IIT JEE 1998 Mathematics Question Paper with Answer and Solution

(B) We are given the sum $S = \sum\limits_{n = 1}^{13} {({i^n} + {i^{n + 1}})} $.
This can be written as $S = \sum\limits_{n = 1}^{13} {i^n} + \sum\limits_{n = 1}^{13} {i^{n + 1}}$.
Both are geometric series with $13$ terms.
For the first series,the sum is $\frac{i(1 - i^{13})}{1 - i}$. Since $i^4 = 1$,$i^{13} = i^{12} \times i = 1 \times i = i$.
So,the first sum is $\frac{i(1 - i)}{1 - i} = i$.
For the second series,the sum is $\frac{i^2(1 - i^{13})}{1 - i} = \frac{-1(1 - i)}{1 - i} = -1$.
Therefore,$S = i + (-1) = i - 1$.

(D) We know that $1 + \omega + \omega^2 = 0$,which implies $1 + \omega = -\omega^2$.
Substituting this into the expression:
$(1 + \omega - \omega^2)^7 = (-\omega^2 - \omega^2)^7$
$= (-2\omega^2)^7$
$= (-2)^7 \times (\omega^2)^7$
$= -128 \times \omega^{14}$
Since $\omega^3 = 1$,we have $\omega^{14} = (\omega^3)^4 \times \omega^2 = 1^4 \times \omega^2 = \omega^2$.
Therefore,the expression equals $-128\omega^2$.

(C) Let the first term be $a$ and the common difference be $d$.
Given $T_m = a + (m - 1)d = \frac{1}{n}$ and $T_n = a + (n - 1)d = \frac{1}{m}$.
Subtracting the two equations: $(m - n)d = \frac{1}{n} - \frac{1}{m} = \frac{m - n}{mn}$.
Thus,$d = \frac{1}{mn}$.
Substituting $d$ into the first equation: $a + (m - 1)\frac{1}{mn} = \frac{1}{n} \implies a + \frac{1}{n} - \frac{1}{mn} = \frac{1}{n} \implies a = \frac{1}{mn}$.
Now,$T_{mn} = a + (mn - 1)d = \frac{1}{mn} + (mn - 1)\frac{1}{mn} = \frac{1 + mn - 1}{mn} = \frac{mn}{mn} = 1$.

(B) Given that $x, y, z$ are in $G.P.$,we have $y^2 = xz$.
Taking the natural logarithm on both sides,we get $2 \ln y = \ln x + \ln z$.
Adding $2$ to both sides,we get $2 + 2 \ln y = 2 + \ln x + \ln z$.
This can be rewritten as $2(1 + \ln y) = (1 + \ln x) + (1 + \ln z)$.
This implies that $(1 + \ln x), (1 + \ln y), (1 + \ln z)$ are in $A.P.$
Since the reciprocals of terms in $A.P.$ are in $H.P.$,it follows that $\frac{1}{1 + \ln x}, \frac{1}{1 + \ln y}, \frac{1}{1 + \ln z}$ are in $H.P.$

(B) Since at any place,any of the digits $2, 5$ and $7$ can be used,the total number of such positive $n$-digit numbers is $3^n$.
We need to form at least $900$ distinct numbers,so we set the inequality $3^n \ge 900$.
Calculating powers of $3$:
$3^1 = 3$
$3^2 = 9$
$3^3 = 27$
$3^4 = 81$
$3^5 = 243$
$3^6 = 729$
$3^7 = 2187$
Since $3^6 = 729 < 900$ and $3^7 = 2187 \ge 900$,the smallest value of $n$ is $7$.

(C) We evaluate each option:
$(a)$ $\sin 15^\circ = \sin(45^\circ - 30^\circ) = \frac{\sqrt{6} - \sqrt{2}}{4}$,which is irrational.
$(b)$ $\cos 15^\circ = \cos(45^\circ - 30^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4}$,which is irrational.
$(c)$ $\sin 15^\circ \cos 15^\circ = \frac{1}{2}(2 \sin 15^\circ \cos 15^\circ) = \frac{1}{2} \sin 30^\circ = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$,which is rational.
$(d)$ $\sin 15^\circ \cos 75^\circ = \sin 15^\circ \sin 15^\circ = \sin^2 15^\circ = \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)^2 = \frac{6 + 2 - 2\sqrt{12}}{16} = \frac{8 - 4\sqrt{3}}{16} = \frac{2 - \sqrt{3}}{4}$,which is irrational.
Thus,the correct option is $(c)$.

(B) Given that $\sin P, \sin Q, \sin R$ are in $A.P.$
By the Sine Rule,$\frac{a}{\sin P} = \frac{b}{\sin Q} = \frac{c}{\sin R} = 2R$,where $R$ is the circumradius.
This implies $\sin P = \frac{a}{2R}, \sin Q = \frac{b}{2R}, \sin R = \frac{c}{2R}$.
Since $\sin P, \sin Q, \sin R$ are in $A.P.$,it follows that $a, b, c$ are in $A.P.$
Let $p_1, p_2, p_3$ be the altitudes from vertices $P, Q, R$ respectively.
The area of the triangle $\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$.
Thus,$p_1 = \frac{2\Delta}{a}, p_2 = \frac{2\Delta}{b}, p_3 = \frac{2\Delta}{c}$.
Since $a, b, c$ are in $A.P.$,their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $H.P.$
Multiplying by $2\Delta$,we get $\frac{2\Delta}{a}, \frac{2\Delta}{b}, \frac{2\Delta}{c}$ are in $H.P.$
Therefore,$p_1, p_2, p_3$ are in $H.P.$

(C) In a parallelogram,the diagonals bisect each other at their midpoint.
Let the midpoint of diagonal $PR$ be $M_1$ and the midpoint of diagonal $QS$ be $M_2$.
Since $M_1 = M_2$,we have:
Midpoint of $PR = (\frac{1+5}{2}, \frac{2+7}{2}) = (3, 4.5)$
Midpoint of $QS = (\frac{4+a}{2}, \frac{6+b}{2})$
Equating the coordinates:
$\frac{4+a}{2} = 3$ $\Rightarrow 4+a = 6$ $\Rightarrow a = 2$
$\frac{6+b}{2} = 4.5$ $\Rightarrow 6+b = 9$ $\Rightarrow b = 3$
Thus,$a = 2$ and $b = 3$.

(A) Let the vertices be $P(x_1, y_1), Q(x_2, y_2),$ and $R(x_3, y_3),$ where $x_i, y_i \in \mathbb{Q}$.
$1$. The Centroid is given by $\left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} \right)$. Since the sum and quotient of rational numbers are rational,the centroid is always a rational point.
$2$. The Circumcentre $(x, y)$ satisfies the equations $(x-x_1)^2 + (y-y_1)^2 = (x-x_2)^2 + (y-y_2)^2$ and $(x-x_2)^2 + (y-y_2)^2 = (x-x_3)^2 + (y-y_3)^2$. These simplify to linear equations in $x$ and $y$ with rational coefficients. Thus,the circumcentre is always a rational point.
$3$. The Incentre is given by $\left( \frac{ax_1+bx_2+cx_3}{a+b+c}, \frac{ay_1+by_2+cy_3}{a+b+c} \right)$,where $a, b, c$ are side lengths. Since side lengths involve square roots (e.g.,$a = \sqrt{(x_2-x_3)^2 + (y_2-y_3)^2}$),they are generally irrational. Thus,the incentre is not necessarily a rational point.
Therefore,both the Centroid and Circumcentre are always rational points. Given the options,the question implies identifying which of the listed points are rational. Since the Centroid and Circumcentre are rational,and the Incentre is not,the most appropriate choice is the one containing the rational points.

(D) Let the equations of the diagonals be $L_1: x + 3y = 4$ and $L_2: 6x - 2y = 7$.
The slope of $L_1$ is $m_1 = -1/3$.
The slope of $L_2$ is $m_2 = -6/(-2) = 3$.
Since $m_1 \times m_2 = (-1/3) \times 3 = -1$,the diagonals are perpendicular to each other.
$A$ parallelogram whose diagonals are perpendicular is a rhombus.

(B) Given circles are $S_1 \equiv x^2 + y^2 = 2^2$ and $S_2 \equiv x^2 + y^2 - 6x - 8y - 24 = 0$.
For $S_2$,the centre is $(3, 4)$ and radius $r_2 = \sqrt{3^2 + 4^2 - (-24)} = \sqrt{9 + 16 + 24} = \sqrt{49} = 7$.
For $S_1$,the centre is $C_1 = (0, 0)$ and radius $r_1 = 2$.
The distance between the centres $C_1(0, 0)$ and $C_2(3, 4)$ is $d = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{9 + 16} = 5$.
Since $d = |r_2 - r_1| = |7 - 2| = 5$,the two circles touch each other internally.
When two circles touch internally,there is exactly $1$ common tangent.

(C) The given equation is $16x^2 + 25y^2 = 400$. Dividing by $400$,we get $\frac{x^2}{25} + \frac{y^2}{16} = 1$,which is an ellipse with $a^2 = 25$ and $b^2 = 16$.
Here,$a = 5$ and $b = 4$. The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
The foci are $(\pm ae, 0) = (\pm 5 \times \frac{3}{5}, 0) = (\pm 3, 0)$,which match $F_1$ and $F_2$.
By the definition of an ellipse,the sum of the distances from any point $P$ on the ellipse to the two foci is equal to the length of the major axis,which is $2a$.
Thus,$PF_1 + PF_2 = 2a = 2 \times 5 = 10$.

(A) The outcome of each coin toss is an independent event.
The appearance of a head on the $5^{th}$ toss does not depend on the outcomes of the first four tosses.
Therefore,the probability of getting a head on the $5^{th}$ toss is $P(\text{Head}) = \frac{1}{2}$.

(B) Let the four machines be $M_1, M_2, F_1, F_2$,where $F$ denotes a faulty machine and $M$ denotes a working machine.
We need to find the probability that both faulty machines are identified in exactly two tests.
This means the first test must result in a faulty machine,and the second test must also result in a faulty machine.
The total number of ways to choose $2$ machines out of $4$ for the first two tests is $^4P_2 = 4 \times 3 = 12$.
The number of ways to choose $2$ faulty machines out of $2$ for the first two tests is $^2P_2 = 2 \times 1 = 2$.
Therefore,the required probability is $\frac{2}{12} = \frac{1}{6}$.
Alternatively,the probability that the first machine tested is faulty is $\frac{2}{4}$.
Given that the first machine was faulty,the probability that the second machine tested is also faulty is $\frac{1}{3}$.
Thus,the required probability is $\frac{2}{4} \times \frac{1}{3} = \frac{1}{6}$.

(B) The total number of ways to arrange $7$ white balls and $3$ black balls in a row is given by the combination formula $\binom{10}{3} = \frac{10!}{7!3!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
To ensure no two black balls are adjacent,we first arrange the $7$ white balls in a row. This creates $8$ possible spaces (gaps) where the $3$ black balls can be placed (one before the first ball,one after the last ball,and $6$ between the balls).
The number of ways to choose $3$ spaces out of $8$ is $\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.
Therefore,the required probability is $\frac{56}{120} = \frac{7}{15}$.

(C) Given $a_n = \sum_{r=0}^n \frac{1}{^nC_r}$.
Let $b_n = \sum_{r=0}^n \frac{r}{^nC_r}$.
Using the property $^nC_r = ^nC_{n-r}$,we can write:
$b_n = \frac{0}{^nC_0} + \frac{1}{^nC_1} + \frac{2}{^nC_2} + \dots + \frac{n}{^nC_n}$.
Also,$b_n = \frac{n}{^nC_n} + \frac{n-1}{^nC_{n-1}} + \dots + \frac{0}{^nC_0} = \sum_{r=0}^n \frac{n-r}{^nC_r}$.
Adding both expressions for $b_n$:
$2b_n = \sum_{r=0}^n \frac{r + (n-r)}{^nC_r} = \sum_{r=0}^n \frac{n}{^nC_r} = n \sum_{r=0}^n \frac{1}{^nC_r}$.
Since $a_n = \sum_{r=0}^n \frac{1}{^nC_r}$,we have $2b_n = na_n$.
Therefore,$b_n = \frac{1}{2}na_n$.

(C) In a regular hexagon inscribed in a circle of unit radius,the side length is equal to the radius,so $A_0 A_1 = 1$.
Since the interior angle of a regular hexagon is $120^\circ$,in $\triangle A_0 A_1 A_2$,by the Law of Cosines:
$A_0 A_2^2 = A_0 A_1^2 + A_1 A_2^2 - 2(A_0 A_1)(A_1 A_2) \cos(120^\circ)$
$A_0 A_2^2 = 1^2 + 1^2 - 2(1)(1)(-\frac{1}{2}) = 1 + 1 + 1 = 3$
Thus,$A_0 A_2 = \sqrt{3}$.
By symmetry,$A_0 A_4 = A_0 A_2 = \sqrt{3}$.
The product of the lengths is $A_0 A_1 \times A_0 A_2 \times A_0 A_4 = 1 \times \sqrt{3} \times \sqrt{3} = 3$.

(C) The line $y = 4x + c$ touches the ellipse $\frac{x^2}{4} + y^2 = 1$.
Substituting $y = 4x + c$ into the equation of the ellipse:
$\frac{x^2}{4} + (4x + c)^2 = 1$
$x^2 + 4(16x^2 + 8cx + c^2) = 4$
$x^2 + 64x^2 + 32cx + 4c^2 - 4 = 0$
$65x^2 + 32cx + (4c^2 - 4) = 0$
For the line to touch the curve,the discriminant $\Delta$ must be $0$:
$\Delta = (32c)^2 - 4(65)(4c^2 - 4) = 0$
$1024c^2 - 16(65)(c^2 - 1) = 0$
Divide by $16$:
$64c^2 - 65(c^2 - 1) = 0$
$64c^2 - 65c^2 + 65 = 0$
$-c^2 + 65 = 0$
$c^2 = 65$
$c = \pm \sqrt{65}$
Thus,there are $2$ possible values for $c$.

(D) Given the equations $x^2 + y^2 = a^2$ and $xy = c^2$.
Substituting $y = \frac{c^2}{x}$ into the circle equation:
$x^2 + \frac{c^4}{x^2} = a^2$
$x^4 - a^2 x^2 + c^4 = 0$
This is a biquadratic equation in $x$. The roots are $x_1, x_2, x_3, x_4$.
From the properties of roots,the sum of roots $x_1 + x_2 + x_3 + x_4 = 0$ (coefficient of $x^3$ is $0$) and the product of roots $x_1 x_2 x_3 x_4 = c^4$.
Since the equations are symmetric with respect to $x$ and $y$,by symmetry,$y_1 + y_2 + y_3 + y_4 = 0$ and $y_1 y_2 y_3 y_4 = c^4$.
Thus,all the given options are correct.

(D) Let $f(x) = \frac{\sqrt{1 - \cos 2(x - 1)}}{x - 1}$.
We know that $1 - \cos 2\theta = 2\sin^2 \theta$,so $1 - \cos 2(x - 1) = 2\sin^2(x - 1)$.
Thus,$f(x) = \frac{\sqrt{2\sin^2(x - 1)}}{x - 1} = \frac{\sqrt{2}|\sin(x - 1)|}{x - 1}$.
Right Hand Limit $(RHL)$: $\mathop {\lim }\limits_{x \to 1^+} f(x) = \mathop {\lim }\limits_{h \to 0^+} \frac{\sqrt{2}|\sin h|}{h} = \mathop {\lim }\limits_{h \to 0^+} \sqrt{2} \frac{\sin h}{h} = \sqrt{2}$.
Left Hand Limit $(LHL)$: $\mathop {\lim }\limits_{x \to 1^-} f(x) = \mathop {\lim }\limits_{h \to 0^+} \frac{\sqrt{2}|\sin(-h)|}{-h} = \mathop {\lim }\limits_{h \to 0^+} \frac{\sqrt{2}\sin h}{-h} = -\sqrt{2}$.
Since $\text{RHL} \neq \text{LHL}$,the limit does not exist.

(A) Let $P(W_i)$ and $P(B_i)$ be the probabilities of drawing one white and one black ball from the $i$-th box where $i = 1, 2, 3$ respectively.
$P(W_1) = \frac{3}{4}, P(B_1) = \frac{1}{4}$
$P(W_2) = \frac{2}{4} = \frac{1}{2}, P(B_2) = \frac{2}{4} = \frac{1}{2}$
$P(W_3) = \frac{1}{4}, P(B_3) = \frac{3}{4}$
Two white and one black ball may be drawn from $3$ boxes in the following three ways:

$Way 1$	$W, W, B$
$Way 2$	$W, B, W$
$Way 3$	$B, W, W$

Required probability $= P(W_1)P(W_2)P(B_3) + P(W_1)P(B_2)P(W_3) + P(B_1)P(W_2)P(W_3)$
$= (\frac{3}{4} \times \frac{2}{4} \times \frac{3}{4}) + (\frac{3}{4} \times \frac{2}{4} \times \frac{1}{4}) + (\frac{1}{4} \times \frac{2}{4} \times \frac{1}{4})$
$= \frac{18}{64} + \frac{6}{64} + \frac{2}{64} = \frac{26}{64} = \frac{13}{32}$.

(C) Let the number of newspapers be $n$.
Since every student reads $5$ newspapers,the total number of readings is $300 \times 5 = 1500$.
Since every newspaper is read by $60$ students,the total number of readings is also $60 \times n$.
Equating the two,we get $60n = 1500$.
Therefore,$n = \frac{1500}{60} = 25$.

(C) Let $n$ be the number of newspapers.
Total number of readings by students = $300 \times 5 = 1500$.
Total number of readings provided by newspapers = $n \times 60$.
Since both represent the total number of readings,we have:
$60n = 1500$
$n = \frac{1500}{60}$
$n = 25$.
Thus,the number of newspapers is $25$.

(B) Given the equation $3 \sin^2 x - 7 \sin x + 2 = 0$.
Let $t = \sin x$. Then the equation becomes $3t^2 - 7t + 2 = 0$.
Factoring the quadratic: $3t^2 - 6t - t + 2 = 0 \implies 3t(t - 2) - 1(t - 2) = 0 \implies (3t - 1)(t - 2) = 0$.
So,$t = \frac{1}{3}$ or $t = 2$.
Since $\sin x$ cannot be $2$,we have $\sin x = \frac{1}{3}$.
In the interval $[0, 2\pi]$,there are $2$ solutions for $\sin x = \frac{1}{3}$ (one in the first quadrant and one in the second quadrant).
In the interval $[2\pi, 4\pi]$,there are another $2$ solutions.
In the interval $[4\pi, 5\pi]$,there is $1$ solution (in the first quadrant relative to $4\pi$).
Total number of solutions = $2 + 2 + 1 = 5$.

(B) Given the determinant: $\Delta = \left| {\begin{array}{*{20}{c}}{6i}&{ - 3i}&1\\4&{3i}&{ - 1}\\{20}&3&i\end{array}} \right| = x + iy$
Expanding along the first row:
$\Delta = 6i((3i)(i) - (3)(-1)) - (-3i)((4)(i) - (20)(-1)) + 1((4)(3) - (20)(3i))$
$\Delta = 6i(3i^2 + 3) + 3i(4i + 20) + (12 - 60i)$
Since $i^2 = -1$,we have $3i^2 + 3 = 3(-1) + 3 = 0$.
$\Delta = 6i(0) + 12i^2 + 60i + 12 - 60i$
$\Delta = 0 + 12(-1) + 60i + 12 - 60i$
$\Delta = -12 + 12 + 60i - 60i = 0$
Thus,$x + iy = 0 + 0i$,which implies $x = 0$ and $y = 0$.

(C) The scalar triple product of three vectors $u, v, w$ is denoted by $[u, v, w] = u \cdot (v \times w)$.
By the cyclic property of the scalar triple product,$[u, v, w] = [v, w, u] = [w, u, v]$.
Option $(a)$ is $u \cdot (v \times w) = [u, v, w]$.
Option $(b)$ is $(v \times w) \cdot u = [v, w, u] = [u, v, w]$.
Option $(d)$ is $(u \times v) \cdot w = [u, v, w]$.
Option $(c)$ is $v \cdot (u \times w) = [v, u, w] = -[u, v, w]$.
Since $[u, v, w] \neq -[u, v, w]$ (in general),option $(c)$ is not equal to the others.

(A) Let $u$,$v$,and $w$ be vectors.
$1$. Consider $u \cdot (v \times w)$: The expression $(v \times w)$ results in a vector. The dot product of a vector $u$ with the resulting vector $(v \times w)$ is defined and results in a scalar. Thus,this expression is meaningful.
$2$. Consider $(u \cdot v) \cdot w$: The expression $(u \cdot v)$ results in a scalar. The dot product of a scalar with a vector $w$ is not defined. Thus,this expression is not meaningful.
$3$. Consider $(u \cdot v) \times w$: The expression $(u \cdot v)$ results in a scalar. The cross product of a scalar with a vector $w$ is not defined. Thus,this expression is not meaningful.
Therefore,only the first expression is meaningful.

(B) Let $f(x) = y$.
Since $f(x) = 3x - 5$,we have $y = 3x - 5$.
To find the inverse,we solve for $x$ in terms of $y$:
$y + 5 = 3x$
$x = \frac{y + 5}{3}$.
By definition,$x = f^{-1}(y)$,so $f^{-1}(y) = \frac{y + 5}{3}$.
Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{x + 5}{3}$.
Since $f(x) = 3x - 5$ is a linear function,it is both one-one (injective) and onto (surjective),therefore $f$ is invertible.

(D) Given $h(x) = \min \{ x, x^2 \}$.
We compare $x$ and $x^2$:
$x \le x^2 \Rightarrow x^2 - x \ge 0 \Rightarrow x(x - 1) \ge 0$.
This inequality holds for $x \le 0$ or $x \ge 1$.
Thus,$h(x) = \begin{cases} x & x \le 0 \\ x^2 & 0 < x < 1 \\ x & x \ge 1 \end{cases}$.
$h(x)$ is continuous for all $x \in \mathbb{R}$ because the pieces match at $x=0$ $(0=0)$ and $x=1$ $(1=1)$.
$h(x)$ is not differentiable at $x=0$ and $x=1$ because the left and right derivatives do not match at these points.
For $x > 1$,$h(x) = x$,so $h'(x) = 1$.
Therefore,all statements $(a)$,$(b)$,and $(c)$ are correct.

(A) Given: $g(f(x)) = |\sin x|$ and $f(g(x)) = (\sin \sqrt{x})^2$.
Let us test option $A$: $f(x) = \sin^2 x$ and $g(x) = \sqrt{x}$.
Step $1$: Calculate $g(f(x))$.
$g(f(x)) = g(\sin^2 x) = \sqrt{\sin^2 x} = |\sin x|$. This matches the given condition.
Step $2$: Calculate $f(g(x))$.
$f(g(x)) = f(\sqrt{x}) = (\sin \sqrt{x})^2$. This also matches the given condition.
Therefore,the correct option is $A$.

(D) Given $f(x) = \frac{x^2 - 1}{x^2 + 1}$.
We can rewrite the function as $f(x) = \frac{x^2 + 1 - 2}{x^2 + 1} = 1 - \frac{2}{x^2 + 1}$.
Since $x^2 \ge 0$ for all real $x$,we have $x^2 + 1 \ge 1$,which implies $0 < \frac{2}{x^2 + 1} \le 2$.
Subtracting this inequality from $1$,we get $1 - 2 \le 1 - \frac{2}{x^2 + 1} < 1 - 0$.
Thus,$-1 \le f(x) < 1$.
The minimum value is attained when $x^2 = 0$,i.e.,$x = 0$.
$f(0) = \frac{0^2 - 1}{0^2 + 1} = -1$.
Therefore,the minimum value of $f$ is $-1$.

(B) Given the function $f(x) = \cos x + \cos (\sqrt{2} x)$.
To find the maxima,we calculate the first derivative: $f'(x) = -\sin x - \sqrt{2} \sin (\sqrt{2} x)$.
Setting $f'(x) = 0$,we get $\sin x + \sqrt{2} \sin (\sqrt{2} x) = 0$.
At $x = 0$,$f'(0) = -\sin(0) - \sqrt{2} \sin(0) = 0$.
Now,check the second derivative: $f''(x) = -\cos x - 2 \cos (\sqrt{2} x)$.
At $x = 0$,$f''(0) = -\cos(0) - 2 \cos(0) = -1 - 2 = -3$.
Since $f''(0) < 0$,the function attains a local maximum at $x = 0$.
Because $\sqrt{2}$ is an irrational number,the function $f(x)$ is not periodic. The values of $\cos x$ and $\cos (\sqrt{2} x)$ can only both be $1$ simultaneously at $x = 0$. For any other $x \neq 0$,the sum $\cos x + \cos (\sqrt{2} x)$ will be strictly less than $2$. Thus,$x = 0$ is the unique point where the maximum value of $2$ is attained.

(A) Given $h(x) = f(x) - (f(x))^2 + (f(x))^3$.
Differentiating with respect to $x$,we get:
$h'(x) = f'(x) - 2f(x)f'(x) + 3(f(x))^2 f'(x)$
Factoring out $f'(x)$:
$h'(x) = f'(x) [1 - 2f(x) + 3(f(x))^2]$
To analyze the quadratic expression $3(f(x))^2 - 2f(x) + 1$,we complete the square:
$3(f(x))^2 - 2f(x) + 1 = 3 \left( (f(x))^2 - \frac{2}{3}f(x) + \frac{1}{3} \right)$
$= 3 \left( (f(x) - \frac{1}{3})^2 - \frac{1}{9} + \frac{1}{3} \right)$
$= 3 \left( (f(x) - \frac{1}{3})^2 + \frac{2}{9} \right)$
Since $(f(x) - \frac{1}{3})^2 \ge 0$,the expression $(f(x) - \frac{1}{3})^2 + \frac{2}{9}$ is always positive.
Therefore,$h'(x) = 3f'(x) \left( (f(x) - \frac{1}{3})^2 + \frac{2}{9} \right)$.
Since the term in the bracket is always positive,the sign of $h'(x)$ is the same as the sign of $f'(x)$.
Thus,$h$ is increasing whenever $f$ is increasing.

(A) Let $I = \int_0^{2a} \frac{f(x)}{f(x) + f(2a - x)} \, dx$ ..... $(i)$
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx$,we have:
$I = \int_0^{2a} \frac{f(2a - x)}{f(2a - x) + f(2a - (2a - x))} \, dx$
$I = \int_0^{2a} \frac{f(2a - x)}{f(2a - x) + f(x)} \, dx$ ..... $(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^{2a} \left( \frac{f(x)}{f(x) + f(2a - x)} + \frac{f(2a - x)}{f(2a - x) + f(x)} \right) \, dx$
$2I = \int_0^{2a} \frac{f(x) + f(2a - x)}{f(x) + f(2a - x)} \, dx$
$2I = \int_0^{2a} 1 \, dx$
$2I = [x]_0^{2a} = 2a$
$I = a$

(A) Given the equation: $\int_0^x {f(t)\,dt} = x + \int_x^1 {t\,f(t)\,dt}$
Rewrite the integral $\int_x^1 {t\,f(t)\,dt}$ as $-\int_1^x {t\,f(t)\,dt}$:
$\int_0^x {f(t)\,dt} = x - \int_1^x {t\,f(t)\,dt}$
Differentiating both sides with respect to $x$ using the Leibniz integral rule:
$\frac{d}{dx} \left( \int_0^x {f(t)\,dt} \right) = \frac{d}{dx} (x) - \frac{d}{dx} \left( \int_1^x {t\,f(t)\,dt} \right)$
Applying the Fundamental Theorem of Calculus:
$f(x) = 1 - xf(x)$
Rearranging the terms to solve for $f(x)$:
$f(x) + xf(x) = 1$
$f(x)(1 + x) = 1$
$f(x) = \frac{1}{1 + x}$
Now,substitute $x = 1$ to find $f(1)$:
$f(1) = \frac{1}{1 + 1} = \frac{1}{2}$.

(D) We know that for any event $A$ and a given event $B$ with $P(B) > 0$,the conditional probability satisfies $P(A/B) + P(\overline{A}/B) = 1$.
For option $(a)$:
$P(E/F) + P(\overline{E}/F) = \frac{P(E \cap F)}{P(F)} + \frac{P(\overline{E} \cap F)}{P(F)} = \frac{P(E \cap F) + P(\overline{E} \cap F)}{P(F)} = \frac{P((E \cup \overline{E}) \cap F)}{P(F)} = \frac{P(S \cap F)}{P(F)} = \frac{P(F)}{P(F)} = 1$.
Thus,$(a)$ is true.
For option $(c)$:
Similarly,replacing $F$ with $\overline{F}$ (given $P(\overline{F}) = 1 - P(F) > 0$),we have $P(E/\overline{F}) + P(\overline{E}/\overline{F}) = 1$.
Thus,$(c)$ is true.
Therefore,both $(a)$ and $(c)$ are correct.

(D) Given $P(E) \le P(F)$ and $P(E \cap F) > 0.$
$P(E) \le P(F)$ does not imply $E \subseteq F.$
$P(E \cap F) > 0$ implies that the intersection of $E$ and $F$ is not empty,but it does not imply that one event is a subset of the other.
For example,let $S = \{1, 2, 3\}$ with equally likely outcomes. Let $E = \{1, 2\}$ and $F = \{2, 3\}$.
Then $P(E) = 2/3$ and $P(F) = 2/3$,so $P(E) \le P(F)$ holds.
Also $P(E \cap F) = P(\{2\}) = 1/3 > 0$.
However,$E \not\subseteq F$ and $F \not\subseteq E$.
Thus,none of the given implications necessarily hold.

(C) We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Given that $P(A \cup B) = P(A \cap B)$,we substitute this into the formula:
$P(A \cap B) = P(A) + P(B) - P(A \cap B)$
$2P(A \cap B) = P(A) + P(B)$
Since $P(A \cap B) = P(A)P\left(\frac{B}{A}\right)$,we have:
$2P(A)P\left(\frac{B}{A}\right) = P(A) + P(B)$.

(D) Given that the vectors $a, b, c$ are linearly dependent,their scalar triple product must be zero: $[a, b, c] = 0$.
This implies the determinant of the matrix formed by their components is zero:
$\begin{vmatrix} 1 & 1 & 1 \\ 4 & 3 & 4 \\ 1 & \alpha & \beta \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1(3\beta - 4\alpha) - 1(4\beta - 4) + 1(4\alpha - 3) = 0$
$3\beta - 4\alpha - 4\beta + 4 + 4\alpha - 3 = 0$
$-\beta + 1 = 0 \Rightarrow \beta = 1$.
Given $|c| = \sqrt{3}$,we have $|c|^2 = 3$.
$1^2 + \alpha^2 + \beta^2 = 3$
$1 + \alpha^2 + 1^2 = 3$
$\alpha^2 + 2 = 3 \Rightarrow \alpha^2 = 1 \Rightarrow \alpha = \pm 1$.
Thus,$\alpha = \pm 1$ and $\beta = 1$.

(A) The function $f(x) = x - [x]$ is the fractional part function,denoted as $\{x\}$.
For the interval $[-1, 1]$,we split the integral at $x = 0$:
$\int_{-1}^{1} f(x) \, dx = \int_{-1}^{0} f(x) \, dx + \int_{0}^{1} f(x) \, dx$.
For $-1 \le x < 0$,$[x] = -1$,so $f(x) = x - (-1) = x + 1$.
For $0 \le x < 1$,$[x] = 0$,so $f(x) = x - 0 = x$.
Now,calculate the integrals:
$\int_{-1}^{0} (x + 1) \, dx = \left[ \frac{x^2}{2} + x \right]_{-1}^{0} = (0 + 0) - \left( \frac{1}{2} - 1 \right) = -(-1/2) = 1/2$.
$\int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1}{2} - 0 = 1/2$.
Adding these results: $1/2 + 1/2 = 1$.

(B) Given the general solution: $y = (c_1 + c_2) \cos (x + c_3) - c_4 e^{x + c_5}$.
We can simplify the expression by grouping the constants:
Let $A = c_1 + c_2$ and $B = c_4 e^{c_5}$.
Then the equation becomes $y = A \cos (x + c_3) - B e^x$.
Expanding the cosine term: $y = A (\cos x \cos c_3 - \sin x \sin c_3) - B e^x$.
$y = (A \cos c_3) \cos x - (A \sin c_3) \sin x - B e^x$.
Let $K_1 = A \cos c_3$,$K_2 = -A \sin c_3$,and $K_3 = -B$.
Thus,$y = K_1 \cos x + K_2 \sin x + K_3 e^x$.
There are $3$ independent arbitrary constants $(K_1, K_2, K_3)$.
The order of a differential equation is equal to the number of independent arbitrary constants in its general solution.
Therefore,the order of the differential equation is $3$.