IIT JEE 2002 Mathematics Question Paper with Answer and Solution

(B) The given equations represent two circles in the complex plane:
Circle $C_1$: Center $O(0, 0)$,radius $r_1 = 12$.
Circle $C_2$: Center $C(3, 4)$,radius $r_2 = 5$.
The distance between the centers is $d = |(3 + 4i) - 0| = \sqrt{3^2 + 4^2} = 5$.
Since $d + r_2 = 5 + 5 = 10 < r_1 = 12$,the circle $C_2$ lies entirely inside the circle $C_1$.
The minimum distance between a point on $C_1$ and a point on $C_2$ is given by $r_1 - (d + r_2) = 12 - (5 + 5) = 12 - 10 = 2$.

(A) By the $AM-GM$ inequality,for $n$ positive real numbers,the arithmetic mean is greater than or equal to the geometric mean.
Consider the $n$ numbers: ${a_1}, {a_2}, ..., {a_{n-1}}, 2{a_n}$.
Their arithmetic mean is $\frac{{a_1} + {a_2} + ... + {a_{n-1}} + 2{a_n}}{n}$.
Their geometric mean is $({a_1} \cdot {a_2} \cdot ... \cdot {a_{n-1}} \cdot 2{a_n})^{1/n} = (2 \cdot {a_1} \cdot {a_2} \cdot ... \cdot {a_n})^{1/n} = (2c)^{1/n}$.
Applying $AM \ge GM$:
$\frac{{a_1} + {a_2} + ... + {a_{n-1}} + 2{a_n}}{n} \ge (2c)^{1/n}$.
Therefore,the minimum value of ${a_1} + {a_2} + ... + {a_{n-1}} + 2{a_n}$ is $n(2c)^{1/n}$.

(D) Given $a, b, c$ are in $A.P.$,let $b = a + d$ and $c = a + 2d$,where $d > 0$ since $a < b < c$.
Given $a^2, b^2, c^2$ are in $G.P.$,we have $(b^2)^2 = a^2 c^2$,which implies $b^4 = a^2 c^2$.
Taking the square root,$b^2 = \pm ac$.
If $b^2 = ac$,then $a, b, c$ are in $G.P.$ Since they are also in $A.P.$,$a = b = c$,which contradicts $a < b < c$.
Thus,$b^2 = -ac$.
Substituting $b = a + d$ and $c = a + 2d$:
$(a + d)^2 = -a(a + 2d)$
$a^2 + d^2 + 2ad = -a^2 - 2ad$
$2a^2 + 4ad + d^2 = 0$.
Given $a + b + c = \frac{3}{2}$,we have $a + (a + d) + (a + 2d) = 3a + 3d = \frac{3}{2}$,so $a + d = \frac{1}{2}$,which means $d = \frac{1}{2} - a$.
Substituting $d$ into the equation $2a^2 + 4ad + d^2 = 0$:
$2a^2 + 4a(\frac{1}{2} - a) + (\frac{1}{2} - a)^2 = 0$
$2a^2 + 2a - 4a^2 + \frac{1}{4} - a + a^2 = 0$
$-a^2 + a + \frac{1}{4} = 0$
$4a^2 - 4a - 1 = 0$.
Using the quadratic formula $a = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(4)(-1)}}{2(4)} = \frac{4 \pm \sqrt{16 + 16}}{8} = \frac{4 \pm 4\sqrt{2}}{8} = \frac{1}{2} \pm \frac{1}{\sqrt{2}}$.
Since $d = \frac{1}{2} - a > 0$,we must have $a < \frac{1}{2}$.
Therefore,$a = \frac{1}{2} - \frac{1}{\sqrt{2}}$.

(B) Case $I$: When $x + 2 \ge 0$ i.e.,$x \ge -2,$
The given inequality becomes ${x^2} - (x + 2) + x > 0 \implies {x^2} - 2 > 0 \implies |x| > \sqrt{2}.$
This implies $x < -\sqrt{2}$ or $x > \sqrt{2}.$
Since $x \ge -2,$ the solution set for this case is $[ -2, -\sqrt{2} ) \cup (\sqrt{2}, \infty ).$
Case $II$: When $x + 2 < 0$ i.e.,$x < -2,$
The given inequality becomes ${x^2} + (x + 2) + x > 0 \implies {x^2} + 2x + 2 > 0.$
Since the discriminant $D = 2^2 - 4(1)(2) = 4 - 8 = -4 < 0$ and the coefficient of ${x^2}$ is positive,the expression ${x^2} + 2x + 2$ is always positive for all real $x.$
Thus,the solution set for this case is $( -\infty, -2).$
Combining the two cases,the total solution set is $( -\infty, -2) \cup [ -2, -\sqrt{2} ) \cup (\sqrt{2}, \infty ) = ( -\infty, -\sqrt{2} ) \cup (\sqrt{2}, \infty ).$

(A) The word $BANANA$ contains $6$ letters: $B, A, N, A, N, A$.
Here,$A$ appears $3$ times and $N$ appears $2$ times.
Total number of arrangements = $\frac{6!}{3! \times 2!} = \frac{720}{6 \times 2} = 60$.
Number of arrangements where two $N$s are together: Treat $(NN)$ as one unit. Now we have $5$ units: $B, A, A, A, (NN)$.
Number of arrangements = $\frac{5!}{3! \times 1!} = \frac{120}{6} = 20$.
Number of arrangements where two $N$s do not appear adjacently = (Total arrangements) - (Arrangements where $N$s are together) = $60 - 20 = 40$.

(B) The given sum is $\sum\limits_{i = 0}^m {\binom{10}{i}} {\binom{20}{m - i}}$.
By Vandermonde's Identity,this sum is equal to the coefficient of $x^m$ in the expansion of $(1+x)^{10} (1+x)^{20} = (1+x)^{30}$.
Thus,the sum is equal to $\binom{30}{m}$.
We know that $\binom{n}{m}$ is maximum when $m = \frac{n}{2}$ if $n$ is even,or $m = \frac{n-1}{2}$ and $m = \frac{n+1}{2}$ if $n$ is odd.
Here,$n = 30$,which is even.
Therefore,the sum $\binom{30}{m}$ is maximum when $m = \frac{30}{2} = 15$.
Hence,the correct option is $B$.

(B) For a system of linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ to have infinitely many solutions,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Given equations: $(k + 1)x + 8y = 4k$ and $kx + (k + 3)y = 3k - 1$.
Applying the condition: $\frac{k + 1}{k} = \frac{8}{k + 3} = \frac{4k}{3k - 1}$.
First,solve $\frac{k + 1}{k} = \frac{8}{k + 3}$:
$(k + 1)(k + 3) = 8k \Rightarrow k^2 + 4k + 3 = 8k \Rightarrow k^2 - 4k + 3 = 0$.
Factoring gives $(k - 1)(k - 3) = 0$,so $k = 1$ or $k = 3$.
Next,check these values in the second equality $\frac{8}{k + 3} = \frac{4k}{3k - 1}$:
If $k = 1$: $\frac{8}{1 + 3} = \frac{8}{4} = 2$ and $\frac{4(1)}{3(1) - 1} = \frac{4}{2} = 2$. Since $2 = 2$,$k = 1$ is a solution.
If $k = 3$: $\frac{8}{3 + 3} = \frac{8}{6} = \frac{4}{3}$ and $\frac{4(3)}{3(3) - 1} = \frac{12}{8} = \frac{3}{2}$. Since $\frac{4}{3} \neq \frac{3}{2}$,$k = 3$ is not a solution.
Thus,there is only $1$ value of $k$ for which the system has infinitely many solutions.

(A) The equation of any tangent to the circle $x^2 + y^2 = b^2$ is given by $y = mx \pm b\sqrt{1 + m^2}$.
Since the given tangent is $y = mx - b\sqrt{1 + m^2}$,it is a tangent to the first circle.
For this line to be a tangent to the second circle $(x - a)^2 + y^2 = b^2$,the perpendicular distance from the center $(a, 0)$ to the line $mx - y - b\sqrt{1 + m^2} = 0$ must be equal to the radius $b$.
Thus,$\frac{|ma - 0 - b\sqrt{1 + m^2}|}{\sqrt{m^2 + 1}} = b$.
Since $a > 2b$,we have $ma > b\sqrt{1 + m^2}$,so $ma - b\sqrt{1 + m^2} = b\sqrt{m^2 + 1}$.
$ma = 2b\sqrt{1 + m^2}$.
Squaring both sides,$m^2 a^2 = 4b^2(1 + m^2) = 4b^2 + 4b^2 m^2$.
$m^2(a^2 - 4b^2) = 4b^2$.
$m^2 = \frac{4b^2}{a^2 - 4b^2}$.
Since we need the positive value of $m$,$m = \frac{2b}{\sqrt{a^2 - 4b^2}}$.

(C) Let the moving point on the parabola $y^2 = 4ax$ be $(at^2, 2at)$.
The focus of the parabola is $S(a, 0)$.
Let $(h, k)$ be the midpoint of the line segment joining $(at^2, 2at)$ and $(a, 0)$.
Then $h = \frac{at^2 + a}{2}$ and $k = \frac{2at + 0}{2} = at$.
From $k = at$,we have $t = \frac{k}{a}$.
Substituting $t$ into the expression for $h$: $h = \frac{a(\frac{k}{a})^2 + a}{2} = \frac{\frac{k^2}{a} + a}{2} = \frac{k^2 + a^2}{2a}$.
Thus,$2ah = k^2 + a^2$,which simplifies to $k^2 = 2a(h - \frac{a}{2})$.
The locus is $y^2 = 2a(x - \frac{a}{2})$.
Comparing this with the standard form $Y^2 = 4AX$,where $Y = y$,$X = x - \frac{a}{2}$,and $4A = 2a$ (so $A = \frac{a}{2}$).
The directrix is given by $X = -A$,which means $x - \frac{a}{2} = -\frac{a}{2}$.
Therefore,$x = 0$.

(D) Let the equation of the tangent to the parabola ${y^2} = 8x$ be $y = mx + \frac{2}{m}$.
Since this line is also a tangent to the hyperbola $xy = -1$,we substitute $x = \frac{-1}{y}$ into the tangent equation:
$y = m(\frac{-1}{y}) + \frac{2}{m}$
$y^2 = -m + \frac{2y}{m}$
$my^2 - 2y + m^2 = 0$.
For the line to be a tangent,the discriminant of this quadratic equation must be zero:
$D = (-2)^2 - 4(m)(m^2) = 0$
$4 - 4m^3 = 0$
$m^3 = 1 \implies m = 1$.
Substituting $m = 1$ into the tangent equation $y = mx + \frac{2}{m}$,we get:
$y = (1)x + \frac{2}{1}$
$y = x + 2$.

(C) Let $L = \mathop {\lim }\limits_{x \to 0} \frac{(\cos x - 1)(\cos x - e^x)}{x^n}$.
Using Taylor series expansions near $x = 0$:
$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$
$e^x = 1 + x + \frac{x^2}{2!} + \dots$
Substituting these into the expression:
$\cos x - 1 = -\frac{x^2}{2} + O(x^4)$
$\cos x - e^x = (1 - \frac{x^2}{2} + \dots) - (1 + x + \frac{x^2}{2} + \dots) = -x - x^2 + O(x^3)$
Thus,the numerator is $(- \frac{x^2}{2} + O(x^4))(-x - x^2 + O(x^3)) = \frac{x^3}{2} + O(x^4)$.
For the limit to be a finite non-zero number,the power of $x$ in the denominator must match the lowest power of $x$ in the numerator.
Therefore,$n = 3$.

(D) Let the probabilities of hitting the plane in the four shots be $p_1 = 0.4$,$p_2 = 0.3$,$p_3 = 0.2$,and $p_4 = 0.1$.
The probability that the gun hits the plane is equal to $1$ minus the probability that the gun misses the plane in all four shots.
Let $q_i = 1 - p_i$ be the probability of missing the shot $i$.
$q_1 = 1 - 0.4 = 0.6$
$q_2 = 1 - 0.3 = 0.7$
$q_3 = 1 - 0.2 = 0.8$
$q_4 = 1 - 0.1 = 0.9$
The probability of missing all four shots is $P(\text{miss}) = q_1 \times q_2 \times q_3 \times q_4 = 0.6 \times 0.7 \times 0.8 \times 0.9 = 0.3024$.
Therefore,the probability that the gun hits the plane is $P(\text{hit}) = 1 - P(\text{miss}) = 1 - 0.3024 = 0.6976$.

(C) The line $QP$ lies along the negative $x$-axis,so its angle with the positive $x$-axis is $\pi$. The line $QR$ passes through $(0, 0)$ and $(3, 3\sqrt{3})$,so its slope is $m = \frac{3\sqrt{3} - 0}{3 - 0} = \sqrt{3}$. The angle $\theta$ that $QR$ makes with the positive $x$-axis is $\tan \theta = \sqrt{3}$,which means $\theta = \frac{\pi}{3}$.
The angle $\angle PQR$ is the angle between the line $QP$ (angle $\pi$) and the line $QR$ (angle $\frac{\pi}{3}$).
The measure of $\angle PQR = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.
The bisector of $\angle PQR$ will make an angle $\alpha$ with the positive $x$-axis,where $\alpha = \frac{\pi + \frac{\pi}{3}}{2} = \frac{4\pi/3}{2} = \frac{2\pi}{3}$.
The slope of the bisector is $\tan(\frac{2\pi}{3}) = -\sqrt{3}$.
Since the bisector passes through the origin $Q(0, 0)$,its equation is $y - 0 = -\sqrt{3}(x - 0)$,which simplifies to $y = -\sqrt{3}x$,or $\sqrt{3}x + y = 0$.

(D) In $\Delta ABC$,the Sine Rule states $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.
For option $(a)$: Given $a, \sin A, \sin B$. From $\frac{a}{\sin A} = 2R$,we find $R$. Then $b = 2R \sin B$ and $c = 2R \sin C = 2R \sin(180^{\circ} - (A+B)) = 2R \sin(A+B)$. Thus,the triangle is uniquely determined.
For option $(b)$: Given $a, b, c$. By the $SSS$ congruence criterion,the triangle is uniquely determined.
For option $(c)$: Given $a, \sin B, R$. We have $b = 2R \sin B$. Now we have two sides $a, b$ and the circum-radius $R$. Since $\sin A = \frac{a}{2R}$,angle $A$ is determined. Thus,the triangle is uniquely determined.
For option $(d)$: Given $a, \sin A, R$. We have $\frac{a}{\sin A} = 2R$. This is an identity that holds for any triangle with side $a$ and angle $A$ having circum-radius $R$. It does not provide enough information to fix the other sides or angles. Therefore,the triangle is not uniquely determined.

(B) The range of the expression $a \cos x + b \sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
For the equation $7 \cos x + 5 \sin x = 2k + 1$,the range of the left side is $[-\sqrt{7^2 + 5^2}, \sqrt{7^2 + 5^2}] = [-\sqrt{74}, \sqrt{74}]$.
Since $\sqrt{74} \approx 8.602$,we have $-8.602 \leq 2k + 1 \leq 8.602$.
Subtracting $1$ from all sides: $-9.602 \leq 2k \leq 7.602$.
Dividing by $2$: $-4.801 \leq k \leq 3.801$.
The integral values of $k$ are $\{-4, -3, -2, -1, 0, 1, 2, 3\}$.
Counting these values,we get $8$ integral values.

(C) The equation of the circle is $S: x^2+y^2+6x+6y-2=0$.
Point $Q$ lies on the $Y$-axis and on the line $5x-2y+6=0$.
Setting $x=0$ in the line equation: $5(0)-2y+6=0$ $\Rightarrow -2y=-6$ $\Rightarrow y=3$.
So,the coordinates of $Q$ are $(0, 3)$.
The length of the tangent from a point $(x_1, y_1)$ to a circle $S=0$ is given by $\sqrt{S_1}$.
Substituting $Q(0, 3)$ into the circle equation $S(x, y) = x^2+y^2+6x+6y-2$:
$S_1 = 0^2 + 3^2 + 6(0) + 6(3) - 2 = 0 + 9 + 0 + 18 - 2 = 25$.
Therefore,the length of the tangent $PQ = \sqrt{S_1} = \sqrt{25} = 5$.

(B) Given $\omega = - \frac{1}{2} + i\frac{\sqrt{3}}{2}$,we know that $\omega$ is a complex cube root of unity,so $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$. Also,$\omega^4 = \omega^3 \cdot \omega = \omega$.
Let $\Delta = \left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & -1 - \omega^2 & \omega^2 \\ 1 & \omega^2 & \omega^4 \end{array} \right|$.
Using the property $1 + \omega + \omega^2 = 0$,we have $-1 - \omega^2 = \omega$.
Substituting this and $\omega^4 = \omega$ into the determinant:
$\Delta = \left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{array} \right|$.
Applying the column operation $C_1 \to C_1 + C_2 + C_3$:
$\Delta = \left| \begin{array}{ccc} 1+1+1 & 1 & 1 \\ 1+\omega+\omega^2 & \omega & \omega^2 \\ 1+\omega^2+\omega & \omega^2 & \omega \end{array} \right| = \left| \begin{array}{ccc} 3 & 1 & 1 \\ 0 & \omega & \omega^2 \\ 0 & \omega^2 & \omega \end{array} \right|$.
Expanding along the first column:
$\Delta = 3(\omega \cdot \omega - \omega^2 \cdot \omega^2) = 3(\omega^2 - \omega^4) = 3(\omega^2 - \omega)$.
Factoring out $-\omega$:
$\Delta = 3\omega(\omega - 1)$.

(B) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$.
Since $(a + 2b)$ and $(5a - 4b)$ are perpendicular,their dot product is zero:
$(a + 2b) \cdot (5a - 4b) = 0$
Expanding the dot product:
$5(a \cdot a) - 4(a \cdot b) + 10(b \cdot a) - 8(b \cdot b) = 0$
Since $a \cdot a = |a|^2 = 1$ and $b \cdot b = |b|^2 = 1$,and $a \cdot b = b \cdot a$:
$5(1) + 6(a \cdot b) - 8(1) = 0$
$5 + 6(a \cdot b) - 8 = 0$
$6(a \cdot b) - 3 = 0$
$6(a \cdot b) = 3$
$a \cdot b = \frac{3}{6} = \frac{1}{2}$
Using the formula $a \cdot b = |a||b| \cos \theta$:
$(1)(1) \cos \theta = \frac{1}{2}$
$\cos \theta = \frac{1}{2}$
$\theta = 60^o$.

(A) Given the function $f(x) = 2x + \sin x$ for $x \in R$.
To check for one-to-one,we find the derivative: $f'(x) = 2 + \cos x$.
Since $-1 \le \cos x \le 1$,we have $f'(x) = 2 + \cos x \ge 2 - 1 = 1 > 0$.
Since $f'(x) > 0$ for all $x \in R$,the function $f(x)$ is strictly increasing,which implies it is one-to-one.
To check for onto,we observe the range of $f(x)$. Since $f(x)$ is a continuous function and $\lim_{x \to \infty} f(x) = \infty$ and $\lim_{x \to -\infty} f(x) = -\infty$,the range of $f(x)$ is $(-\infty, \infty) = R$.
Since the range equals the codomain,the function is onto.
Therefore,$f$ is one-to-one and onto.

(C) The function $f(x)$ can be rewritten by expanding the absolute value as:
$f(x) = \begin{cases} \frac{1}{2}(-x - 1), & x < -1 \\ \tan^{-1}x, & -1 \le x \le 1 \\ \frac{1}{2}(x - 1), & x > 1 \end{cases}$
To find the domain of the derivative $f'(x)$,we check the differentiability at the transition points $x = -1$ and $x = 1$.
For $x < -1$,$f'(x) = -\frac{1}{2}$.
For $-1 < x < 1$,$f'(x) = \frac{1}{1+x^2}$.
For $x > 1$,$f'(x) = \frac{1}{2}$.
At $x = -1$:
Left-hand derivative $f'(-1^-) = -\frac{1}{2}$.
Right-hand derivative $f'(-1^+) = \frac{1}{1+(-1)^2} = \frac{1}{2}$.
Since $f'(-1^-) \neq f'(-1^+)$,the function is not differentiable at $x = -1$.
At $x = 1$:
Left-hand derivative $f'(1^-) = \frac{1}{1+(1)^2} = \frac{1}{2}$.
Right-hand derivative $f'(1^+) = \frac{1}{2}$.
Since $f'(1^-) = f'(1^+)$,the function is differentiable at $x = 1$.
Thus,the derivative $f'(x)$ exists for all $x \in R$ except $x = -1$.
Therefore,the domain of $f'(x)$ is $R - \{-1\}$.

(A) Given function is $f(x) = 3\sin x - 4\sin^3 x$.
Using the trigonometric identity $\sin(3x) = 3\sin x - 4\sin^3 x$,we can write $f(x) = \sin(3x)$.
For the function to be increasing,its derivative must be positive: $f'(x) > 0$.
$f'(x) = \frac{d}{dx}(\sin(3x)) = 3\cos(3x)$.
Setting $3\cos(3x) > 0$,we get $\cos(3x) > 0$.
The cosine function is positive in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$ for its argument.
Thus,$-\frac{\pi}{2} < 3x < \frac{\pi}{2}$.
Dividing by $3$,we get $-\frac{\pi}{6} < x < \frac{\pi}{6}$.
The length of this interval is $\frac{\pi}{6} - (-\frac{\pi}{6}) = \frac{2\pi}{6} = \frac{\pi}{3}$.

(D) Given the curve equation: $y^3 + 3x^2 = 12y$.
Differentiating both sides with respect to $x$:
$3y^2 \frac{dy}{dx} + 6x = 12 \frac{dy}{dx}$.
Rearranging to find $\frac{dy}{dx}$:
$\frac{dy}{dx}(3y^2 - 12) = -6x \implies \frac{dy}{dx} = \frac{-6x}{3y^2 - 12} = \frac{2x}{4 - y^2}$.
For the tangent to be vertical (parallel to the $y$-axis),the slope $\frac{dy}{dx}$ must be undefined,which occurs when the denominator is zero:
$4 - y^2 = 0 \implies y^2 = 4 \implies y = \pm 2$.
Case $1$: If $y = 2$,then $2^3 + 3x^2 = 12(2) \implies 8 + 3x^2 = 24 \implies 3x^2 = 16 \implies x^2 = \frac{16}{3} \implies x = \pm \frac{4}{\sqrt{3}}$.
Case $2$: If $y = -2$,then $(-2)^3 + 3x^2 = 12(-2) \implies -8 + 3x^2 = -24 \implies 3x^2 = -16$,which has no real solution for $x$.
Thus,the points are $\left( \pm \frac{4}{\sqrt{3}}, 2 \right)$.

(B) Let $I = \int_{1/e}^{\tan x} \frac{t \, dt}{1 + t^2} + \int_{1/e}^{\cot x} \frac{dt}{t(1 + t^2)}$.
For the first integral,let $u = 1 + t^2$,then $du = 2t \, dt$,so $\int \frac{t \, dt}{1 + t^2} = \frac{1}{2} \ln(1 + t^2)$.
Evaluating from $1/e$ to $\tan x$: $\frac{1}{2} [\ln(1 + \tan^2 x) - \ln(1 + 1/e^2)] = \frac{1}{2} [\ln(\sec^2 x) - \ln(1 + 1/e^2)]$.
For the second integral,$\int \frac{dt}{t(1 + t^2)} = \int (\frac{1}{t} - \frac{t}{1 + t^2}) \, dt = \ln|t| - \frac{1}{2} \ln(1 + t^2)$.
Evaluating from $1/e$ to $\cot x$: $[\ln(\cot x) - \frac{1}{2} \ln(1 + \cot^2 x)] - [\ln(1/e) - \frac{1}{2} \ln(1 + 1/e^2)]$.
Since $1 + \cot^2 x = \csc^2 x$,this becomes $[\ln(\cot x) - \frac{1}{2} \ln(\csc^2 x)] - [\ln(1/e) - \frac{1}{2} \ln(1 + 1/e^2)]$.
Combining both parts:
$I = \frac{1}{2} \ln(\sec^2 x) - \frac{1}{2} \ln(1 + 1/e^2) + \ln(\cot x) - \frac{1}{2} \ln(\csc^2 x) - \ln(1/e) + \frac{1}{2} \ln(1 + 1/e^2)$.
$I = \ln(\sec x) + \ln(\cot x) - \ln(\csc x) - \ln(1/e)$.
$I = \ln(\frac{\sec x \cdot \cot x}{\csc x}) - \ln(1/e) = \ln(1) - \ln(1/e) = 0 - (-1) = 1$.

(NONE) Let $I = \int_{-\pi/4}^{\pi/4} \sin^4 x \, dx$. Since $f(x) = \sin^4 x$ is an even function,$I = 2 \int_0^{\pi/4} \sin^4 x \, dx$.
Using the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$,we have $\sin^4 x = (\frac{1 - \cos 2x}{2})^2 = \frac{1}{4}(1 - 2\cos 2x + \cos^2 2x)$.
Further,$\cos^2 2x = \frac{1 + \cos 4x}{2}$,so $\sin^4 x = \frac{1}{4}(1 - 2\cos 2x + \frac{1 + \cos 4x}{2}) = \frac{1}{4}(\frac{3}{2} - 2\cos 2x + \frac{1}{2}\cos 4x) = \frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$.
Integrating this,$I = 2 \int_0^{\pi/4} (\frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x) \, dx = 2 [\frac{3}{8}x - \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x]_0^{\pi/4}$.
Evaluating at the limits: $I = 2 [(\frac{3}{8} \cdot \frac{\pi}{4} - \frac{1}{4}\sin \frac{\pi}{2} + \frac{1}{32}\sin \pi) - (0)] = 2 [\frac{3\pi}{32} - \frac{1}{4}] = \frac{3\pi}{16} - \frac{1}{2}$.

(D) Let $I = \int_1^5 (|x - 3| + |1 - x|) \, dx$.
Since $x \in [1, 5]$,$|1 - x| = x - 1$.
So,$I = \int_1^5 |x - 3| \, dx + \int_1^5 (x - 1) \, dx$.
For the first part,$\int_1^5 |x - 3| \, dx = \int_1^3 -(x - 3) \, dx + \int_3^5 (x - 3) \, dx$.
$= [-\frac{(x - 3)^2}{2}]_1^3 + [\frac{(x - 3)^2}{2}]_3^5 = (0 - (-2)) + (2 - 0) = 2 + 2 = 4$.
For the second part,$\int_1^5 (x - 1) \, dx = [\frac{x^2}{2} - x]_1^5 = (\frac{25}{2} - 5) - (\frac{1}{2} - 1) = 7.5 - (-0.5) = 8$.
Therefore,$I = 4 + 8 = 12$.

(A) Let $I = \int_{-1/2}^{1/2} [x] dx + \int_{-1/2}^{1/2} \log \left( \frac{1+x}{1-x} \right) dx$.
Consider the function $f(x) = \log \left( \frac{1+x}{1-x} \right)$.
Since $f(-x) = \log \left( \frac{1-x}{1+x} \right) = -\log \left( \frac{1+x}{1-x} \right) = -f(x)$,$f(x)$ is an odd function.
Therefore,$\int_{-1/2}^{1/2} \log \left( \frac{1+x}{1-x} \right) dx = 0$.
Now,$I = \int_{-1/2}^{1/2} [x] dx$.
In the interval $[-1/2, 0)$,$[x] = -1$.
In the interval $[0, 1/2]$,$[x] = 0$.
Thus,$I = \int_{-1/2}^{0} (-1) dx + \int_{0}^{1/2} (0) dx = -[x]_{-1/2}^{0} = -(0 - (-1/2)) = -1/2$.

(A) First,evaluate the integrals for ${F_1}(x)$ and ${F_2}(x)$.
${F_1}(x) = \int_2^x (2t - 5) dt = [t^2 - 5t]_2^x = (x^2 - 5x) - (2^2 - 5(2)) = x^2 - 5x + 6$.
${F_2}(x) = \int_0^x 2t dt = [t^2]_0^x = x^2 - 0^2 = x^2$.
To find the point of intersection,set ${F_1}(x) = {F_2}(x)$:
$x^2 - 5x + 6 = x^2$.
Subtracting $x^2$ from both sides,we get $-5x + 6 = 0$,which implies $5x = 6$,so $x = \frac{6}{5}$.
Now,substitute $x = \frac{6}{5}$ into ${F_2}(x)$ to find the $y$-coordinate:
$y = x^2 = \left( \frac{6}{5} \right)^2 = \frac{36}{25}$.
Thus,the point of intersection is $\left( \frac{6}{5}, \frac{36}{25} \right)$.

(C) The function is defined as $F(x) = \int_1^x {|t| \, dt}$.
By the Fundamental Theorem of Calculus,$F'(x) = |x|$.
Since $|x| \ge 0$ for all $x \in \left[ -\frac{1}{2}, \frac{1}{2} \right]$,the function $F(x)$ is non-decreasing on the given interval.
Therefore,the maximum value occurs at the right endpoint $x = \frac{1}{2}$.
$F\left( \frac{1}{2} \right) = \int_1^{1/2} {|t| \, dt} = -\int_{1/2}^1 {|t| \, dt}$.
Since $t > 0$ in the interval $\left[ \frac{1}{2}, 1 \right]$,we have $|t| = t$.
$F\left( \frac{1}{2} \right) = -\int_{1/2}^1 {t \, dt} = -\left[ \frac{t^2}{2} \right]_{1/2}^1 = -\left( \frac{1}{2} - \frac{1}{8} \right) = -\left( \frac{4-1}{8} \right) = -\frac{3}{8}$.

(A) Given $f(x) = \int_{1}^{x} \sqrt{2 - t^2} dt$.
Using the Leibniz integral rule,we differentiate $f(x)$ with respect to $x$:
$f'(x) = \frac{d}{dx} \int_{1}^{x} \sqrt{2 - t^2} dt = \sqrt{2 - x^2}$.
The given equation is $x^2 - f'(x) = 0$,which implies $x^2 = f'(x)$.
Substituting $f'(x)$,we get $x^2 = \sqrt{2 - x^2}$.
Squaring both sides,we obtain $x^4 = 2 - x^2$,which simplifies to $x^4 + x^2 - 2 = 0$.
Let $u = x^2$. Then the equation becomes $u^2 + u - 2 = 0$.
Factoring the quadratic,we get $(u + 2)(u - 1) = 0$.
Since $u = x^2$,we have $x^2 = -2$ (which gives no real roots) or $x^2 = 1$.
Thus,$x = \pm 1$.

(B) The given curves are $y = |x| - 1$ and $y = -|x| + 1$.
These curves represent a square with vertices at $(1, 0), (0, 1), (-1, 0),$ and $(0, -1)$.
Alternatively,we can analyze the regions:
For $x \ge 0$,the curves are $y = x - 1$ and $y = -x + 1$.
For $x < 0$,the curves are $y = -x - 1$ and $y = x + 1$.
The region is symmetric about both axes.
The area in the first quadrant is bounded by $y = x - 1$ (for $x \ge 1$,but here the intersection is at $x=1, y=0$) and $y = -x + 1$.
Actually,the region is a square formed by the lines $y = x - 1, y = -x - 1, y = -x + 1,$ and $y = x + 1$.
The area of this square is $4 \times (\text{Area of triangle in one quadrant})$.
Each triangle in a quadrant has base $1$ and height $1$.
Area $= 4 \times (\frac{1}{2} \times 1 \times 1) = 2$ square units.

(A) Since $A$ and $B$ are independent events,$A$ and $B'$ are also independent,and $A'$ and $B$ are independent.
Given $P(A \cap B') = P(A) \times P(B') = P(A)(1 - P(B)) = \frac{3}{25}$ .....$(i)$
Given $P(A' \cap B) = P(A') \times P(B) = (1 - P(A))P(B) = \frac{8}{25}$ .....$(ii)$
Let $P(A) = x$ and $P(B) = y$. Then the equations become:
$x(1 - y) = \frac{3}{25} \Rightarrow x - xy = \frac{3}{25}$
$y(1 - x) = \frac{8}{25} \Rightarrow y - xy = \frac{8}{25}$
Subtracting the two equations:
$x - y = \frac{3}{25} - \frac{8}{25} = -\frac{5}{25} = -\frac{1}{5}$
$y = x + \frac{1}{5}$
Substitute $y$ into the first equation:
$x(1 - (x + \frac{1}{5})) = \frac{3}{25}$
$x(\frac{4}{5} - x) = \frac{3}{25}$
$\frac{4}{5}x - x^2 = \frac{3}{25}$
$25x^2 - 20x + 3 = 0$
$(5x - 1)(5x - 3) = 0$
Thus,$P(A) = \frac{1}{5}$ or $P(A) = \frac{3}{5}$.
Comparing with the given options,the correct value is $\frac{1}{5}$.

(D) The event that at most one of $A$ and $B$ occurs means either none of them occur,or exactly one of them occurs.
This is represented by the set $(A' \cap B') \cup (A \cap B') \cup (A' \cap B)$.
Since these are mutually exclusive events,the probability is $P(A' \cap B') + P(A \cap B') + P(A' \cap B)$,which matches option $A$.
Also,the event 'at most one occurs' is the complement of the event 'both occur',which is $1 - P(A \cap B)$,matching option $B$.
Option $C$ is equivalent to $1 - P(A \cap B)$ as well,since $P(A') + P(B') + P(A \cup B) - 1 = (1 - P(A)) + (1 - P(B)) + (P(A) + P(B) - P(A \cap B)) - 1 = 1 - P(A \cap B)$.
Therefore,all options are correct.

(B) For a system of linear equations to have no solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and at least one of the determinants $\Delta_1$ or $\Delta_2$ must be non-zero.
$\Delta = \begin{vmatrix} k+1 & 8 \\ k & k+3 \end{vmatrix} = (k+1)(k+3) - 8k = k^2 + 4k + 3 - 8k = k^2 - 4k + 3 = (k-3)(k-1)$.
Setting $\Delta = 0$,we get $k = 1$ or $k = 3$.
Now,calculate $\Delta_1$ and $\Delta_2$ for these values:
$\Delta_1 = \begin{vmatrix} 4k & 8 \\ 3k-1 & k+3 \end{vmatrix} = 4k(k+3) - 8(3k-1) = 4k^2 + 12k - 24k + 8 = 4k^2 - 12k + 8 = 4(k-1)(k-2)$.
$\Delta_2 = \begin{vmatrix} k+1 & 4k \\ k & 3k-1 \end{vmatrix} = (k+1)(3k-1) - 4k^2 = 3k^2 + 2k - 1 - 4k^2 = -k^2 + 2k - 1 = -(k-1)^2$.
Case $k=1$: $\Delta = 0, \Delta_1 = 0, \Delta_2 = 0$. This leads to infinitely many solutions (or consistency).
Case $k=3$: $\Delta = 0, \Delta_1 = 4(3-1)(3-2) = 8 \neq 0, \Delta_2 = -(3-1)^2 = -4 \neq 0$.
Since $\Delta = 0$ and $\Delta_1, \Delta_2 \neq 0$ for $k=3$,the system has no solutions.
Thus,there is only $1$ value of $k$.

(C) The scalar triple product is defined as $[U V W] = U \cdot (V \times W)$.
First,we calculate the cross product $V \times W$:
$V \times W = \begin{vmatrix} i & j & k \\ 2 & 1 & -1 \\ 1 & 0 & 3 \end{vmatrix} = i(3 - 0) - j(6 - (-1)) + k(0 - 1) = 3i - 7j - k$.
Let $A = V \times W = 3i - 7j - k$.
The scalar triple product is $[U V W] = U \cdot A = |U| |A| \cos \theta$,where $\theta$ is the angle between $U$ and $A$.
Since $U$ is a unit vector,$|U| = 1$.
Thus,$[U V W] = |A| \cos \theta$.
The maximum value occurs when $\cos \theta = 1$,which is $|A|$.
$|A| = \sqrt{3^2 + (-7)^2 + (-1)^2} = \sqrt{9 + 49 + 1} = \sqrt{59}$.
Therefore,the maximum value is $\sqrt{59}$.

(D) The graph of $f(x)$ is given by the equation $y = (x + 1)^2$ for $x \ge -1$.
To find the reflection of the graph of $f(x)$ with respect to the line $y = x$,we interchange $x$ and $y$.
Thus,the equation for the graph of $g(x)$ becomes $x = (y + 1)^2$ for $y \ge -1$.
Solving for $y$:
$y + 1 = \pm\sqrt{x}$
Since $y \ge -1$,we have $y + 1 \ge 0$,so we take the positive root:
$y + 1 = \sqrt{x}$
$y = \sqrt{x} - 1$
Since the domain of $f(x)$ is $x \ge -1$ and its range is $y \ge 0$,the domain of $g(x)$ is $x \ge 0$ and its range is $y \ge -1$.
Therefore,$g(x) = \sqrt{x} - 1$ for $x \ge 0$.

(C) Let $L = \lim_{x \to 0} \left\{ \frac{f(1 + x)}{f(1)} \right\}^{\frac{1}{x}}$.
Taking natural logarithm on both sides,$\ln L = \lim_{x \to 0} \frac{1}{x} \ln \left( \frac{f(1 + x)}{f(1)} \right)$.
Using the definition of the derivative or $L$'Hopital's rule,$\ln L = \lim_{x \to 0} \frac{\ln f(1 + x) - \ln f(1)}{x}$.
Applying $L$'Hopital's rule,$\ln L = \lim_{x \to 0} \frac{f'(1 + x)}{f(1 + x)} = \frac{f'(1)}{f(1)}$.
Given $f(1) = 3$ and $f'(1) = 6$,we have $\ln L = \frac{6}{3} = 2$.
Therefore,$L = e^2$.

(D) Given the equation: $\int_{\pi /2}^{\alpha} \sin x \, dx = \sin 2\alpha$
Evaluating the integral: $[-\cos x]_{\pi /2}^{\alpha} = \sin 2\alpha$
$-(\cos \alpha - \cos(\pi /2)) = \sin 2\alpha$
$-\cos \alpha = \sin 2\alpha$
Using the identity $\sin 2\alpha = 2 \sin \alpha \cos \alpha$:
$-\cos \alpha = 2 \sin \alpha \cos \alpha$
$\cos \alpha (1 + 2 \sin \alpha) = 0$
Case $1$: $\cos \alpha = 0$. In the interval $[0, 2\pi]$,$\alpha = \frac{\pi}{2}, \frac{3\pi}{2}$.
Case $2$: $1 + 2 \sin \alpha = 0 \Rightarrow \sin \alpha = -1/2$. In the interval $[0, 2\pi]$,$\alpha = \frac{7\pi}{6}, \frac{11\pi}{6}$.
Comparing with the options provided,the values $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ satisfy the equation.

(C) Let $J = \int_{3}^{3 + 3T} f(2x) dx$. Substitute $u = 2x$,so $du = 2 dx$ or $dx = \frac{1}{2} du$.
When $x = 3, u = 6$. When $x = 3 + 3T, u = 6 + 6T$.
Thus,$J = \frac{1}{2} \int_{6}^{6 + 6T} f(u) du$.
Since $f$ is periodic with period $T$,$\int_{a}^{a + nT} f(x) dx = n \int_{0}^{T} f(x) dx$ for any $a \in \mathbb{R}$ and $n \in \mathbb{Z}$.
Here,the integral is over an interval of length $6T$,which is $6$ times the period $T$.
Therefore,$\int_{6}^{6 + 6T} f(u) du = 6 \int_{0}^{T} f(u) du = 6I$.
Substituting this back,$J = \frac{1}{2} \times 6I = 3I$.

(D) Let $F(x) = \int_0^{x^2} \frac{t^2 - 5t + 4}{2 + e^t} \,dt$.
Using the Leibniz integral rule,we differentiate $F(x)$ with respect to $x$:
$F'(x) = \frac{(x^2)^2 - 5(x^2) + 4}{2 + e^{x^2}} \cdot \frac{d}{dx}(x^2)$
$F'(x) = \frac{x^4 - 5x^2 + 4}{2 + e^{x^2}} \cdot 2x$
For extremum points,we set $F'(x) = 0$:
$\frac{(x^2 - 1)(x^2 - 4)}{2 + e^{x^2}} \cdot 2x = 0$
This implies $x = 0$ or $x^2 = 1$ or $x^2 = 4$.
Solving these,we get $x = 0$,$x = \pm 1$,and $x = \pm 2$.
Since all these values are included in the options provided,the correct answer is $D$.

(D) The probability of getting $k$ heads in $n$ tosses of a fair coin is given by the binomial distribution: $P(X = k) = \binom{n}{k} (\frac{1}{2})^n$.
Given that $P(X = 4), P(X = 5), P(X = 6)$ are in $H.P.$,their reciprocals must be in $A.P.$
Thus,$\frac{1}{P(X = 4)}, \frac{1}{P(X = 5)}, \frac{1}{P(X = 6)}$ are in $A.P.$
This implies $\frac{2}{P(X = 5)} = \frac{1}{P(X = 4)} + \frac{1}{P(X = 6)}$.
Substituting the values: $\frac{2}{\binom{n}{5}} = \frac{1}{\binom{n}{4}} + \frac{1}{\binom{n}{6}}$.
Using $\binom{n}{k} = \frac{n!}{k!(n-k)!}$,we get $\frac{2 \times 5!(n-5)!}{n!} = \frac{4!(n-4)!}{n!} + \frac{6!(n-6)!}{n!}$.
Multiplying by $\frac{n!}{4!(n-6)!}$,we obtain: $2 \times 5 \times (n-5) = (n-4)(n-5) + 6 \times 5$.
$10n - 50 = n^2 - 9n + 20 + 30$.
$n^2 - 19n + 100 = 0$.
The discriminant $D = (-19)^2 - 4(1)(100) = 361 - 400 = -39$.
Since $D < 0$,there is no real integer solution for $n$. Thus,the correct option is $D$.