IIT JEE 2000 Mathematics Question Paper with Answer and Solution

(B) Using the change of base formula $\log _{a}b = \frac{\log b}{\log a}$:
$\log _{3}4 \cdot \log _{4}5 \cdot \log _{5}6 \cdot \log _{6}7 \cdot \log _{7}8 \cdot \log _{8}9$
$= \frac{\log 4}{\log 3} \cdot \frac{\log 5}{\log 4} \cdot \frac{\log 6}{\log 5} \cdot \frac{\log 7}{\log 6} \cdot \frac{\log 8}{\log 7} \cdot \frac{\log 9}{\log 8}$
$= \frac{\log 9}{\log 3}$
$= \log _{3}9 = \log _{3}(3^{2}) = 2 \cdot \log _{3}3 = 2 \cdot 1 = 2$.

(A) Given that $|{z_1}| = |{z_2}| = |{z_3}| = 1$.
Since $|{z_i}| = 1$,we have $|{z_i}|^2 = {z_i} \overline{z_i} = 1$,which implies $\frac{1}{z_i} = \overline{z_i}$ for $i = 1, 2, 3$.
Given $\left| \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \right| = 1$.
Substituting $\frac{1}{z_i} = \overline{z_i}$,we get $|\overline{z_1} + \overline{z_2} + \overline{z_3}| = 1$.
Using the property of conjugates $|\overline{z}| = |z|$,we have $|\overline{z_1 + z_2 + z_3}| = |z_1 + z_2 + z_3| = 1$.
Therefore,$|{z_1} + {z_2} + {z_3}| = 1$.

(D) For an infinite $G.P.$,the sum $S = \frac{a}{1-r} = 4$ and the second term $ar = \frac{3}{4}$.
From the first equation,$a = 4(1-r)$.
Substituting $a$ into the second equation: $4(1-r)r = \frac{3}{4}$.
$r(1-r) = \frac{3}{16} \implies r - r^2 = \frac{3}{16} \implies 16r^2 - 16r + 3 = 0$.
Factoring the quadratic: $(4r - 3)(4r - 1) = 0$.
This gives $r = \frac{3}{4}$ or $r = \frac{1}{4}$.
If $r = \frac{1}{4}$,then $a = 4(1 - \frac{1}{4}) = 4(\frac{3}{4}) = 3$.
If $r = \frac{3}{4}$,then $a = 4(1 - \frac{3}{4}) = 4(\frac{1}{4}) = 1$.
The possible pairs $(a, r)$ are $(3, \frac{1}{4})$ and $(1, \frac{3}{4})$.
Comparing with the options,$(a, r) = (3, \frac{1}{4})$ is correct.

(A) Given $a + b + c + d = 2$. Let $x = a + b$ and $y = c + d$. Then $x + y = 2$ and $M = xy$.
By the Arithmetic Mean-Geometric Mean inequality $(AM \ge GM)$,we have $\frac{x + y}{2} \ge \sqrt{xy}$.
Substituting the values,$\frac{2}{2} \ge \sqrt{M}$,which implies $1 \ge \sqrt{M}$,or $M \le 1$.
Since $a, b, c, d$ are positive real numbers,$x > 0$ and $y > 0$,so $M = xy > 0$.
Thus,the relation is $0 < M \le 1$.

(B) Given the quadratic equation $x^2 + bx + c = 0$ with $c < 0 < b$.
First,calculate the discriminant $D = b^2 - 4c$. Since $b^2 > 0$ and $c < 0$,$-4c > 0$,so $D = b^2 - 4c > 0$. Thus,the roots are real and distinct.
From the relation between roots and coefficients:
Sum of roots: $\alpha + \beta = -b$. Since $b > 0$,$\alpha + \beta < 0$.
Product of roots: $\alpha \beta = c$. Since $c < 0$,$\alpha \beta < 0$.
Since the product $\alpha \beta < 0$,one root must be positive and the other negative.
Since the sum $\alpha + \beta < 0$,the negative root must have a larger absolute value than the positive root.
Given $\alpha < \beta$,$\alpha$ is the negative root and $\beta$ is the positive root.
Therefore,$|\alpha| > \beta$,which implies $\alpha < 0 < \beta < |\alpha|$.

(D) Let $f(x) = (x - a)(x - b) - 1 = 0$.
$f(a) = (a - a)(a - b) - 1 = -1 < 0$.
$f(b) = (b - a)(b - b) - 1 = -1 < 0$.
Since $f(x)$ is a parabola opening upwards,and $f(a) < 0$ and $f(b) < 0$,the vertex of the parabola lies between $a$ and $b$.
Because the parabola opens upwards and $f(a) < 0$ and $f(b) < 0$,the graph must cross the $x$-axis at one point less than $a$ and one point greater than $b$.
Thus,one root lies in $(-\infty, a)$ and the other root lies in $(b, +\infty)$.

(C) Let the roots be $\alpha$ and $\alpha^2$.
From the relation between roots and coefficients for $3x^2 + px + 3 = 0$:
Sum of roots: $\alpha + \alpha^2 = -p/3$
Product of roots: $\alpha \cdot \alpha^2 = \alpha^3 = 3/3 = 1$
Since $\alpha^3 = 1$,the possible values for $\alpha$ are $1, \omega, \omega^2$,where $\omega$ is a complex cube root of unity.
If $\alpha = 1$,then $\alpha + \alpha^2 = 1 + 1 = 2$,so $2 = -p/3 \implies p = -6$. But we are given $p > 0$,so this case is rejected.
If $\alpha = \omega$ or $\alpha = \omega^2$,then $\alpha + \alpha^2 = \omega + \omega^2 = -1$.
Substituting this into the sum of roots equation: $-1 = -p/3 \implies p = 3$.
Thus,$p = 3$.

(D) The given expression is $\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2}$.
We can rewrite $2\binom{n}{r-1}$ as $\binom{n}{r-1} + \binom{n}{r-1}$.
So,the expression becomes $\binom{n}{r} + \binom{n}{r-1} + \binom{n}{r-1} + \binom{n}{r-2}$.
Using the Pascal's identity $\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$,we get:
$(\binom{n}{r} + \binom{n}{r-1}) + (\binom{n}{r-1} + \binom{n}{r-2}) = \binom{n+1}{r} + \binom{n+1}{r-1}$.
Applying the identity again,$\binom{n+1}{r} + \binom{n+1}{r-1} = \binom{n+2}{r}$.
Thus,the correct option is $D$.

(C) The given number is $223355888$. The digits are $2, 2, 3, 3, 5, 5, 8, 8, 8$.
There are $9$ digits in total. The positions are $1, 2, 3, 4, 5, 6, 7, 8, 9$.
The even positions are $2, 4, 6, 8$,which are $4$ in number.
The odd digits in the given number are $3, 3, 5, 5$.
These $4$ odd digits must occupy the $4$ even positions. The number of ways to arrange these is $\frac{4!}{2! \times 2!} = \frac{24}{4} = 6$.
The remaining $5$ digits are $2, 2, 8, 8, 8$,which must occupy the $5$ odd positions $(1, 3, 5, 7, 9)$.
The number of ways to arrange these is $\frac{5!}{2! \times 3!} = \frac{120}{2 \times 6} = 10$.
Therefore,the total number of ways is $6 \times 10 = 60$.

(B) We know that in a $\Delta ABC,$ $A + B + C = \pi,$ so $A + C = \pi - B.$
Substituting this into the expression: $\frac{A - B + C}{2} = \frac{(A + C) - B}{2} = \frac{(\pi - B) - B}{2} = \frac{\pi - 2B}{2} = \frac{\pi}{2} - B.$
Thus,$2ac \sin \left( \frac{\pi}{2} - B \right) = 2ac \cos B.$
Using the Law of Cosines,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}.$
Therefore,$2ac \cos B = 2ac \left( \frac{a^2 + c^2 - b^2}{2ac} \right) = a^2 + c^2 - b^2.$

(A) In a right-angled triangle $\Delta ABC$ with $\angle C = \frac{\pi}{2},$ the hypotenuse is $c = AB.$
The circumradius $R$ is half of the hypotenuse,so $R = \frac{c}{2}.$
The inradius $r$ is given by $r = \frac{\Delta}{s},$ where $\Delta = \frac{1}{2}ab$ and $s = \frac{a + b + c}{2}.$
Thus,$r = \frac{\frac{1}{2}ab}{\frac{1}{2}(a + b + c)} = \frac{ab}{a + b + c}.$
Now,$r + R = \frac{ab}{a + b + c} + \frac{c}{2} = \frac{2ab + c(a + b + c)}{2(a + b + c)}.$
Since $c^2 = a^2 + b^2,$ we have $2ab + c(a + b + c) = 2ab + ca + cb + c^2 = 2ab + ca + cb + a^2 + b^2 = (a + b)^2 + c(a + b) = (a + b)(a + b + c).$
Therefore,$r + R = \frac{(a + b)(a + b + c)}{2(a + b + c)} = \frac{a + b}{2}.$
Hence,$2(r + R) = a + b.$

(C) The circle is given by $x^2 + y^2 = 25$,which has its center at the origin $O(0, 0)$ and radius $r = 5$.
The coordinates of $Q$ are $(3, 4)$ and $R$ are $(-4, 3)$.
The slope of $OQ$ is $m_1 = \frac{4 - 0}{3 - 0} = \frac{4}{3}$.
The slope of $OR$ is $m_2 = \frac{3 - 0}{-4 - 0} = -\frac{3}{4}$.
Since $m_1 \times m_2 = \left(\frac{4}{3}\right) \times \left(-\frac{3}{4}\right) = -1$,the lines $OQ$ and $OR$ are perpendicular to each other.
Therefore,the central angle $\angle QOR = \frac{\pi}{2}$.
According to the circle theorem,the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
Thus,$\angle QPR = \frac{1}{2} \angle QOR = \frac{1}{2} \times \frac{\pi}{2} = \frac{\pi}{4}$.

(D) The median $PS$ connects vertex $P(2, 2)$ to the midpoint $S$ of side $QR$.
$S = \left( \frac{6 + 7}{2}, \frac{-1 + 3}{2} \right) = \left( \frac{13}{2}, 1 \right)$.
The slope $m$ of $PS$ is given by $m = \frac{1 - 2}{\frac{13}{2} - 2} = \frac{-1}{\frac{9}{2}} = -\frac{2}{9}$.
Since the required line is parallel to $PS$,its slope is also $-\frac{2}{9}$.
The equation of the line passing through $(1, -1)$ with slope $-\frac{2}{9}$ is:
$y - (-1) = -\frac{2}{9}(x - 1)$
$9(y + 1) = -2(x - 1)$
$9y + 9 = -2x + 2$
$2x + 9y + 7 = 0$.

(B) The equation of the parabola is ${y^2} = 12x$,which is of the form ${y^2} = 4ax$,where $a = 3$.
The equation of a normal to the parabola ${y^2} = 4ax$ at point $t$ is given by $y + tx = 2at + at^3$.
Substituting $a = 3$,the normal equation becomes $y + tx = 6t + 3t^3$.
Comparing this with the given normal $x + y = k$,we have $tx + y = 6t + 3t^3$ and $x + y = k$.
Since these represent the same line,the coefficients must be proportional:
$\frac{t}{1} = \frac{1}{1} = \frac{6t + 3t^3}{k}$.
From $\frac{t}{1} = 1$,we get $t = 1$.
Substituting $t = 1$ into the ratio $\frac{1}{1} = \frac{6(1) + 3(1)^3}{k}$,we get $1 = \frac{6 + 3}{k}$,which implies $k = 9$.

(C) The given equation of the parabola is ${y^2} - kx + 8 = 0$,which can be rewritten as ${y^2} = k(x - 8/k)$.
This is in the form ${Y^2} = 4AX$,where $Y = y$,$X = x - 8/k$,and $4A = k$,so $A = k/4$.
The directrix of the parabola ${Y^2} = 4AX$ is $X + A = 0$.
Substituting the values of $X$ and $A$,we get $(x - 8/k) + k/4 = 0$,which simplifies to $x = 8/k - k/4$.
We are given that the directrix is $x - 1 = 0$,or $x = 1$.
Equating the two expressions for $x$,we have $8/k - k/4 = 1$.
Multiplying by $4k$,we get $32 - k^2 = 4k$,or ${k^2} + 4k - 32 = 0$.
Factoring the quadratic equation,we get $(k + 8)(k - 4) = 0$.
Thus,the values of $k$ are $k = -8$ or $k = 4$.

(C) Given $f(\theta) = \sin \theta (\sin \theta + \sin 3\theta)$.
Using the identity $\sin 3\theta = 3\sin \theta - 4\sin^3 \theta$,we have:
$f(\theta) = \sin \theta (\sin \theta + 3\sin \theta - 4\sin^3 \theta)$
$f(\theta) = \sin \theta (4\sin \theta - 4\sin^3 \theta)$
$f(\theta) = 4\sin^2 \theta (1 - \sin^2 \theta)$
Since $1 - \sin^2 \theta = \cos^2 \theta$,we get:
$f(\theta) = 4\sin^2 \theta \cos^2 \theta$
$f(\theta) = (2 \sin \theta \cos \theta)^2$
$f(\theta) = (\sin 2\theta)^2$
Since the square of any real number is always non-negative,$(\sin 2\theta)^2 \ge 0$ for all real $\theta$.
Therefore,$f(\theta) \ge 0$ for all real $\theta$.

(C) We use the standard limit formula $\mathop {\lim }\limits_{x \to \infty } {(1 + \frac{a}{x})^x} = e^a$.
Given expression: $\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x - 3}}{{x + 2}}} \right)^x} = \mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 2 - 5}}{{x + 2}}} \right)^x} = \mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{5}{{x + 2}}} \right)^x}$.
Rewrite the expression as: $\mathop {\lim }\limits_{x \to \infty } {\left[ {{{\left( {1 - \frac{5}{{x + 2}}} \right)}^{\frac{{x + 2}}{{-5}}}}} \right]^{\frac{-5x}{x + 2}}}$.
Since $\mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{5}{{x + 2}}} \right)^{\frac{x + 2}{-5}}} = e$ and $\mathop {\lim }\limits_{x \to \infty } \frac{-5x}{x + 2} = \mathop {\lim }\limits_{x \to \infty } \frac{-5}{1 + \frac{2}{x}} = -5$.
The limit is $e^{-5}$.

(D) The median $PS$ connects vertex $P(2,2)$ to the midpoint $S$ of side $QR$.
The coordinates of $S$ are $\left(\frac{6+7}{2}, \frac{-1+3}{2}\right) = \left(\frac{13}{2}, 1\right)$.
The slope of $PS$ is $m = \frac{1-2}{\frac{13}{2}-2} = \frac{-1}{\frac{9}{2}} = -\frac{2}{9}$.
Since the required line is parallel to $PS$,its slope is also $m = -\frac{2}{9}$.
The equation of the line passing through $(1,-1)$ with slope $m = -\frac{2}{9}$ is given by $y - y_1 = m(x - x_1)$.
$y - (-1) = -\frac{2}{9}(x - 1)$
$9(y + 1) = -2(x - 1)$
$9y + 9 = -2x + 2$
$2x + 9y + 7 = 0$.

(C) Given $\frac{z_1 - z_3}{z_2 - z_3} = \frac{1}{2} - i\frac{\sqrt{3}}{2}$.
Taking the modulus on both sides,we get $\left| \frac{z_1 - z_3}{z_2 - z_3} \right| = \sqrt{\left( \frac{1}{2} \right)^2 + \left( -\frac{\sqrt{3}}{2} \right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1$.
This implies $|z_1 - z_3| = |z_2 - z_3|$,meaning two sides of the triangle are equal.
Now,the argument of the complex number is $\text{amp}\left( \frac{z_1 - z_3}{z_2 - z_3} \right) = \tan^{-1}\left( \frac{-\sqrt{3}/2}{1/2} \right) = \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$.
This means the angle between the sides $z_1 - z_3$ and $z_2 - z_3$ is $60^\circ$.
Since two sides are equal and the included angle is $60^\circ$,the triangle is equilateral.

(B) Let the height of the pole be $h$ and the foot of the pole be $P$. Let the corners of the park be $A, B,$ and $C$.
Given that the angle of elevation $\theta$ of the top of the pole from each corner is the same.
In the right-angled triangles formed by the pole and the distance from the foot $P$ to each corner,we have $\tan(\theta) = \frac{h}{PA} = \frac{h}{PB} = \frac{h}{PC}$.
Since $h$ and $\theta$ are constant,it follows that $PA = PB = PC$.
The point that is equidistant from all the vertices of a triangle is the circumcentre of the triangle.

(C) We know that $arg(-z) = arg(-1 \times z)$.
Using the property $arg(z_1 z_2) = arg(z_1) + arg(z_2)$,we get $arg(-z) = arg(-1) + arg(z)$.
Since $arg(-1) = \pi$,we have $arg(-z) = \pi + arg(z)$.
Therefore,$arg(-z) - arg(z) = (\pi + arg(z)) - arg(z) = \pi$.

(A) Two circles $x^{2}+y^{2}+2g_{1}x+2f_{1}y+c_{1}=0$ and $x^{2}+y^{2}+2g_{2}x+2f_{2}y+c_{2}=0$ intersect orthogonally if and only if $2(g_{1}g_{2}+f_{1}f_{2})=c_{1}+c_{2}$.
For the given circles:
Circle $1$: $g_{1}=1, f_{1}=k, c_{1}=6$
Circle $2$: $g_{2}=0, f_{2}=k, c_{2}=k$
Substituting these values into the condition:
$2((1)(0) + (k)(k)) = 6 + k$
$2k^{2} = 6 + k$
$2k^{2} - k - 6 = 0$
Factoring the quadratic equation:
$2k^{2} - 4k + 3k - 6 = 0$
$2k(k-2) + 3(k-2) = 0$
$(k-2)(2k+3) = 0$
Thus,$k = 2$ or $k = -\frac{3}{2}$.

(D) For a system of linear equations to have a non-zero (non-trivial) solution,the determinant of the coefficient matrix must be equal to zero,i.e.,$\Delta = 0$.
The system is given by:
$x - ky - z = 0$
$kx - y - z = 0$
$x + y - z = 0$
The determinant $\Delta$ is:
$\Delta = \begin{vmatrix} 1 & -k & -1 \\ k & -1 & -1 \\ 1 & 1 & -1 \end{vmatrix} = 0$
Expanding along the first row:
$1((-1)(-1) - (-1)(1)) - (-k)((k)(-1) - (-1)(1)) + (-1)((k)(1) - (-1)(1)) = 0$
$1(1 + 1) + k(-k + 1) - 1(k + 1) = 0$
$2 - k^2 + k - k - 1 = 0$
$1 - k^2 = 0$
$k^2 = 1$
$k = \pm 1$
Thus,the possible values of $k$ are $1$ and $-1$.

(B) In $\Delta ABC$,the vectors representing the sides are $BC = a$,$CA = b$,and $AB = c$.
By the triangle law of vector addition,the sum of vectors along the perimeter of a triangle in order is zero: $a + b + c = 0$.
Taking the cross product with $a$ on both sides: $a \times (a + b + c) = a \times 0 \implies a \times a + a \times b + a \times c = 0$.
Since $a \times a = 0$,we get $a \times b = c \times a$ (as $a \times c = -c \times a$).
Similarly,taking the cross product with $b$: $b \times (a + b + c) = b \times 0 \implies b \times a + b \times b + b \times c = 0$.
Since $b \times b = 0$,we get $b \times a + b \times c = 0 \implies b \times c = a \times b$.
Combining these,we have $a \times b = b \times c = c \times a$.

(A) The scalar triple product of three vectors $a, b,$ and $c$ is defined as $[a, b, c] = a \cdot (b \times c)$.
If three vectors are coplanar,they lie in the same plane.
For any three coplanar vectors,the volume of the parallelepiped formed by them is zero.
Since the scalar triple product represents the volume of the parallelepiped,it follows that $[a, b, c] = 0$ for any set of coplanar vectors.
Therefore,if $a, b,$ and $c$ are unit coplanar vectors,their scalar triple product is $0$.

(A) vector perpendicular to the plane $P_1$ formed by vectors $a$ and $b$ is given by $n_1 = a \times b$.
$A$ vector perpendicular to the plane $P_2$ formed by vectors $c$ and $d$ is given by $n_2 = c \times d$.
Given the condition $(a \times b) \times (c \times d) = 0$,it implies that the vector $n_1$ is parallel to the vector $n_2$ (i.e.,$n_1 \parallel n_2$).
Since the normal vectors to the planes are parallel,the planes $P_1$ and $P_2$ are parallel to each other.
Therefore,the angle between the planes $P_1$ and $P_2$ is $0^o$.

(D) Given the equation $2^x + 2^y = 2$.
We can rewrite this as $2^y = 2 - 2^x$.
For $y$ to be a real number,the value of $2^y$ must be strictly greater than $0$.
Therefore,$2 - 2^x > 0$.
This implies $2 > 2^x$.
Since the base $2 > 1$,the inequality holds when the exponents satisfy $1 > x$.
Thus,the domain of $x$ is $( - \infty , 1)$.

(D) Statement $S$: In the interval $\left( \frac{\pi}{2}, \pi \right)$,$\sin x$ decreases from $1$ to $0$,and $\cos x$ decreases from $0$ to $-1$. Therefore,both functions are decreasing in this interval. Thus,statement $S$ is correct.
Statement $R$: If a function $f(x)$ is decreasing in $(a, b)$,it implies $f'(x) \le 0$. It does not necessarily imply that $f'(x)$ is a decreasing function. For example,consider $f(x) = -x^3$ on $(-1, 1)$. Here $f'(x) = -3x^2$,which is not a decreasing function on $(-1, 1)$. The provided graph also illustrates a case where a function is decreasing,but its slope (derivative) is increasing. Thus,statement $R$ is incorrect.
Conclusion: $S$ is correct and $R$ is wrong. The correct option is $D$.

(B) Consider the function $f(x) = x - \log_e(1 + x)$ for $x \in (0, 1)$.
Taking the derivative,we get $f'(x) = 1 - \frac{1}{1 + x} = \frac{1 + x - 1}{1 + x} = \frac{x}{1 + x}$.
Since $x > 0$,$f'(x) > 0$ for all $x \in (0, 1)$.
This implies that $f(x)$ is a strictly increasing function on $(0, 1)$.
Since $f(0) = 0 - \log_e(1) = 0$,and $f(x)$ is strictly increasing,$f(x) > f(0)$ for all $x > 0$.
Therefore,$x - \log_e(1 + x) > 0$,which means $\log_e(1 + x) < x$ for all $x \in (0, 1)$.
Thus,option $(b)$ is correct.

(B) Let $I = \int_{e^{-1}}^{e^2} \left| \frac{\log_e x}{x} \right| dx$.
Since $\log_e x < 0$ for $x \in [e^{-1}, 1)$ and $\log_e x \ge 0$ for $x \in [1, e^2]$,we split the integral:
$I = \int_{e^{-1}}^{1} -\frac{\log_e x}{x} dx + \int_{1}^{e^2} \frac{\log_e x}{x} dx$.
Let $z = \log_e x$,then $dz = \frac{1}{x} dx$.
When $x = e^{-1}, z = -1$. When $x = 1, z = 0$. When $x = e^2, z = 2$.
$I = \int_{-1}^{0} -z dz + \int_{0}^{2} z dz$.
$I = \left[ -\frac{z^2}{2} \right]_{-1}^{0} + \left[ \frac{z^2}{2} \right]_{0}^{2}$.
$I = (0 - (-1/2)) + (4/2 - 0) = \frac{1}{2} + 2 = \frac{5}{2}$.

(C) We are given the function $f(x) = \begin{cases} e^{\cos x}\sin x, & |x| \le 2 \\ 2, & \text{otherwise} \end{cases}$.
We need to evaluate the integral $I = \int_{-2}^{3} f(x) dx$.
We can split the integral at $x = 2$ as follows:
$I = \int_{-2}^{2} f(x) dx + \int_{2}^{3} f(x) dx$.
For the interval $[-2, 2]$,$f(x) = e^{\cos x}\sin x$. Let $g(x) = e^{\cos x}\sin x$. Then $g(-x) = e^{\cos(-x)}\sin(-x) = e^{\cos x}(-\sin x) = -g(x)$. Thus,$g(x)$ is an odd function.
By the property of definite integrals,if $g(x)$ is an odd function,then $\int_{-a}^{a} g(x) dx = 0$. Therefore,$\int_{-2}^{2} e^{\cos x}\sin x dx = 0$.
For the interval $(2, 3]$,$f(x) = 2$. Thus,$\int_{2}^{3} 2 dx = [2x]_{2}^{3} = 2(3 - 2) = 2$.
Adding these results,$I = 0 + 2 = 2$.

(B) Given $g(2) = \int_0^2 f(t) \, dt = \int_0^1 f(t) \, dt + \int_1^2 f(t) \, dt$.
For $t \in [0, 1]$,we have $\frac{1}{2} \le f(t) \le 1$. Integrating this over $[0, 1]$:
$\int_0^1 \frac{1}{2} \, dt \le \int_0^1 f(t) \, dt \le \int_0^1 1 \, dt$
$\frac{1}{2} \le \int_0^1 f(t) \, dt \le 1 \quad \dots (i)$
For $t \in (1, 2]$,we have $0 \le f(t) \le \frac{1}{2}$. Integrating this over $(1, 2]$:
$\int_1^2 0 \, dt \le \int_1^2 f(t) \, dt \le \int_1^2 \frac{1}{2} \, dt$
$0 \le \int_1^2 f(t) \, dt \le \frac{1}{2} \quad \dots (ii)$
Adding $(i)$ and $(ii)$:
$\frac{1}{2} + 0 \le \int_0^1 f(t) \, dt + \int_1^2 f(t) \, dt \le 1 + \frac{1}{2}$
$\frac{1}{2} \le g(2) \le \frac{3}{2}$.
Since $\frac{1}{2} \le g(2) \le \frac{3}{2}$,the value of $g(2)$ lies within the interval $[0, 2)$. Thus,$0 \le g(2) < 2$ is the correct inequality.

(B) Given the equation: ${x^2} + {y^2} = 1$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}({x^2} + {y^2}) = \frac{d}{dx}(1)$
$2x + 2y \frac{dy}{dx} = 0$
$x + y y' = 0$
Differentiating again with respect to $x$ using the product rule on $y y'$:
$\frac{d}{dx}(x) + \frac{d}{dx}(y y') = 0$
$1 + (y \cdot y'' + y' \cdot y') = 0$
$1 + y y'' + (y')^2 = 0$
Thus,the relation is $yy'' + (y')^2 + 1 = 0$.

(C) The correct option is $(c)$.
$1$. $g(x) = |f(x)| \ge 0$ for all $x \in R$. Since the range of $g$ is a subset of $[0, \infty)$,$g$ cannot be onto if the codomain is $R$.
$2$. If $f(x)$ is one-one,$g(x)$ is not necessarily one-one. For example,if $f(x) = x$,then $f$ is one-one,but $g(x) = |x|$ is not one-one because $g(1) = g(-1) = 1$.
$3$. If $f(x)$ is continuous,then $g(x) = |f(x)|$ is also continuous. This is a standard property of continuous functions: the composition of a continuous function $f(x)$ with the continuous function $h(u) = |u|$ is continuous.
$4$. If $f(x)$ is differentiable,$g(x) = |f(x)|$ is not necessarily differentiable. As shown in the graph,if $f(x)$ crosses the $x$-axis at a point $P$ (where $f(P) = 0$),the function $g(x) = |f(x)|$ will have a sharp corner at $P$,making it non-differentiable at that point.

(A) The function is defined as $f(x) = |x|$ for $x \in [-2, 0) \cup (0, 2]$ and $f(0) = 1$.
For any small neighborhood $(0 - h, 0 + h)$ around $x = 0$ (where $h > 0$ is very small),we have:
$f(0) = 1$
For $x \neq 0$ in this neighborhood,$f(x) = |x|$. Since $x$ is very close to $0$,$|x| < 1$.
Thus,$f(x) < f(0)$ for all $x$ in the neighborhood of $0$.
By the definition of a local maximum,if $f(x) \le f(a)$ for all $x$ in some neighborhood of $a$,then $f$ has a local maximum at $x = a$.
Since $f(x) < f(0)$ for all $x \in (-h, h) \setminus \{0\}$,the function $f$ has a local maximum at $x = 0$.

(B) Given $f(x) = \int {e^x}(x - 1)(x - 2)dx$.
By the Fundamental Theorem of Calculus,$f'(x) = \frac{d}{dx} \int {e^x}(x - 1)(x - 2)dx = {e^x}(x - 1)(x - 2)$.
For the function $f$ to be decreasing,we must have $f'(x) < 0$.
So,${e^x}(x - 1)(x - 2) < 0$.
Since ${e^x} > 0$ for all real $x$,the inequality simplifies to $(x - 1)(x - 2) < 0$.
Using the sign scheme for the quadratic expression $(x - 1)(x - 2)$,the product is negative between the roots $x = 1$ and $x = 2$.
Thus,$f$ decreases in the interval $(1, 2)$.

(B) The slope of the normal to the curve $y=f(x)$ at a point is given by $m_n = \tan(\theta)$,where $\theta$ is the angle made with the positive $X$-axis.
Given $\theta = \frac{3 \pi}{4}$,the slope of the normal is $m_n = \tan\left(\frac{3 \pi}{4}\right) = -1$.
We know that the slope of the normal is also related to the derivative of the function by the formula $m_n = -\frac{1}{f^{\prime}(x)}$.
At the point $(3,4)$,we have $m_n = -\frac{1}{f^{\prime}(3)}$.
Equating the two expressions for the slope of the normal:
$-1 = -\frac{1}{f^{\prime}(3)}$
$f^{\prime}(3) = 1$.