearranging: $x^2 - 2xy + y^2 - 2x - 2y + 4 = 0$.
Comparing with the general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we have $a = 1, h = -1, b = 1$.
Since $h^2 - ab = (-1)^2 - (1)(1) = 1 - 1 = 0$,the curve represents a parabola.

(C) We evaluate the limit: $\lim _{x \rightarrow 0} \frac{x \tan 2x - 2x \tan x}{(1 - \cos 2x)^2}$.
Using the identity $1 - \cos 2x = 2 \sin^2 x$,the denominator becomes $(2 \sin^2 x)^2 = 4 \sin^4 x$.
So,the expression is $\lim _{x \rightarrow 0} \frac{x(\tan 2x - 2 \tan x)}{4 \sin^4 x}$.
Using Taylor series expansions: $\tan \theta = \theta + \frac{\theta^3}{3} + \dots$ and $\sin x \approx x$.
$\tan 2x = 2x + \frac{(2x)^3}{3} + O(x^5) = 2x + \frac{8x^3}{3} + O(x^5)$.
$2 \tan x = 2(x + \frac{x^3}{3} + O(x^5)) = 2x + \frac{2x^3}{3} + O(x^5)$.
Numerator: $x[(2x + \frac{8x^3}{3}) - (2x + \frac{2x^3}{3})] = x[\frac{6x^3}{3}] = 2x^4$.
Denominator: $4 \sin^4 x \approx 4x^4$.
Limit: $\lim _{x \rightarrow 0} \frac{2x^4}{4x^4} = \frac{2}{4} = \frac{1}{2}$.

(C) The given equation is $\tan^{-1}\sqrt{x(x + 1)} + \sin^{-1}\sqrt{x^2 + x + 1} = \frac{\pi}{2}$.
For $\tan^{-1}\sqrt{x(x + 1)}$ to be defined,we must have $x(x + 1) \ge 0$.
For $\sin^{-1}\sqrt{x^2 + x + 1}$ to be defined,the argument must satisfy $0 \le \sqrt{x^2 + x + 1} \le 1$,which implies $0 \le x^2 + x + 1 \le 1$.
Since $x^2 + x + 1 = (x + \frac{1}{2})^2 + \frac{3}{4}$,the minimum value of $x^2 + x + 1$ is $\frac{3}{4}$.
Thus,the condition $0 \le x^2 + x + 1 \le 1$ simplifies to $\frac{3}{4} \le x^2 + x + 1 \le 1$,which implies $x^2 + x \le 0$.
Combining $x(x + 1) \ge 0$ and $x(x + 1) \le 0$,we must have $x(x + 1) = 0$.
This gives $x = 0$ or $x = -1$.
Checking $x = 0$: $\tan^{-1}(0) + \sin^{-1}(1) = 0 + \frac{\pi}{2} = \frac{\pi}{2}$. This is a solution.
Checking $x = -1$: $\tan^{-1}(0) + \sin^{-1}(1) = 0 + \frac{\pi}{2} = \frac{\pi}{2}$. This is a solution.
Therefore,there are $2$ real solutions.

(D) Let $\theta$ be the angle between $a$ and $b$. Since $a$ and $b$ are unit vectors,$|a| = 1$ and $|b| = 1$.
$v = a \times b$,so $|v| = |a||b| \sin \theta = \sin \theta$.
Now,$u = a - (a \cdot b)b = a - (\cos \theta)b$.
Calculating $|u|^2 = u \cdot u = (a - \cos \theta \, b) \cdot (a - \cos \theta \, b) = |a|^2 + \cos^2 \theta |b|^2 - 2 \cos \theta (a \cdot b) = 1 + \cos^2 \theta - 2 \cos^2 \theta = 1 - \cos^2 \theta = \sin^2 \theta$.
Thus,$|u| = \sin \theta$.
Comparing $|v|$ and $|u|$,we get $|v| = |u|$.
Also,$u \cdot b = (a - \cos \theta \, b) \cdot b = a \cdot b - \cos \theta (b \cdot b) = \cos \theta - \cos \theta = 0$.
Therefore,$|u| + |u \cdot b| = |u| + 0 = |u| = |v|$.
Both options $(A)$ and $(C)$ are correct.

(B) First,calculate the cross product $a \times b$:
$a \times b = \begin{vmatrix} i & j & k \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = i(0 - (-2)) - j(0 - (-2)) + k(2 - 1) = 2i - 2j + k$.
Next,find the magnitude of $a \times b$:
$|a \times b| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
Given $|c - a| = 2\sqrt{2}$,square both sides:
$|c - a|^2 = (2\sqrt{2})^2 = 8$.
$|c|^2 - 2(a \cdot c) + |a|^2 = 8$.
Since $a \cdot c = |c|$ and $|a|^2 = 2^2 + 1^2 + (-2)^2 = 9$:
$|c|^2 - 2|c| + 9 = 8 \Rightarrow |c|^2 - 2|c| + 1 = 0 \Rightarrow (|c| - 1)^2 = 0 \Rightarrow |c| = 1$.
Finally,calculate the magnitude of the cross product $|(a \times b) \times c|$ using the formula $|u \times v| = |u||v|\sin\theta$:
$|(a \times b) \times c| = |a \times b| |c| \sin 30^\circ = 3 \times 1 \times \frac{1}{2} = \frac{3}{2}$.

(A) Since $c$ is coplanar with $a$ and $b$,we can write $c = xa + yb$ for some scalars $x$ and $y$.
Substituting the given vectors:
$c = x(2i + j + k) + y(i + 2j - k) = (2x + y)i + (x + 2y)j + (x - y)k$.
Given that $c$ is perpendicular to $a$,their dot product must be zero:
$a \cdot c = 0 \implies 2(2x + y) + 1(x + 2y) + 1(x - y) = 0$.
$4x + 2y + x + 2y + x - y = 0 \implies 6x + 3y = 0 \implies y = -2x$.
Substituting $y = -2x$ into the expression for $c$:
$c = (2x - 2x)i + (x - 4x)j + (x + 2x)k = -3xj + 3xk = 3x(-j + k)$.
Since $c$ is a unit vector,$|c| = 1$:
$|3x(-j + k)| = 1 \implies |3x| \sqrt{(-1)^2 + 1^2} = 1 \implies |3x| \sqrt{2} = 1 \implies x = \pm \frac{1}{3\sqrt{2}}$.
Thus,$c = 3(\pm \frac{1}{3\sqrt{2}})(-j + k) = \pm \frac{1}{\sqrt{2}}(-j + k)$.
Comparing with the given options,the correct answer is $\frac{1}{\sqrt{2}}(-j + k)$.

(B) Given $f(x) = 2^{x(x - 1)}$.
To find the inverse,let $y = f(x) = 2^{x(x - 1)}$.
Taking $\log_2$ on both sides,we get $\log_2 y = x(x - 1)$.
This simplifies to the quadratic equation $x^2 - x - \log_2 y = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{1 \pm \sqrt{1 + 4\log_2 y}}{2}$.
Since the domain of $f$ is $[1, \infty)$,we must have $x \ge 1$.
If we choose the negative sign,$x = \frac{1 - \sqrt{1 + 4\log_2 y}}{2} < 1$ for $y \ge 1$.
Thus,we must choose the positive sign: $x = \frac{1 + \sqrt{1 + 4\log_2 y}}{2}$.
Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{1}{2}(1 + \sqrt{1 + 4\log_2 x})$.

(D) The function $f(x) = (x^2 - 1)|(x - 1)(x - 2)| + \cos(|x|)$.
First,consider the term $\cos(|x|)$. The function $|x|$ is not differentiable at $x = 0$,so $\cos(|x|)$ is not differentiable at $x = 0$.
Next,consider the term $(x^2 - 1)|(x - 1)(x - 2)|$. The expression $|(x - 1)(x - 2)|$ is not differentiable at the roots of $(x - 1)(x - 2) = 0$,which are $x = 1$ and $x = 2$.
At $x = 1$,$f(x) = (x^2 - 1)|(x - 1)(x - 2)| + \cos(|x|)$. Since $(x^2 - 1)$ has a factor $(x - 1)$,the product $(x^2 - 1)|(x - 1)(x - 2)|$ becomes differentiable at $x = 1$ because the zero of the polynomial cancels the non-differentiability of the absolute value.
At $x = 2$,the term $(x^2 - 1)$ is $3 \neq 0$. Thus,the non-differentiability of $|(x - 1)(x - 2)|$ at $x = 2$ persists.
Checking $x = 2$:
$Lf'(2) = \lim_{h \to 0^-} \frac{f(2+h) - f(2)}{h} = -3 - \sin(2)$
$Rf'(2) = \lim_{h \to 0^+} \frac{f(2+h) - f(2)}{h} = 3 - \sin(2)$
Since $Lf'(2) \neq Rf'(2)$,the function is not differentiable at $x = 2$.

(D) Given $f(x) = [x]^2 - [x^2]$.
For $-1 < x < 0$,$[x] = -1$ and $[x^2] = 0$,so $f(x) = (-1)^2 - 0 = 1$.
For $x = 0$,$f(0) = 0^2 - 0 = 0$.
For $0 < x < 1$,$[x] = 0$ and $[x^2] = 0$,so $f(x) = 0^2 - 0 = 0$.
For $x = 1$,$f(1) = 1^2 - 1 = 0$.
For $1 < x < \sqrt{2}$,$[x] = 1$ and $[x^2] = 1$,so $f(x) = 1^2 - 1 = 0$.
For $x = \sqrt{2}$,$f(\sqrt{2}) = 1^2 - 2 = -1$.
For $\sqrt{2} < x < \sqrt{3}$,$[x] = 1$ and $[x^2] = 2$,so $f(x) = 1^2 - 2 = -1$.
For $x = \sqrt{3}$,$f(\sqrt{3}) = 1^2 - 3 = -2$.
For $\sqrt{3} < x < 2$,$[x] = 1$ and $[x^2] = 3$,so $f(x) = 1^2 - 3 = -2$.
For $x = 2$,$f(2) = 2^2 - 4 = 0$.
For $2 < x < \sqrt{5}$,$[x] = 2$ and $[x^2] = 4$,so $f(x) = 2^2 - 4 = 0$.
For $x = \sqrt{5}$,$f(\sqrt{5}) = 2^2 - 5 = -1$.
By analyzing the behavior at integers $n$,we find that the function is discontinuous at all integers except $x = 1$.

(B) Given $f(x) = \int_{-1}^x {t({e^t} - 1)(t - 1){(t - 2)}^3{(t - 3)}^5} dt$.
By the Fundamental Theorem of Calculus,$f'(x) = x({e^x} - 1)(x - 1){(x - 2)}^3{(x - 3)}^5$.
For local extrema,we set $f'(x) = 0$,which gives critical points $x = 0, 1, 2, 3$.
We examine the sign change of $f'(x)$ around these points:
$1$. At $x = 0$: $f'(x)$ changes from negative to negative (no extremum).
$2$. At $x = 1$: $f'(x)$ changes from negative to positive (local minimum).
$3$. At $x = 2$: $f'(x)$ changes from positive to negative (local maximum).
$4$. At $x = 3$: $f'(x)$ changes from negative to positive (local minimum).
Thus,the function has local minima at $x = 1$ and $x = 3$. Since $1$ is an option,the correct answer is $1$.

(B) Given $f(x) = \sin^4 x + \cos^4 x$.
We can rewrite this as $f(x) = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x$.
Since $\sin^2 x + \cos^2 x = 1$,we have $f(x) = 1 - 2\sin^2 x \cos^2 x$.
Using the identity $\sin 2x = 2\sin x \cos x$,we get $f(x) = 1 - \frac{1}{2}(2\sin x \cos x)^2 = 1 - \frac{1}{2}\sin^2 2x$.
Using $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$,we have $f(x) = 1 - \frac{1}{2} \left( \frac{1 - \cos 4x}{2} \right) = 1 - \frac{1}{4} + \frac{1}{4}\cos 4x = \frac{3}{4} + \frac{1}{4}\cos 4x$.
For the function to be increasing,$f'(x) > 0$.
$f'(x) = \frac{d}{dx} \left( \frac{3}{4} + \frac{1}{4}\cos 4x \right) = -\sin 4x$.
Setting $f'(x) > 0$,we get $-\sin 4x > 0$,which implies $\sin 4x < 0$.
The sine function is negative in the interval $(\pi, 2\pi)$.
Thus,$\pi < 4x < 2\pi$.
Dividing by $4$,we get $\frac{\pi}{4} < x < \frac{\pi}{2}$.
Comparing this with the given options,the interval $\frac{\pi}{4} < x < \frac{3\pi}{8}$ is a subset of this range,so the function is increasing in this interval.

(A) Let $I = \int_{\pi /4}^{3\pi /4} \frac{dx}{1 + \cos x}$.
Using the identity $1 + \cos x = 2\cos^2(x/2)$,we have:
$I = \int_{\pi /4}^{3\pi /4} \frac{dx}{2\cos^2(x/2)} = \frac{1}{2} \int_{\pi /4}^{3\pi /4} \sec^2(x/2) dx$.
Integrating $\sec^2(x/2)$,we get:
$I = \frac{1}{2} [2 \tan(x/2)]_{\pi /4}^{3\pi /4} = [\tan(x/2)]_{\pi /4}^{3\pi /4}$.
Evaluating at the limits:
$I = \tan(3\pi /8) - \tan(\pi /8)$.
Using $\tan(3\pi /8) = \cot(\pi /8)$:
$I = \cot(\pi /8) - \tan(\pi /8) = \frac{\cos(\pi /8)}{\sin(\pi /8)} - \frac{\sin(\pi /8)}{\cos(\pi /8)} = \frac{\cos^2(\pi /8) - \sin^2(\pi /8)}{\sin(\pi /8)\cos(\pi /8)}$.
Using $\cos(2\theta) = \cos^2\theta - \sin^2\theta$ and $2\sin\theta\cos\theta = \sin(2\theta)$:
$I = \frac{\cos(\pi /4)}{\frac{1}{2}\sin(\pi /4)} = 2 \cot(\pi /4) = 2(1) = 2$.

(B) The equation of the curve is $y = x - x^2$.
The points of intersection of the curve $y = x - x^2$ and the line $y = mx$ are found by setting $x - x^2 = mx$,which gives $x(1 - x - m) = 0$. Thus,$x = 0$ or $x = 1 - m$.
The area $A$ of the region bounded by the curve and the line is given by the integral:
$A = \int_{0}^{1-m} (x - x^2 - mx) dx = \int_{0}^{1-m} ((1 - m)x - x^2) dx$
Evaluating the integral:
$A = \left[ (1 - m)\frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1-m}$
$A = (1 - m)\frac{(1 - m)^2}{2} - \frac{(1 - m)^3}{3} = \frac{(1 - m)^3}{2} - \frac{(1 - m)^3}{3} = \frac{(1 - m)^3}{6}$
Given that the area is $\frac{9}{2}$,we have:
$\frac{(1 - m)^3}{6} = \frac{9}{2}$
$(1 - m)^3 = 27$
$1 - m = 3$
$m = -2$
Thus,the correct value of $m$ is $-2$.

(C) We know that for $x \in [\pi/2, 3\pi/2]$,$\sin x$ ranges from $-1$ to $1$,so $2\sin x$ ranges from $-2$ to $2$.
We break the integral at points where $[2\sin x]$ changes its value:
$I = \int_{\pi/2}^{3\pi/2} [2\sin x] \, dx$
$= \int_{\pi/2}^{5\pi/6} [2\sin x] \, dx + \int_{5\pi/6}^{\pi} [2\sin x] \, dx + \int_{\pi}^{7\pi/6} [2\sin x] \, dx + \int_{7\pi/6}^{3\pi/2} [2\sin x] \, dx$
In the interval $[\pi/2, 5\pi/6]$,$1 \le 2\sin x < 2$,so $[2\sin x] = 1$.
In the interval $[5\pi/6, \pi]$,$0 \le 2\sin x < 1$,so $[2\sin x] = 0$.
In the interval $[\pi, 7\pi/6]$,$-1 \le 2\sin x < 0$,so $[2\sin x] = -1$.
In the interval $[7\pi/6, 3\pi/2]$,$-2 \le 2\sin x < -1$,so $[2\sin x] = -2$.
Substituting these values:
$I = \int_{\pi/2}^{5\pi/6} (1) \, dx + \int_{5\pi/6}^{\pi} (0) \, dx + \int_{\pi}^{7\pi/6} (-1) \, dx + \int_{7\pi/6}^{3\pi/2} (-2) \, dx$
$I = (5\pi/6 - \pi/2) + 0 - (7\pi/6 - \pi) - 2(3\pi/2 - 7\pi/6)$
$I = (2\pi/6) - (\pi/6) - 2(2\pi/6) = \pi/6 - 4\pi/6 = -3\pi/6 = -\pi/2$.

(D) Given curve is ${y^2} = 2c(x + \sqrt{c}).$
Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 2c$,which implies $c = y \frac{dy}{dx}.$
Substituting $c$ into the original equation:
${y^2} = 2 \left( y \frac{dy}{dx} \right) \left( x + \sqrt{y \frac{dy}{dx}} \right).$
Rearranging the terms:
$\frac{y}{2(dy/dx)} - x = \sqrt{y \frac{dy}{dx}}.$
Squaring both sides:
$\left( \frac{y}{2(dy/dx)} - x \right)^2 = y \frac{dy}{dx}.$
Multiplying by $4(dy/dx)^2$:
$(y - 2x(dy/dx))^2 = 4y(dy/dx)^3.$
Expanding the square:
${y^2} - 4xy \frac{dy}{dx} + 4{x^2} \left( \frac{dy}{dx} \right)^2 = 4y \left( \frac{dy}{dx} \right)^3.$
Rearranging into the standard form:
$4y \left( \frac{dy}{dx} \right)^3 - 4{x^2} \left( \frac{dy}{dx} \right)^2 + 4xy \frac{dy}{dx} - {y^2} = 0.$
The highest order derivative present is $\frac{dy}{dx}$,so the order is $1$. The highest power of the highest order derivative is $3$,so the degree is $3$.
Thus,the differential equation is of order $1$ and degree $3$.

(C) The given differential equation is ${\left( {\frac{{dy}}{{dx}}} \right)^2} - x\frac{{dy}}{{dx}} + y = 0$.
This can be rewritten as $y = x\frac{{dy}}{{dx}} - {\left( {\frac{{dy}}{{dx}}} \right)^2}$.
Let $\frac{{dy}}{{dx}} = p$. Then the equation becomes $y = px - {p^2}$.
Differentiating both sides with respect to $x$,we get:
$\frac{{dy}}{{dx}} = p + x\frac{{dp}}{{dx}} - 2p\frac{{dp}}{{dx}}$.
Since $\frac{{dy}}{{dx}} = p$,we have $p = p + (x - 2p)\frac{{dp}}{{dx}}$.
This simplifies to $(x - 2p)\frac{{dp}}{{dx}} = 0$.
Case $1$: $\frac{{dp}}{{dx}} = 0$,which implies $p = c$ (a constant).
Substituting $p = c$ into $y = px - {p^2}$,we get the general solution $y = cx - {c^2}$.
For $c = 2$,we get $y = 2x - {2^2}$,which is $y = 2x - 4$.
Thus,$y = 2x - 4$ is a solution.

(A) The last digit of $7^k$ follows a cycle of length $4$: $7^1=7, 7^2=9, 7^3=3, 7^4=1, 7^5=7, \dots$
For $7^m + 7^n$ to be divisible by $5$,the last digit of the sum must be $0$ or $5$.
Since the last digits of powers of $7$ are ${1, 3, 7, 9}$,the possible sums of last digits are:
$1+1=2, 1+3=4, 1+7=8, 1+9=10$ (divisible by $5$)
$3+1=4, 3+3=6, 3+7=10$ (divisible by $5$),$3+9=12$
$7+1=8, 7+3=10$ (divisible by $5$),$7+7=14, 7+9=16$
$9+1=10$ (divisible by $5$),$9+3=12, 9+7=16, 9+9=18$
In each cycle of $4$ for $m$ and $n$,there are $4 \times 4 = 16$ pairs. The pairs $(m, n) \pmod 4$ that result in a sum divisible by $5$ are $(1, 3), (3, 1), (2, 4), (4, 2)$.
There are $4$ such pairs out of $16$ total possibilities in each $4 \times 4$ block.
Since $100$ is a multiple of $4$,the probability is $\frac{4}{16} = \frac{1}{4}$.

(A) Given determinant is $f(x) = \left| \begin{array}{ccc} 1 & x & x + 1 \\ 2x & x(x - 1) & (x + 1)x \\ 3x(x - 1) & x(x - 1)(x - 2) & (x + 1)x(x - 1) \end{array} \right|$.
Taking $x$ common from $C_2$ and $(x+1)$ common from $C_3$ is not directly possible,but we can factor out $x$ from the second column and $x(x+1)$ from the third row.
Alternatively,observe the columns. Notice that $C_3 = C_2 + C_1$ is not immediately obvious,but let's simplify by taking common factors:
$f(x) = x(x-1) \cdot x \cdot \left| \begin{array}{ccc} 1 & 1 & x+1 \\ 2x & x-1 & x(x+1) \\ 3x(x-1) & (x-2) & (x+1)x \end{array} \right|$.
Actually,a simpler way is to notice that for any $x$,the rows are linearly dependent. Let $R_3 = R_3 - (x-2)R_2$.
Performing row operations or expanding the determinant shows that the expression simplifies to $0$ for all $x$.
Specifically,$f(x) = 0$ for all $x \in \mathbb{R}$.
Therefore,$f(100) = 0$.