IIT JEE 2004 Mathematics Question Paper with Answer and Solution

(D) Given the equation: $(1 + \omega^2)^m = (1 + \omega^4)^m$.
Since $\omega^3 = 1$,we have $\omega^4 = \omega$.
Substituting this into the equation: $(1 + \omega^2)^m = (1 + \omega)^m$.
Using the property $1 + \omega + \omega^2 = 0$,we know $1 + \omega^2 = -\omega$ and $1 + \omega = -\omega^2$.
So,$(-\omega)^m = (-\omega^2)^m$.
Dividing both sides by $(-\omega)^m$ (assuming $\omega \neq 0$): $1 = \frac{(-\omega^2)^m}{(-\omega)^m} = (\frac{-\omega^2}{-\omega})^m = (\omega)^m$.
For $(\omega)^m = 1$,the smallest positive integer $m$ must be $3$ because $\omega^3 = 1$.

(B) The sum of an infinite geometric progression is given by $S = \frac{a}{1-r}$,where $a$ is the first term and $r$ is the common ratio,with the condition $|r| < 1$.
Given $a = x$ and $S = 5$,we have $5 = \frac{x}{1-r}$.
Rearranging for $r$,we get $1 - r = \frac{x}{5}$,which implies $r = 1 - \frac{x}{5}$.
Since $|r| < 1$,we have $|1 - \frac{x}{5}| < 1$.
This inequality can be written as $-1 < 1 - \frac{x}{5} < 1$.
Subtracting $1$ from all parts,we get $-2 < -\frac{x}{5} < 0$.
Multiplying by $-5$ (and reversing the inequality signs),we get $10 > x > 0$,or $0 < x < 10$.

(D) Let the roots of the given equation $x^2 + px + q = 0$ be $\alpha$ and $\alpha^2$.
From the relation between roots and coefficients,we have:
$\alpha \cdot \alpha^2 = \alpha^3 = q$
$\alpha + \alpha^2 = -p$
Cubing both sides of the second equation:
$(\alpha + \alpha^2)^3 = (-p)^3$
$\alpha^3 + (\alpha^2)^3 + 3\alpha \cdot \alpha^2(\alpha + \alpha^2) = -p^3$
Substitute $\alpha^3 = q$ and $\alpha + \alpha^2 = -p$ into the equation:
$q + q^2 + 3q(-p) = -p^3$
$p^3 + q^2 + q - 3pq = 0$
$p^3 + q^2 + q(1 - 3p) = 0$.

(A) For a quadratic expression $Ax^2 + Bx + C > 0$ to be true for all $x \in R$,the discriminant $D$ must be less than $0$ and $A > 0$.
Here,$A = 1$,which is $> 0$.
The discriminant $D = B^2 - 4AC = (2a)^2 - 4(1)(10 - 3a) < 0$.
$4a^2 - 40 + 12a < 0$.
Dividing by $4$,we get $a^2 + 3a - 10 < 0$.
Factoring the quadratic,we get $(a + 5)(a - 2) < 0$.
This inequality holds when $a$ lies between the roots,so $-5 < a < 2$.

(D) Given the equation: $^{n-1}C_r = (k^2 - 3) \cdot ^nC_{r+1}$
Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$,we have:
$\frac{(n-1)!}{r!(n-r-1)!} = (k^2 - 3) \cdot \frac{n!}{(r+1)!(n-r-1)!}$
Simplifying the factorials:
$\frac{1}{r!} = (k^2 - 3) \cdot \frac{n}{(r+1)r!}$
$1 = (k^2 - 3) \cdot \frac{n}{r+1}$
$k^2 - 3 = \frac{r+1}{n}$
$k^2 = \frac{r+1}{n} + 3$
Since $0 \le r \le n-1$,we have $1 \le r+1 \le n$,which implies $\frac{1}{n} \le \frac{r+1}{n} \le 1$.
Thus,$k^2 \in [\frac{1}{n} + 3, 4]$.
For $n \ge 2$,the value of $\frac{1}{n} + 3$ lies in the interval $(3, 3.5]$.
Therefore,$k \in [-2, -\sqrt{\frac{1}{n} + 3}] \cup [\sqrt{\frac{1}{n} + 3}, 2]$.
Comparing with the given options,the interval $(\sqrt{3}, 2)$ is the most appropriate subset.

(D) Given the ratio of sides $a:b:c = 1:\sqrt{3}:2$.
Let $a = \lambda$,$b = \sqrt{3}\lambda$,and $c = 2\lambda$.
Using the Law of Cosines,$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{3\lambda^2 + 4\lambda^2 - \lambda^2}{2(\sqrt{3}\lambda)(2\lambda)} = \frac{6\lambda^2}{4\sqrt{3}\lambda^2} = \frac{\sqrt{3}}{2}$.
Thus,$A = 30^\circ$.
Similarly,$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{\lambda^2 + 4\lambda^2 - 3\lambda^2}{2(\lambda)(2\lambda)} = \frac{2\lambda^2}{4\lambda^2} = \frac{1}{2}$.
Thus,$B = 60^\circ$.
Finally,$C = 180^\circ - (30^\circ + 60^\circ) = 90^\circ$.
The ratio $A:B:C = 30^\circ:60^\circ:90^\circ = 1:2:3$.

(A) The given equation is ${x^2} - ({y^2} - 2y + 1) = 0$,which simplifies to ${x^2} - {(y - 1)^2} = 0$.
This represents the pair of lines $(x - (y - 1))(x + (y - 1)) = 0$,or $(x - y + 1)(x + y - 1) = 0$.
The lines are $x - y + 1 = 0$ and $x + y - 1 = 0$.
The angle bisectors of these lines are given by $\frac{x - y + 1}{\sqrt{1^2 + (-1)^2}} = \pm \frac{x + y - 1}{\sqrt{1^2 + 1^2}}$.
This simplifies to $x - y + 1 = \pm (x + y - 1)$.
Case $1$: $x - y + 1 = x + y - 1 \implies 2y = 2 \implies y = 1$.
Case $2$: $x - y + 1 = -(x + y - 1) \implies x - y + 1 = -x - y + 1 \implies 2x = 0 \implies x = 0$.
Thus,the angle bisectors are the lines $x = 0$ and $y = 1$.
The third line is $x + y = 3$.
The vertices of the triangle formed by $x = 0$,$y = 1$,and $x + y = 3$ are found by intersection:
Intersection of $x = 0$ and $y = 1$ is $(0, 1)$.
Intersection of $x = 0$ and $x + y = 3$ is $(0, 3)$.
Intersection of $y = 1$ and $x + y = 3$ is $(2, 1)$.
The triangle is a right-angled triangle with vertices $(0, 1)$,$(0, 3)$,and $(2, 1)$.
The base length is $2 - 0 = 2$ and the height is $3 - 1 = 2$.
Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 2 = 2$.

(C) Given circle is $x^2 + y^2 - 2x - 6y + 6 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1, f = -3, c = 6$.
The centre of this circle is $C(-g, -f) = (1, 3)$ and its radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + (-3)^2 - 6} = \sqrt{1 + 9 - 6} = \sqrt{4} = 2$.
Let the required circle have centre $D(2, 1)$.
Since a chord of the required circle is a diameter of the given circle,the length of this chord is equal to the diameter of the given circle,which is $2 \times r = 2 \times 2 = 4$.
Let $AB$ be the chord of the required circle,which is a diameter of the given circle. The length of $AB = 4$,so $AC = CB = 2$.
In the right-angled triangle $\triangle ACD$,$AD$ is the radius of the required circle $(R)$.
$AD^2 = AC^2 + CD^2$.
$CD$ is the distance between $D(2, 1)$ and $C(1, 3)$.
$CD = \sqrt{(2 - 1)^2 + (1 - 3)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$.
$R^2 = 2^2 + (\sqrt{5})^2 = 4 + 5 = 9$.
$R = 3$.

(C) The given ellipse is $x^2 + 2y^2 = 2$,which can be written as $\frac{x^2}{2} + \frac{y^2}{1} = 1$. Here $a^2 = 2$ and $b^2 = 1$.
Let the point of contact be $R \equiv (\sqrt{2} \cos \theta, \sin \theta)$.
The equation of the tangent at $R$ is $\frac{x \cos \theta}{\sqrt{2}} + y \sin \theta = 1$.
The tangent intersects the $x$-axis at $A \equiv (\sqrt{2} \sec \theta, 0)$ and the $y$-axis at $B \equiv (0, \csc \theta)$.
Let $Q(h, k)$ be the midpoint of $AB$. Then $h = \frac{\sqrt{2} \sec \theta}{2} = \frac{\sec \theta}{\sqrt{2}}$ and $k = \frac{\csc \theta}{2}$.
This implies $\cos \theta = \frac{1}{h\sqrt{2}}$ and $\sin \theta = \frac{1}{2k}$.
Using $\cos^2 \theta + \sin^2 \theta = 1$,we get $\left(\frac{1}{h\sqrt{2}}\right)^2 + \left(\frac{1}{2k}\right)^2 = 1$.
Thus,the locus is $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$.

(C) Given the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{f({x^2}) - f(x)}}{{f(x) - f(0)}}$.
This is a $\frac{0}{0}$ indeterminate form.
Applying $L'\text{Hospital's rule}$,we differentiate the numerator and denominator with respect to $x$:
$= \mathop {\lim }\limits_{x \to 0} \frac{{2x \cdot f'({x^2}) - f'(x)}}{{f'(x)}}$.
$= \mathop {\lim }\limits_{x \to 0} \left( \frac{2x \cdot f'({x^2})}{f'(x)} - \frac{f'(x)}{f'(x)} \right)$.
$= \mathop {\lim }\limits_{x \to 0} \left( \frac{2x \cdot f'({x^2})}{f'(x)} - 1 \right)$.
Since $f$ is strictly increasing,$f'(x) > 0$. Assuming $f'(0) \neq 0$,the term $\mathop {\lim }\limits_{x \to 0} \frac{2x \cdot f'({x^2})}{f'(x)} = \frac{0 \cdot f'(0)}{f'(0)} = 0$.
Therefore,the limit is $0 - 1 = -1$.

(B) Given $\sin \theta = \frac{1}{2}$,since $\theta$ is acute,$\theta = \frac{\pi}{6}$.
Given $\cos \phi = \frac{1}{3}$,since $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\cos \frac{\pi}{2} = 0$,and $\frac{1}{3}$ lies between $0$ and $\frac{1}{2}$,we have $\frac{\pi}{3} < \phi < \frac{\pi}{2}$.
Adding $\theta = \frac{\pi}{6}$ to the inequality $\frac{\pi}{3} < \phi < \frac{\pi}{2}$ gives $\frac{\pi}{6} + \frac{\pi}{3} < \theta + \phi < \frac{\pi}{6} + \frac{\pi}{2}$.
This simplifies to $\frac{\pi}{2} < \theta + \phi < \frac{2\pi}{3}$.

(A) The equation of the curve is $x^2 - 2y^2 = 4$.
Differentiating with respect to $x$,we get $2x - 4y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = \frac{x}{2y}$.
The slope of the tangent line $2x + \sqrt{6}y = 2$ is $-\frac{2}{\sqrt{6}} = -\frac{\sqrt{6}}{3}$.
Let the point of contact be $(x_1, y_1)$. Then $\frac{x_1}{2y_1} = -\frac{\sqrt{6}}{3}$,so $3x_1 = -2\sqrt{6}y_1$,or $y_1 = -\frac{3x_1}{2\sqrt{6}} = -\frac{\sqrt{6}x_1}{4}$.
Since the point $(x_1, y_1)$ lies on the curve,$x_1^2 - 2(-\frac{\sqrt{6}x_1}{4})^2 = 4$.
$x_1^2 - 2(\frac{6x_1^2}{16}) = 4 \implies x_1^2 - \frac{3}{4}x_1^2 = 4 \implies \frac{1}{4}x_1^2 = 4 \implies x_1^2 = 16$.
Thus $x_1 = 4$ or $x_1 = -4$.
If $x_1 = 4$,then $y_1 = -\frac{\sqrt{6}(4)}{4} = -\sqrt{6}$.
Checking the line equation: $2(4) + \sqrt{6}(-\sqrt{6}) = 8 - 6 = 2$. This satisfies the equation.
Thus,the point of contact is $(4, -\sqrt{6})$.

(D) number is divisible by both $2$ and $3$ if and only if it is divisible by their least common multiple,which is $6$.
In the first $100$ natural numbers,the numbers divisible by $6$ are $6, 12, 18, \dots, 96$.
Using the formula $a_n = a + (n-1)d$,we have $96 = 6 + (n-1)6$,which gives $n = 16$.
There are $16$ such numbers.
The total number of ways to select $3$ distinct numbers from $100$ is $^{100}C_3$.
The number of ways to select $3$ numbers divisible by $6$ is $^{16}C_3$.
The required probability is $\frac{^{16}C_3}{^{100}C_3} = \frac{16 \times 15 \times 14}{100 \times 99 \times 98} = \frac{3360}{970200} = \frac{4}{1155}$.

(C) The equation of the parabola is $y^2 = 4ax$,where $a = 1$.
The point is $(x_1, y_1) = (1, 4)$.
The locus of the point of intersection of perpendicular tangents to a parabola is its directrix.
The directrix of the parabola $y^2 = 4x$ is $x = -a$,which is $x = -1$.
Since the point $(1, 4)$ does not lie on the directrix $x = -1$,the tangents are not perpendicular.
The angle $\theta$ between the tangents from a point $(x_1, y_1)$ to the parabola $y^2 = 4ax$ is given by $\tan \theta = \left| \frac{\sqrt{y_1^2 - 4ax_1}}{x_1 + a} \right|$.
Substituting the values: $a = 1, x_1 = 1, y_1 = 4$.
$\tan \theta = \left| \frac{\sqrt{4^2 - 4(1)(1)}}{1 + 1} \right| = \left| \frac{\sqrt{16 - 4}}{2} \right| = \frac{\sqrt{12}}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}$.
Therefore,$\theta = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$.

(D) The system of equations has no solution if the determinant of the coefficient matrix $D = 0$ and at least one of the Cramer's rule determinants $(D_x, D_y, D_z)$ is non-zero.
First, calculate the determinant $D$ of the coefficient matrix:
$D = \begin{vmatrix} 2 & -1 & -1 \\ 1 & -2 & 1 \\ 1 & 1 & \lambda \end{vmatrix}$
$D = 2(-2\lambda - 1) - (-1)(\lambda - 1) + (-1)(1 - (-2))$
$D = -4\lambda - 2 + \lambda - 1 - 3$
$D = -3\lambda - 6$
For the system to have no solution or infinitely many solutions, we set $D = 0$:
$-3\lambda - 6 = 0 \Rightarrow \lambda = -2$
Now, check for $\lambda = -2$ using $D_x$:
$D_x = \begin{vmatrix} 12 & -1 & -1 \\ -4 & -2 & 1 \\ 4 & 1 & -2 \end{vmatrix} = 12(4 - 1) - (-1)(8 - 4) + (-1)(-4 + 8)$
$D_x = 12(3) + 1(4) - 1(4) = 36 \neq 0$
Since $D = 0$ and $D_x \neq 0$, the system has no solution for $\lambda = -2$.

(A) Given that $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$.
The determinant of matrix $A$ is $|A| = \alpha^2 - (2 \times 2) = \alpha^2 - 4$.
We are given that $|A^3| = 125$.
Using the property of determinants $|A^n| = |A|^n$,we have $|A|^3 = 125$.
Since $125 = 5^3$,we get $|A|^3 = 5^3$,which implies $|A| = 5$.
Substituting the value of $|A|$,we get $\alpha^2 - 4 = 5$.
$\alpha^2 = 9$.
Therefore,$\alpha = \pm 3$.

(A) Given the equation: $\sin (\cot ^{ - 1}(x + 1)) = \cos (\tan ^{ - 1}x)$.
First,convert $\cot ^{ - 1}(x + 1)$ to $\sin ^{ - 1}$ form. Let $\theta = \cot ^{ - 1}(x + 1)$,then $\cot \theta = x + 1$. Using the identity $\sin \theta = \frac{1}{\sqrt{1 + \cot ^2 \theta}}$,we get $\sin \theta = \frac{1}{\sqrt{1 + (x + 1)^2}} = \frac{1}{\sqrt{x^2 + 2x + 2}}$.
Next,convert $\tan ^{ - 1}x$ to $\cos ^{ - 1}$ form. Let $\phi = \tan ^{ - 1}x$,then $\tan \phi = x$. Using the identity $\cos \phi = \frac{1}{\sqrt{1 + \tan ^2 \phi}}$,we get $\cos \phi = \frac{1}{\sqrt{1 + x^2}}$.
Equating the two sides: $\frac{1}{\sqrt{x^2 + 2x + 2}} = \frac{1}{\sqrt{1 + x^2}}$.
Squaring both sides: $x^2 + 2x + 2 = 1 + x^2$.
Subtracting $x^2$ from both sides: $2x + 2 = 1$.
Solving for $x$: $2x = -1$,which gives $x = -\frac{1}{2}$.

(B) Let the required unit vector be $\hat{a}$. Since $\hat{a}$ lies in the plane of $\vec{u} = 2i + j + k$ and $\vec{v} = i - j + k$,it can be expressed as $\hat{a} = \alpha(2i + j + k) + \beta(i - j + k)$.
Simplifying,we get $\hat{a} = (2\alpha + \beta)i + (\alpha - \beta)j + (\alpha + \beta)k$.
Since $\hat{a}$ is a unit vector,its magnitude is $1$,so $(2\alpha + \beta)^2 + (\alpha - \beta)^2 + (\alpha + \beta)^2 = 1$.
Expanding this,we get $6\alpha^2 + 4\alpha\beta + 3\beta^2 = 1$ ... $(i)$.
Since $\hat{a}$ is orthogonal to $\vec{w} = 5i + 2j + 6k$,their dot product is zero: $5(2\alpha + \beta) + 2(\alpha - \beta) + 6(\alpha + \beta) = 0$.
This simplifies to $18\alpha + 9\beta = 0$,which gives $\beta = -2\alpha$.
Substituting $\beta = -2\alpha$ into equation $(i)$,we get $6\alpha^2 + 4\alpha(-2\alpha) + 3(-2\alpha)^2 = 1$.
$6\alpha^2 - 8\alpha^2 + 12\alpha^2 = 1$,which implies $10\alpha^2 = 1$,so $\alpha = \pm \frac{1}{\sqrt{10}}$.
Then $\beta = -2(\pm \frac{1}{\sqrt{10}}) = \mp \frac{2}{\sqrt{10}}$.
Substituting these values into the expression for $\hat{a}$,we get $\hat{a} = \pm \frac{1}{\sqrt{10}} [ (2 - 2)i + (1 + 2)j + (1 - 2)k ] = \pm \frac{3j - k}{\sqrt{10}}$.

(A) Let $b = b_1 i + b_2 j + b_3 k$.
Given $a \times b = j - k$,we have:
$j - k = \begin{vmatrix} i & j & k \\ 1 & 1 & 1 \\ b_1 & b_2 & b_3 \end{vmatrix}$
$j - k = i(b_3 - b_2) - j(b_3 - b_1) + k(b_2 - b_1)$
Comparing the components:
$b_3 - b_2 = 0 \Rightarrow b_3 = b_2$
$b_1 - b_3 = 1 \Rightarrow b_1 = b_3 + 1 = b_2 + 1$
$b_2 - b_1 = -1 \Rightarrow b_2 - (b_2 + 1) = -1$ (This is consistent).
Given $a \cdot b = 1$,we have:
$(i + j + k) \cdot (b_1 i + b_2 j + b_3 k) = 1$
$b_1 + b_2 + b_3 = 1$
Substituting $b_1 = b_2 + 1$ and $b_3 = b_2$:
$(b_2 + 1) + b_2 + b_2 = 1$
$3b_2 + 1 = 1$
$3b_2 = 0 \Rightarrow b_2 = 0$
Thus,$b_3 = 0$ and $b_1 = 0 + 1 = 1$.
Therefore,$b = 1i + 0j + 0k = i$.

(B) Let any point on the first line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4} = \lambda$ be $(2\lambda + 1, 3\lambda - 1, 4\lambda + 1)$.
Let any point on the second line $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1} = \mu$ be $(\mu + 3, 2\mu + k, \mu)$.
The lines intersect if there exist $\lambda$ and $\mu$ such that the coordinates are equal:
$2\lambda + 1 = \mu + 3 \implies 2\lambda - \mu = 2$ $(i)$
$3\lambda - 1 = 2\mu + k \implies 3\lambda - 2\mu = k + 1$ (ii)
$4\lambda + 1 = \mu \implies 4\lambda - \mu = -1$ (iii)
Subtracting $(i)$ from (iii):
$(4\lambda - \mu) - (2\lambda - \mu) = -1 - 2$
$2\lambda = -3 \implies \lambda = -\frac{3}{2}$.
Substitute $\lambda = -\frac{3}{2}$ into (iii):
$4(-\frac{3}{2}) + 1 = \mu \implies -6 + 1 = \mu \implies \mu = -5$.
Substitute $\lambda = -\frac{3}{2}$ and $\mu = -5$ into (ii):
$3(-\frac{3}{2}) - 2(-5) = k + 1$
$-\frac{9}{2} + 10 = k + 1$
$k = -\frac{9}{2} + 9 = \frac{9}{2}$.

(A) The function is given by $f(x) = \sin x - \sqrt{3} \cos x + 1$.
We know that the expression $a \sin x + b \cos x$ lies in the interval $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here,$a = 1$ and $b = -\sqrt{3}$.
Thus,the range of $\sin x - \sqrt{3} \cos x$ is $[-\sqrt{1^2 + (-\sqrt{3})^2}, \sqrt{1^2 + (-\sqrt{3})^2}] = [-\sqrt{1+3}, \sqrt{1+3}] = [-2, 2]$.
Adding $1$ to the entire inequality,we get:
$-2 + 1 \le \sin x - \sqrt{3} \cos x + 1 \le 2 + 1$.
$-1 \le f(x) \le 3$.
Since the function $f$ is onto,the codomain $S$ must be equal to the range of the function.
Therefore,$S = [-1, 3]$.

(C) Given $f(x) = \sin x + \cos x$ and $g(x) = x^2 - 1$.
First,we find the composition $g(f(x))$:
$g(f(x)) = (\sin x + \cos x)^2 - 1$
$= (\sin^2 x + \cos^2 x + 2 \sin x \cos x) - 1$
$= (1 + \sin 2x) - 1 = \sin 2x$.
For a function to be invertible,it must be bijective (one-to-one and onto).
The function $h(x) = \sin 2x$ is invertible in the interval where the argument $2x$ lies within the principal branch of the inverse sine function,i.e.,$[ -\frac{\pi}{2}, \frac{\pi}{2} ]$.
Thus,we require:
$-\frac{\pi}{2} \le 2x \le \frac{\pi}{2}$
Dividing by $2$,we get:
$-\frac{\pi}{4} \le x \le \frac{\pi}{4}$.
Therefore,$g(f(x))$ is invertible for $x \in [ -\frac{\pi}{4}, \frac{\pi}{4} ]$.

(A) Given the equation $\ln (x + y) = 2xy$.
First,find the value of $y$ when $x = 0$:
$\ln (0 + y) = 2(0)y \implies \ln (y) = 0 \implies y = e^0 = 1$.
Now,differentiate both sides with respect to $x$:
$\frac{d}{dx} [\ln (x + y)] = \frac{d}{dx} [2xy]$
$\frac{1}{x + y} \cdot (1 + y') = 2(y + xy')$
Substitute $x = 0$ and $y = 1$ into the differentiated equation:
$\frac{1}{0 + 1} \cdot (1 + y'(0)) = 2(1 + 0 \cdot y'(0))$
$1 \cdot (1 + y'(0)) = 2(1)$
$1 + y'(0) = 2$
$y'(0) = 2 - 1 = 1$.

(D) Given the function $f(x) = x^3 + bx^2 + cx + d$.
First,we find the derivative of the function:
$f'(x) = 3x^2 + 2bx + c$.
To determine the nature of the function,we examine the discriminant $D$ of the quadratic expression $f'(x)$:
$D = (2b)^2 - 4(3)(c) = 4b^2 - 12c$.
We are given that $0 < b^2 < c$.
Since $b^2 < c$,it follows that $4b^2 < 4c$.
Therefore,$D = 4b^2 - 12c < 4c - 12c = -8c$.
Since $b^2 > 0$,$c$ must also be greater than $0$ (because $c > b^2$).
Thus,$D < -8c < 0$.
Since the discriminant $D < 0$ and the coefficient of $x^2$ (which is $3$) is positive,$f'(x) > 0$ for all $x \in \mathbb{R}$.
Since the derivative $f'(x)$ is strictly positive for all real $x$,the function $f(x)$ is strictly increasing.

(D) For Rolle's theorem to be applicable to $f(x)$ on $[0, 1]$,the following conditions must be satisfied:
$1$. $f(x)$ must be continuous on $[0, 1]$.
$2$. $f(x)$ must be differentiable on $(0, 1)$.
$3$. $f(0) = f(1)$.
Checking condition $3$: $f(1) = 1^\alpha \ln(1) = 0$ and $f(0) = 0$. Thus,$f(0) = f(1) = 0$ holds for any $\alpha$.
Checking condition $1$ (Continuity at $x=0$): We require $\lim_{x \to 0^+} x^\alpha \ln x = 0$. This limit exists and equals $0$ if and only if $\alpha > 0$.
Checking condition $2$ (Differentiability on $(0, 1)$): $f'(x) = \alpha x^{\alpha-1} \ln x + x^{\alpha-1} = x^{\alpha-1}(\alpha \ln x + 1)$. This is defined for all $x \in (0, 1)$ for any $\alpha$.
Among the given options,only $\alpha = 0.5$ satisfies the condition $\alpha > 0$.

(A) Let $I = \int_0^1 \sqrt{\frac{1-x}{1+x}} \,dx$.
Rationalizing the integrand,we get:
$I = \int_0^1 \frac{\sqrt{1-x}}{\sqrt{1+x}} \cdot \frac{\sqrt{1-x}}{\sqrt{1-x}} \,dx = \int_0^1 \frac{1-x}{\sqrt{1-x^2}} \,dx$.
Splitting the integral:
$I = \int_0^1 \frac{1}{\sqrt{1-x^2}} \,dx - \int_0^1 \frac{x}{\sqrt{1-x^2}} \,dx$.
For the first part,$\int \frac{1}{\sqrt{1-x^2}} \,dx = \sin^{-1}(x)$.
For the second part,let $u = 1-x^2$,then $du = -2x \,dx$,so $\int \frac{x}{\sqrt{1-x^2}} \,dx = -\sqrt{1-x^2}$.
Evaluating the definite integral:
$I = [\sin^{-1}(x)]_0^1 + [\sqrt{1-x^2}]_0^1$.
$I = (\sin^{-1}(1) - \sin^{-1}(0)) + (\sqrt{1-1^2} - \sqrt{1-0^2})$.
$I = (\frac{\pi}{2} - 0) + (0 - 1) = \frac{\pi}{2} - 1$.

(B) The curves are $y = ax^2$ and $x = ay^2$ (or $y = \sqrt{x/a}$ for the upper branch).
To find the intersection points,substitute $y = ax^2$ into $x = ay^2$:
$x = a(ax^2)^2 = a^3x^4$
$x(a^3x^3 - 1) = 0$
So,$x = 0$ or $x^3 = 1/a^3$,which gives $x = 1/a$.
The intersection point $A$ is $(1/a, 1/a)$.
The area $A$ is given by:
$A = \int_0^{1/a} (\sqrt{x/a} - ax^2) dx = 1$
$\int_0^{1/a} (\frac{1}{\sqrt{a}} x^{1/2} - ax^2) dx = 1$
$[\frac{1}{\sqrt{a}} \cdot \frac{2}{3} x^{3/2} - \frac{a}{3} x^3]_0^{1/a} = 1$
$\frac{1}{\sqrt{a}} \cdot \frac{2}{3} (\frac{1}{a})^{3/2} - \frac{a}{3} (\frac{1}{a})^3 = 1$
$\frac{2}{3a^2} - \frac{1}{3a^2} = 1$
$\frac{1}{3a^2} = 1$
$a^2 = 1/3$
Since $a > 0$,$a = \frac{1}{\sqrt{3}}$.

(A) Given the equation $\int_0^{t^2} xf(x)dx = \frac{2}{5}t^5$.
Applying the Leibniz rule for differentiation under the integral sign with respect to $t$:
$\frac{d}{dt} \left( \int_0^{t^2} xf(x)dx \right) = \frac{d}{dt} \left( \frac{2}{5}t^5 \right)$.
Using the chain rule,we get:
$(t^2)f(t^2) \cdot \frac{d}{dt}(t^2) = \frac{2}{5} \cdot 5t^4$.
$(t^2)f(t^2) \cdot (2t) = 2t^4$.
$2t^3 f(t^2) = 2t^4$.
For $t > 0$,we can divide by $2t^3$:
$f(t^2) = t$.
We need to find $f\left( \frac{4}{25} \right)$.
Let $t^2 = \frac{4}{25}$,which implies $t = \frac{2}{5}$ (since $t > 0$).
Substituting $t = \frac{2}{5}$ into $f(t^2) = t$,we get:
$f\left( \frac{4}{25} \right) = \frac{2}{5}$.

(C) The given differential equation is $\left( \frac{2 + \sin x}{1 + y} \right) \frac{dy}{dx} = - \cos x$.
Separating the variables,we get:
$\frac{dy}{1 + y} = - \frac{\cos x}{2 + \sin x} dx$.
Integrating both sides:
$\int \frac{dy}{1 + y} = - \int \frac{\cos x}{2 + \sin x} dx$.
Let $u = 2 + \sin x$,then $du = \cos x dx$.
So,$\ln|1 + y| = - \ln|2 + \sin x| + C$.
This can be written as $\ln|1 + y| + \ln|2 + \sin x| = C$,or $\ln|(1 + y)(2 + \sin x)| = C$.
Thus,$(1 + y)(2 + \sin x) = K$ (where $K = e^C$).
Given $y(0) = 1$,we substitute $x = 0$ and $y = 1$:
$(1 + 1)(2 + \sin 0) = K \implies 2(2 + 0) = K \implies K = 4$.
So,$(1 + y)(2 + \sin x) = 4$.
To find $y\left( \frac{\pi}{2} \right)$,substitute $x = \frac{\pi}{2}$:
$(1 + y)(2 + \sin \frac{\pi}{2}) = 4
(1 + y)(2 + 1) = 4
3(1 + y) = 4
1 + y = \frac{4}{3}
y = \frac{4}{3} - 1 = \frac{1}{3}$.

(C) Let the two lines be represented by parameters $\lambda$ and $\mu$ respectively:
$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda \implies x=2\lambda+1, y=3\lambda-1, z=4\lambda+1$
$\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mu \implies x=\mu+3, y=2\mu+k, z=\mu$
If the lines intersect,there must exist a point common to both,so:
$2\lambda+1 = \mu+3 \implies 2\lambda - \mu = 2 \quad (i)$
$3\lambda-1 = 2\mu+k \implies 3\lambda - 2\mu = k+1 \quad (ii)$
$4\lambda+1 = \mu \implies 4\lambda - \mu = -1 \quad (iii)$
Subtracting $(i)$ from $(iii)$:
$(4\lambda - \mu) - (2\lambda - \mu) = -1 - 2
\implies 2\lambda = -3 \implies \lambda = -\frac{3}{2}$
Substituting $\lambda = -\frac{3}{2}$ into $(iii)$:
$4(-\frac{3}{2}) + 1 = \mu \implies -6 + 1 = \mu \implies \mu = -5$
Substituting $\lambda = -\frac{3}{2}$ and $\mu = -5$ into $(ii)$:
$3(-\frac{3}{2}) - 2(-5) = k+1
\implies -\frac{9}{2} + 10 = k+1
\implies k = 9 - \frac{9}{2} = \frac{9}{2}$