IIT JEE 2001 Mathematics Question Paper with Answer and Solution

(D) The $n^{th}$ roots of unity are given by ${z_r} = \cos \frac{2r\pi}{n} + i\sin \frac{2r\pi}{n}$ for $r = 0, 1, \dots, n-1$.
Let ${z_1} = \cos \frac{2r_1\pi}{n} + i\sin \frac{2r_1\pi}{n}$ and ${z_2} = \cos \frac{2r_2\pi}{n} + i\sin \frac{2r_2\pi}{n}$.
The angle subtended by the segment joining ${z_1}$ and ${z_2}$ at the origin is given by the argument of the ratio $\frac{z_1}{z_2}$.
$\text{arg}\left(\frac{z_1}{z_2}\right) = \text{arg}(z_1) - \text{arg}(z_2) = \frac{2(r_1 - r_2)\pi}{n}$.
Since the angle is a right angle,we have $\frac{2(r_1 - r_2)\pi}{n} = \pm \frac{\pi}{2}$.
This implies $\frac{2(r_1 - r_2)}{n} = \pm \frac{1}{2}$,which simplifies to $n = \pm 4(r_1 - r_2)$.
Since $r_1$ and $r_2$ are integers,$n$ must be a multiple of $4$,i.e.,$n = 4k$ for some integer $k$.

(C) For the first arithmetic progression $2, 5, 8, \dots$,the first term $a_1 = 2$ and common difference $d_1 = 3$. The sum of the first $2n$ terms is $S_{2n} = \frac{2n}{2} [2(2) + (2n - 1)3] = n[4 + 6n - 3] = n(6n + 1)$.
For the second arithmetic progression $57, 59, 61, \dots$,the first term $a_2 = 57$ and common difference $d_2 = 2$. The sum of the first $n$ terms is $S_n = \frac{n}{2} [2(57) + (n - 1)2] = \frac{n}{2} [114 + 2n - 2] = \frac{n}{2} [112 + 2n] = n(56 + n)$.
Given that $S_{2n} = S_n$,we have $n(6n + 1) = n(56 + n)$.
Since $n \neq 0$,we divide by $n$: $6n + 1 = 56 + n$.
$5n = 55$,which gives $n = 11$.

(D) Given that $a, b, c, d$ are in $A.P.$
Dividing each term by the product $abcd$,we get:
$\frac{a}{abcd}, \frac{b}{abcd}, \frac{c}{abcd}, \frac{d}{abcd}$ are in $A.P.$
This simplifies to:
$\frac{1}{bcd}, \frac{1}{acd}, \frac{1}{abd}, \frac{1}{abc}$ are in $A.P.$
By the definition of $H.P.$,the reciprocals of terms in $A.P.$ are in $H.P.$
Therefore,$bcd, acd, abd, abc$ are in $H.P.$
Reversing the order,$abc, abd, acd, bcd$ are also in $H.P.$

(B) Given equation: $\log_{4}(x - 1) = \log_{2}(x - 3)$
Using the base change formula $\log_{a^n}(b) = \frac{1}{n} \log_{a}(b)$,we have $\log_{4}(x - 1) = \frac{1}{2} \log_{2}(x - 1)$.
So,$\frac{1}{2} \log_{2}(x - 1) = \log_{2}(x - 3)$
$\log_{2}(x - 1) = 2 \log_{2}(x - 3)$
$\log_{2}(x - 1) = \log_{2}((x - 3)^2)$
$x - 1 = (x - 3)^2$
$x - 1 = x^2 - 6x + 9$
$x^2 - 7x + 10 = 0$
$(x - 5)(x - 2) = 0$
Thus,$x = 5$ or $x = 2$.
Check the domain: For $\log_{2}(x - 3)$ to be defined,$x - 3 > 0$,so $x > 3$.
For $x = 2$,$x - 3 = -1$,which is not allowed.
For $x = 5$,$x - 3 = 2 > 0$,which is valid.
Therefore,there is only $1$ solution.

(A) Let $r$ be the common ratio of the $G.P.$ $\alpha, \beta, \gamma, \delta$. Then $\beta = \alpha r, \gamma = \alpha r^2, \delta = \alpha r^3$.
From the sum and product of roots:
$\alpha + \beta = 1 \Rightarrow \alpha(1 + r) = 1$ $(i)$
$\alpha \beta = p \Rightarrow \alpha^2 r = p$ $(ii)$
$\gamma + \delta = 4 \Rightarrow \alpha r^2(1 + r) = 4$ $(iii)$
$\gamma \delta = q \Rightarrow \alpha^2 r^5 = q$ $(iv)$
Dividing $(iii)$ by $(i)$,we get $r^2 = 4$,so $r = \pm 2$.
If $r = 2$,then $\alpha(1+2) = 1 \Rightarrow \alpha = 1/3$ (not an integer).
If $r = -2$,then $\alpha(1-2) = 1 \Rightarrow \alpha = -1$.
Substituting $r = -2$ and $\alpha = -1$ into $(ii)$ and $(iv)$:
$p = (-1)^2(-2) = -2$
$q = (-1)^2(-2)^5 = -32$
Thus,$(p, q) = (-2, -32)$.

(B) The number of triangles formed by $n$ vertices is given by $T_n = ^nC_3$.
Given the equation $T_{n+1} - T_n = 21$,we substitute the formula:
$^{n+1}C_3 - ^nC_3 = 21$.
Using the Pascal identity property,$^nC_r + ^nC_{r-1} = ^{n+1}C_r$,we can write $^{n+1}C_3 = ^nC_3 + ^nC_2$.
Substituting this into the equation:
$(^nC_3 + ^nC_2) - ^nC_3 = 21$
$^nC_2 = 21$.
Expanding the combination formula:
$\frac{n(n-1)}{2} = 21$
$n(n-1) = 42$
$n(n-1) = 7 \times 6$.
Thus,$n = 7$.

(B) Let the height of the tower be $PQ = 100 \ m$. Let the initial position of the car be $R$ and the final position be $S$. The distance travelled by the car is $x = RS$.
In $\triangle PQS$,$\tan{60^\circ} = \frac{PQ}{QS} \implies \sqrt{3} = \frac{100}{QS} \implies QS = \frac{100}{\sqrt{3}} \ m$.
In $\triangle PQR$,$\tan{30^\circ} = \frac{PQ}{QR} \implies \frac{1}{\sqrt{3}} = \frac{100}{QR} \implies QR = 100\sqrt{3} \ m$.
The distance travelled by the car is $x = QR - QS = 100\sqrt{3} - \frac{100}{\sqrt{3}} = \frac{300 - 100}{\sqrt{3}} = \frac{200}{\sqrt{3}} = \frac{200\sqrt{3}}{3} \ m$.

(A) Given the lines $3x + 4y = 9$ and $y = mx + 1$.
Substitute $y$ from the second equation into the first:
$3x + 4(mx + 1) = 9$
$3x + 4mx + 4 = 9$
$x(3 + 4m) = 5$
$x = \frac{5}{3 + 4m}$
For $x$ to be an integer,$(3 + 4m)$ must be a divisor of $5$. The divisors of $5$ are $\pm 1$ and $\pm 5$.
Case $1$: $3 + 4m = 1 \implies 4m = -2 \implies m = -0.5$ (not an integer).
Case $2$: $3 + 4m = -1 \implies 4m = -4 \implies m = -1$ (integer).
Case $3$: $3 + 4m = 5 \implies 4m = 2 \implies m = 0.5$ (not an integer).
Case $4$: $3 + 4m = -5 \implies 4m = -8 \implies m = -2$ (integer).
The integral values of $m$ are $-1$ and $-2$. Thus,there are $2$ such values.

(B) Let the coordinates of $A$ be $(r, 0)$ and $B$ be $(0, r)$ such that $\angle AOB = 90^\circ$. Let $P$ be a point on the circle given by $(r \cos \theta, r \sin \theta)$.
The centroid $G(\alpha, \beta)$ of $\Delta PAB$ is given by:
$\alpha = \frac{r \cos \theta + r + 0}{3} = \frac{r(\cos \theta + 1)}{3}$
$\beta = \frac{r \sin \theta + 0 + r}{3} = \frac{r(\sin \theta + 1)}{3}$
Rearranging,we get:
$\cos \theta = \frac{3\alpha}{r} - 1$
$\sin \theta = \frac{3\beta}{r} - 1$
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$:
$\left(\frac{3\alpha}{r} - 1\right)^2 + \left(\frac{3\beta}{r} - 1\right)^2 = 1$
$\frac{9}{r^2} \left(\alpha - \frac{r}{3}\right)^2 + \frac{9}{r^2} \left(\beta - \frac{r}{3}\right)^2 = 1$
$\left(\alpha - \frac{r}{3}\right)^2 + \left(\beta - \frac{r}{3}\right)^2 = \left(\frac{r}{3}\right)^2$
Thus,the locus is $(x - \frac{r}{3})^2 + (y - \frac{r}{3})^2 = (\frac{r}{3})^2$,which represents a circle.

(C) Any tangent to the parabola $y^2 = 4x$ is of the form $y = mx + \frac{1}{m}$.
This line touches the circle $(x - 3)^2 + y^2 = 9$ (with center $(3, 0)$ and radius $3$) if the perpendicular distance from the center to the line equals the radius:
$3 = \left| \frac{m(3) - 0 + \frac{1}{m}}{\sqrt{m^2 + 1}} \right|$
Squaring both sides:
$9(m^2 + 1) = (3m + \frac{1}{m})^2$
$9m^2 + 9 = 9m^2 + 6 + \frac{1}{m^2}$
$3 = \frac{1}{m^2} \implies m^2 = \frac{1}{3} \implies m = \pm \frac{1}{\sqrt{3}}$.
Since the tangent is above the $x$-axis,we choose the positive slope $m = \frac{1}{\sqrt{3}}$.
Substituting $m$ into the tangent equation:
$y = \frac{1}{\sqrt{3}}x + \sqrt{3}$
Multiplying by $\sqrt{3}$:
$\sqrt{3}y = x + 3$.

(B) We need to evaluate the limit $L = \mathop {\lim }\limits_{x \to 0} \frac{{\sin (\pi {{\cos }^2}x)}}{{{x^2}}}$.
Since $\cos^2 x = 1 - \sin^2 x$,we can rewrite the expression as:
$L = \mathop {\lim }\limits_{x \to 0} \frac{{\sin (\pi (1 - \sin^2 x))}}{{{x^2}}}$
$L = \mathop {\lim }\limits_{x \to 0} \frac{{\sin (\pi - \pi \sin^2 x)}}{{{x^2}}}$
Using the identity $\sin(\pi - \theta) = \sin \theta$,we get:
$L = \mathop {\lim }\limits_{x \to 0} \frac{{\sin (\pi \sin^2 x)}}{{{x^2}}}$
Multiply and divide by $\pi \sin^2 x$:
$L = \mathop {\lim }\limits_{x \to 0} \left( \frac{{\sin (\pi \sin^2 x)}}{{\pi \sin^2 x}} \right) \times \frac{{\pi \sin^2 x}}{{{x^2}}}$
As $x \to 0$,$\pi \sin^2 x \to 0$,so $\frac{{\sin (\pi \sin^2 x)}}{{\pi \sin^2 x}} \to 1$.
Also,$\mathop {\lim }\limits_{x \to 0} \frac{{\sin^2 x}}{{{x^2}}} = 1$.
Therefore,$L = 1 \times \pi \times 1 = \pi $.

(D) The given function is a quadratic in $x$ of the form $f(x) = Ax^2 + Bx + C$,where $A = 1 + b^2$,$B = 2b$,and $C = 1$.
Since $A = 1 + b^2 > 0$ for all real $b$,the function has a minimum value at $x = -\frac{B}{2A}$.
The minimum value $m(b)$ is given by $f(-\frac{B}{2A}) = C - \frac{B^2}{4A}$.
Substituting the values: $m(b) = 1 - \frac{(2b)^2}{4(1 + b^2)} = 1 - \frac{4b^2}{4(1 + b^2)} = 1 - \frac{b^2}{1 + b^2}$.
Simplifying this,we get $m(b) = \frac{1 + b^2 - b^2}{1 + b^2} = \frac{1}{1 + b^2}$.
Since $b^2 \ge 0$,we have $1 + b^2 \ge 1$,which implies $0 < \frac{1}{1 + b^2} \le 1$.
Thus,the range of $m(b)$ is $(0, 1]$.

(B) The general term in the expansion of $(a - b)^n$ is given by $T_{r+1} = ^nC_r a^{n-r} (-b)^r$.
For the $5^{th}$ term $(r=4)$: $T_5 = ^nC_4 a^{n-4} (-b)^4 = ^nC_4 a^{n-4} b^4$.
For the $6^{th}$ term $(r=5)$: $T_6 = ^nC_5 a^{n-5} (-b)^5 = -^nC_5 a^{n-5} b^5$.
Given $T_5 + T_6 = 0$,we have:
$^nC_4 a^{n-4} b^4 - ^nC_5 a^{n-5} b^5 = 0$.
Rearranging the terms:
$^nC_4 a^{n-4} b^4 = ^nC_5 a^{n-5} b^5$.
Dividing both sides by $a^{n-5} b^4$:
$\frac{a}{b} = \frac{^nC_5}{^nC_4}$.
Using the formula $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$:
$\frac{a}{b} = \frac{n-5+1}{5} = \frac{n-4}{5}$.

(B) The general term in the expansion of $(a - b)^n$ is given by $T_{r + 1} = {}^nC_r a^{n - r} (-b)^r$.
The $5^{th}$ term is $T_5 = T_{4 + 1} = {}^nC_4 a^{n - 4} (-b)^4 = {}^nC_4 a^{n - 4} b^4$.
The $6^{th}$ term is $T_6 = T_{5 + 1} = {}^nC_5 a^{n - 5} (-b)^5 = -{}^nC_5 a^{n - 5} b^5$.
Given that $T_5 + T_6 = 0$,we have:
${}^nC_4 a^{n - 4} b^4 - {}^nC_5 a^{n - 5} b^5 = 0$
Rearranging the terms:
${}^nC_4 a^{n - 4} b^4 = {}^nC_5 a^{n - 5} b^5$
Dividing both sides by $a^{n - 5} b^4$:
$\frac{a}{b} = \frac{{}^nC_5}{{}^nC_4}$
Using the formula ${}^nC_r = \frac{n!}{r!(n - r)!}$:
$\frac{a}{b} = \frac{n!}{5!(n - 5)!} \times \frac{4!(n - 4)!}{n!}$
$\frac{a}{b} = \frac{(n - 4)!}{(n - 5)!} \times \frac{4!}{5!}$
$\frac{a}{b} = \frac{n - 4}{5}$.

(A) Given $\cot \alpha_1 \cdot \cot \alpha_2 \cdots \cot \alpha_n = 1$.
This implies $\frac{\cos \alpha_1}{\sin \alpha_1} \cdot \frac{\cos \alpha_2}{\sin \alpha_2} \cdots \frac{\cos \alpha_n}{\sin \alpha_n} = 1$,so $\cos \alpha_1 \cdot \cos \alpha_2 \cdots \cos \alpha_n = \sin \alpha_1 \cdot \sin \alpha_2 \cdots \sin \alpha_n$.
Let $P = \cos \alpha_1 \cdot \cos \alpha_2 \cdots \cos \alpha_n$.
Then $P^2 = (\cos \alpha_1 \cdot \cos \alpha_2 \cdots \cos \alpha_n) \cdot (\sin \alpha_1 \cdot \sin \alpha_2 \cdots \sin \alpha_n)$.
Using the identity $\sin \theta \cos \theta = \frac{1}{2} \sin 2\theta$,we get $P^2 = \frac{1}{2^n} \sin 2\alpha_1 \cdot \sin 2\alpha_2 \cdots \sin 2\alpha_n$.
Since $\sin 2\alpha_i \le 1$ for all $i$,we have $P^2 \le \frac{1}{2^n}$.
Taking the square root,$P \le \sqrt{\frac{1}{2^n}} = \frac{1}{2^{n/2}}$.
The maximum value is $\frac{1}{2^{n/2}}$.

(D) The given lines are $y - mx = 0$,$y - mx - 1 = 0$,$y - nx = 0$,and $y - nx - 1 = 0$.
These lines form a parallelogram.
The area of a parallelogram formed by lines $a_1x + b_1y + c_1 = 0$,$a_1x + b_1y + c_2 = 0$,$a_2x + b_2y + d_1 = 0$,and $a_2x + b_2y + d_2 = 0$ is given by the formula:
$\text{Area} = \left| \frac{(c_1 - c_2)(d_1 - d_2)}{a_1b_2 - a_2b_1} \right|$
Here,the equations are:
$mx - y + 0 = 0$
$mx - y + 1 = 0$
$nx - y + 0 = 0$
$nx - y + 1 = 0$
Comparing with the formula,we have $c_1 = 0, c_2 = 1, d_1 = 0, d_2 = 1, a_1 = m, b_1 = -1, a_2 = n, b_2 = -1$.
$\text{Area} = \left| \frac{(0 - 1)(0 - 1)}{m(-1) - n(-1)} \right| = \left| \frac{(-1)(-1)}{-m + n} \right| = \left| \frac{1}{n - m} \right| = \frac{1}{|m - n|}$.

(D) Given the equation of the parabola: $y^2+4y+4x+2=0$.
Rearranging the terms to complete the square for $y$:
$y^2+4y = -4x-2$.
Adding $4$ to both sides:
$y^2+4y+4 = -4x-2+4$.
$(y+2)^2 = -4x+2$.
$(y+2)^2 = -4(x-\frac{1}{2})$.
Comparing this with the standard form $(y-k)^2 = -4a(x-h)$,we get:
$h = \frac{1}{2}$,$k = -2$,and $4a = 4 \implies a = 1$.
The equation of the directrix for a left-opening parabola is $x = h+a$.
Substituting the values: $x = \frac{1}{2} + 1 = \frac{3}{2}$.
Thus,the correct option is $D$.

(C) Given,$\alpha = \beta + \gamma$.
Since $\alpha + \beta = \frac{\pi}{2}$,we have $\beta = \frac{\pi}{2} - \alpha$.
Substituting $\beta$ in the first equation: $\alpha = (\frac{\pi}{2} - \alpha) + \gamma$,which implies $\gamma = 2\alpha - \frac{\pi}{2}$.
Alternatively,using $\gamma = \alpha - \beta$:
$\tan \gamma = \tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$.
Since $\beta = \frac{\pi}{2} - \alpha$,then $\tan \beta = \cot \alpha = \frac{1}{\tan \alpha}$.
Substituting this: $\tan \gamma = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha (\frac{1}{\tan \alpha})} = \frac{\tan \alpha - \tan \beta}{1 + 1} = \frac{\tan \alpha - \tan \beta}{2}$.
Therefore,$2 \tan \gamma = \tan \alpha - \tan \beta$,which gives $\tan \alpha = \tan \beta + 2 \tan \gamma$.

(A) Let the diameter be $PR = 2r$. Since $PQ$ and $RS$ are tangents at $P$ and $R$ respectively,$PQ \perp PR$ and $RS \perp PR$.
Let $\angle PRQ = \theta$. In $\triangle PQR$,$\tan \theta = \frac{PQ}{PR}$,so $PR = PQ \cot \theta$.
Since $X$ lies on the circle and $PR$ is the diameter,$\angle PXR = 90^{\circ}$.
In $\triangle PXR$,$\angle XPR = 90^{\circ} - \theta$ and $\angle XRP = \theta$.
In $\triangle PXS$,$\angle XPS = 90^{\circ} - \theta$ and $\angle XSP = \theta$. Thus,$\tan \theta = \frac{RS}{PR}$,so $PR = RS \tan \theta$.
Equating the two expressions for $PR$:
$PQ \cot \theta = RS \tan \theta$
$\tan^2 \theta = \frac{PQ}{RS} \Rightarrow \tan \theta = \sqrt{\frac{PQ}{RS}}$.
Substituting this back into $PR = RS \tan \theta$:
$PR = RS \cdot \sqrt{\frac{PQ}{RS}} = \sqrt{PQ \cdot RS}$.
Since $PR = 2r$,we have $2r = \sqrt{PQ \cdot RS}$.

(C) The scalar triple product $[a\,b\,c]$ is given by the determinant of the components of the vectors $a$, $b$, and $c$:
$[a\,b\,c] = \begin{vmatrix} 1 & 0 & -1 \\ x & 1 & 1 - x \\ y & x & 1 + x - y \end{vmatrix}$
Applying the column operation $C_3 \to C_3 + C_1$:
$[a\,b\,c] = \begin{vmatrix} 1 & 0 & 0 \\ x & 1 & 1 \\ y & x & 1 + x \end{vmatrix}$
Expanding along the first row:
$[a\,b\,c] = 1 \times \begin{vmatrix} 1 & 1 \\ x & 1 + x \end{vmatrix} - 0 + 0 = (1 + x) - x = 1$.
Since the result is a constant $1$, the value of $[a\,b\,c]$ does not depend on $x$ or $y$.

(B) Given that $a$,$b$,and $c$ are unit vectors,so $|a| = |b| = |c| = 1$.
We know that $|a - b|^2 = |a|^2 + |b|^2 - 2(a \cdot b) = 1 + 1 - 2(a \cdot b) = 2 - 2(a \cdot b)$.
Similarly,$|b - c|^2 = 2 - 2(b \cdot c)$ and $|c - a|^2 = 2 - 2(c \cdot a)$.
Adding these three expressions:
$|a - b|^2 + |b - c|^2 + |c - a|^2 = 6 - 2(a \cdot b + b \cdot c + c \cdot a)$.
We also know that $|a + b + c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a) = 3 + 2(a \cdot b + b \cdot c + c \cdot a)$.
Therefore,$2(a \cdot b + b \cdot c + c \cdot a) = |a + b + c|^2 - 3$.
Substituting this into the sum:
$|a - b|^2 + |b - c|^2 + |c - a|^2 = 6 - (|a + b + c|^2 - 3) = 9 - |a + b + c|^2$.
Since $|a + b + c|^2 \ge 0$,the maximum value of the expression is $9 - 0 = 9$.
Thus,the expression does not exceed $9$.

(D) For the function $f(x) = \frac{\log_2(x + 3)}{x^2 + 3x + 2}$ to be defined:
$1$. The argument of the logarithm must be positive: $x + 3 > 0 \implies x > -3$.
$2$. The denominator must not be zero: $x^2 + 3x + 2 \neq 0$.
Factorizing the denominator: $(x + 1)(x + 2) \neq 0$,which implies $x \neq -1$ and $x \neq -2$.
Combining these conditions,the domain is $x \in (-3, \infty)$ excluding the points $\{-1, -2\}$.
Thus,the domain is $(-3, \infty) - \{-1, -2\}$.

(B) Given $g(x) = 1 + x - [x]$.
We know that $x - [x] = \{x\}$,where $\{x\}$ is the fractional part of $x$.
Thus,$g(x) = 1 + \{x\}$.
Since $0 \le \{x\} < 1$,it follows that $1 \le g(x) < 2$.
Therefore,$g(x) > 0$ for all $x \in \mathbb{R}$.
Now,$f(g(x))$ is defined as:
$f(g(x)) = \begin{cases} -1, & g(x) < 0 \\ 0, & g(x) = 0 \\ 1, & g(x) > 0 \end{cases}$.
Since $g(x) > 0$ for all $x$,we have $f(g(x)) = 1$ for all $x$.

(D) Given $f(x) = \frac{\alpha x}{x + 1}$.
We need to find $\alpha$ such that $f(f(x)) = x$.
First,calculate $f(f(x)) = \frac{\alpha f(x)}{f(x) + 1}$.
Substitute $f(x) = \frac{\alpha x}{x + 1}$ into the expression:
$f(f(x)) = \frac{\alpha \left( \frac{\alpha x}{x + 1} \right)}{\frac{\alpha x}{x + 1} + 1} = \frac{\frac{\alpha^2 x}{x + 1}}{\frac{\alpha x + x + 1}{x + 1}} = \frac{\alpha^2 x}{(\alpha + 1)x + 1}$.
We are given $f(f(x)) = x$,so:
$\frac{\alpha^2 x}{(\alpha + 1)x + 1} = x$.
Assuming $x \neq 0$,we have $\frac{\alpha^2}{(\alpha + 1)x + 1} = 1$.
$\alpha^2 = (\alpha + 1)x + 1$.
For this to hold for all $x$ in the domain,the coefficients of $x$ must be zero and the constant terms must be equal:
$\alpha + 1 = 0 \implies \alpha = -1$.
Checking the constant term: $\alpha^2 = 1$. If $\alpha = -1$,then $(-1)^2 = 1$,which is true.
Thus,the value of $\alpha$ is $-1$.

(A) The left-hand derivative at $x = k$ is defined as $f'(k^-) = \lim_{h \to 0^+} \frac{f(k - h) - f(k)}{-h}$.
Given $f(x) = [x]\sin(\pi x)$,we have $f(k) = [k]\sin(\pi k) = k \times 0 = 0$.
For small $h > 0$,$[k - h] = k - 1$.
Thus,$f(k - h) = (k - 1)\sin(\pi(k - h)) = (k - 1)\sin(\pi k - \pi h) = (k - 1)(\sin(\pi k)\cos(\pi h) - \cos(\pi k)\sin(\pi h))$.
Since $\sin(\pi k) = 0$ and $\cos(\pi k) = (-1)^k$,we get $f(k - h) = (k - 1)(0 - (-1)^k \sin(\pi h)) = -(k - 1)(-1)^k \sin(\pi h) = (k - 1)(-1)^{k+1} \sin(\pi h)$.
Substituting this into the limit: $f'(k^-) = \lim_{h \to 0^+} \frac{(k - 1)(-1)^{k+1} \sin(\pi h) - 0}{-h}$.
Using $\lim_{h \to 0} \frac{\sin(\pi h)}{\pi h} = 1$,we get $f'(k^-) = (k - 1)(-1)^{k+1} \times (-\pi) = (k - 1)(-1)^{k+1} (-1) \pi = (k - 1)(-1)^{k+2} \pi = (k - 1)(-1)^k \pi$.

(A) Given $f(x) = x e^{x(1 - x)}$.
Applying the product rule for differentiation,we get:
$f'(x) = 1 \cdot e^{x(1 - x)} + x \cdot e^{x(1 - x)} \cdot \frac{d}{dx}(x - x^2)$
$f'(x) = e^{x(1 - x)} + x \cdot e^{x(1 - x)} \cdot (1 - 2x)$
$f'(x) = e^{x(1 - x)} \{1 + x(1 - 2x)\}$
$f'(x) = e^{x(1 - x)} \cdot (1 + x - 2x^2)$
$f'(x) = e^{x(1 - x)} \cdot (1 - x)(1 + 2x)$
Since $e^{x(1 - x)}$ is always positive for all real $x$,the sign of $f'(x)$ depends on the quadratic expression $(1 - x)(1 + 2x)$.
The roots of the quadratic are $x = 1$ and $x = -\frac{1}{2}$.
Testing the intervals:
For $x \in \left[ -\frac{1}{2}, 1 \right]$,$f'(x) \ge 0$.
Thus,$f(x)$ is increasing on the interval $\left[ -\frac{1}{2}, 1 \right]$.

(C) Given $F(x) = \int_0^{x^2} f(t) dt = x^2(1 + x)$.
Applying the Leibniz rule for differentiation under the integral sign,we differentiate both sides with respect to $x$:
$\frac{d}{dx} \left( \int_0^{x^2} f(t) dt \right) = \frac{d}{dx} (x^2 + x^3)$.
Using the chain rule,$f(x^2) \cdot \frac{d}{dx}(x^2) = 2x + 3x^2$.
$f(x^2) \cdot 2x = 2x + 3x^2$.
For $x > 0$,we can divide by $2x$:
$f(x^2) = \frac{2x + 3x^2}{2x} = 1 + \frac{3}{2}x$.
To find $f(4)$,we set $x^2 = 4$,which implies $x = 2$ (since $x > 0$):
$f(4) = 1 + \frac{3}{2}(2) = 1 + 3 = 4$.

(C) Let $I = \int_{-\pi}^{\pi} \frac{\cos^2 x}{1 + a^x} dx$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we replace $x$ with $-\pi + \pi - x = -x$:
$I = \int_{-\pi}^{\pi} \frac{\cos^2(-x)}{1 + a^{-x}} dx = \int_{-\pi}^{\pi} \frac{\cos^2 x}{1 + \frac{1}{a^x}} dx = \int_{-\pi}^{\pi} \frac{a^x \cos^2 x}{a^x + 1} dx$.
Adding the two expressions for $I$:
$2I = \int_{-\pi}^{\pi} \frac{\cos^2 x}{1 + a^x} dx + \int_{-\pi}^{\pi} \frac{a^x \cos^2 x}{1 + a^x} dx = \int_{-\pi}^{\pi} \frac{(1 + a^x) \cos^2 x}{1 + a^x} dx = \int_{-\pi}^{\pi} \cos^2 x dx$.
Since $\cos^2 x$ is an even function,$2I = 2 \int_{0}^{\pi} \cos^2 x dx = \int_{0}^{\pi} (1 + \cos 2x) dx$.
$2I = [x + \frac{\sin 2x}{2}]_{0}^{\pi} = (\pi + 0) - (0 + 0) = \pi$.
Therefore,$I = \frac{\pi}{2}$.

(C) Let the given determinant be $\Delta$. Adding $R_2$ and $R_3$ to $R_1$,we get:
$\Delta = \left| {\begin{array}{*{20}{c}}{\sin x + 2\cos x}&{\sin x + 2\cos x}&{\sin x + 2\cos x}\\{\cos x}&{\sin x}&{\cos x}\\{\cos x}&{\cos x}&{\sin x}\end{array}} \right| = 0$
Taking $(\sin x + 2\cos x)$ common from $R_1$:
$\Delta = (\sin x + 2\cos x) \left| {\begin{array}{*{20}{c}}1&1&1\\{\cos x}&{\sin x}&{\cos x}\\{\cos x}&{\cos x}&{\sin x}\end{array}} \right| = 0$
Applying $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$:
$\Delta = (\sin x + 2\cos x) \left| {\begin{array}{*{20}{c}}1&0&0\\{\cos x}&{\sin x - \cos x}&0\\{\cos x}&0&{\sin x - \cos x}\end{array}} \right| = 0$
$\Delta = (\sin x + 2\cos x)(\sin x - \cos x)^2 = 0$
This implies $\sin x + 2\cos x = 0$ or $(\sin x - \cos x)^2 = 0$.
Case $1$: $\tan x = -2$. In the interval $[-\frac{\pi}{4}, \frac{\pi}{4}]$,$\tan x$ ranges from $-1$ to $1$. Thus,$\tan x = -2$ has no solution.
Case $2$: $\sin x = \cos x \implies \tan x = 1$. In the interval $[-\frac{\pi}{4}, \frac{\pi}{4}]$,$\tan x = 1$ at $x = \frac{\pi}{4}$.
Thus,there is only $1$ distinct real root.

(B) We know that ${\sin ^{ - 1}}y + {\cos ^{ - 1}}y = \frac{\pi }{2}$ for $|y| \le 1$.
Given the equation ${\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - \dots} \right) + {\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - \dots} \right) = \frac{\pi }{2}$,it implies that the arguments of the inverse trigonometric functions must be equal.
Let $y = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - \dots$. This is a geometric series with first term $a = x$ and common ratio $r = -x/2$. The sum is $\frac{x}{1 - (-x/2)} = \frac{x}{1 + x/2} = \frac{2x}{2 + x}$.
Similarly,the second argument is $x^2 - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - \dots$,which is a geometric series with $a = x^2$ and $r = -x^2/2$. The sum is $\frac{x^2}{1 + x^2/2} = \frac{2x^2}{2 + x^2}$.
Equating the two: $\frac{2x}{2 + x} = \frac{2x^2}{2 + x^2}$.
Since $x \neq 0$,we can divide by $2x$: $\frac{1}{2 + x} = \frac{x}{2 + x^2}$.
Cross-multiplying gives $2 + x^2 = 2x + x^2$,which simplifies to $2 = 2x$,so $x = 1$.

(A) The total number of functions from a set $E$ with $n$ elements to a set $F$ with $m$ elements is $m^n$.
Here,$n = |E| = 4$ and $m = |F| = 2$.
So,the total number of functions is $2^4 = 16$.
An onto function (surjective function) means every element in the codomain $F$ must have at least one pre-image in the domain $E$.
The only functions that are not onto are those where all elements of $E$ map to either ${1}$ or ${2}$.
There are exactly $2$ such constant functions: $f(x) = 1$ for all $x \in E$ and $f(x) = 2$ for all $x \in E$.
Therefore,the number of onto functions is $16 - 2 = 14$.

(A) Given $y = f(x) = x + \frac{1}{x}$ where $x \ge 1$ and $y \ge 2$.
Multiplying by $x$,we get $yx = x^2 + 1$,which rearranges to the quadratic equation $x^2 - yx + 1 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{y \pm \sqrt{y^2 - 4}}{2}$.
Since the domain of $f$ is $[1, +\infty)$,we must have $x \ge 1$.
If we take the negative sign,$x = \frac{y - \sqrt{y^2 - 4}}{2}$. For $y \ge 2$,this value is $\le 1$. Specifically,as $y \to \infty$,this expression approaches $0$.
Thus,to satisfy $x \ge 1$,we must take the positive sign: $x = \frac{y + \sqrt{y^2 - 4}}{2}$.
Replacing $y$ with $x$,we obtain ${f^{-1}}(x) = \frac{x + \sqrt{x^2 - 4}}{2}$.

(D) To find the points where $f(x) = \max(x, x^3)$ is not differentiable,we analyze the behavior of $x$ and $x^3$ in different intervals:
$1$. If $x < -1$,then $x^3 < x$,so $f(x) = x$.
$2$. If $-1 < x < 0$,then $x^3 > x$,so $f(x) = x^3$.
$3$. If $0 < x < 1$,then $x > x^3$,so $f(x) = x$.
$4$. If $x > 1$,then $x^3 > x$,so $f(x) = x^3$.
At the transition points $x = -1, 0, 1$,we check the left-hand and right-hand derivatives:
At $x = -1$: $f'( -1^-) = 1$ and $f'( -1^+) = 3(-1)^2 = 3$. Since $1 \neq 3$,it is not differentiable.
At $x = 0$: $f'( 0^-) = 3(0)^2 = 0$ and $f'( 0^+) = 1$. Since $0 \neq 1$,it is not differentiable.
At $x = 1$: $f'( 1^-) = 1$ and $f'( 1^+) = 3(1)^2 = 3$. Since $1 \neq 3$,it is not differentiable.
Thus,the set of points where $f(x)$ is not differentiable is $\{ -1, 0, 1\}$.

(D) Let $f(x) = \sin(|x|) - |x|$. We check the differentiability at $x = 0$ by calculating the Left Hand Derivative $(LHD)$ and Right Hand Derivative $(RHD)$.
For $x < 0$,$|x| = -x$,so $f(x) = \sin(-x) - (-x) = -\sin x + x$. The derivative is $f'(x) = -\cos x + 1$. Thus,$LHD = \lim_{x \to 0^-} (-\cos x + 1) = -\cos(0) + 1 = -1 + 1 = 0$.
For $x > 0$,$|x| = x$,so $f(x) = \sin x - x$. The derivative is $f'(x) = \cos x - 1$. Thus,$RHD = \lim_{x \to 0^+} (\cos x - 1) = \cos(0) - 1 = 1 - 1 = 0$.
Since $LHD = RHD = 0$,the function $\sin(|x|) - |x|$ is differentiable at $x = 0$.
For the other options,$\cos(|x|)$ is differentiable at $x=0$ but $|x|$ is not,so their sum or difference is not differentiable at $x=0$. Similarly,$\sin(|x|) + |x|$ has $LHD = -2$ and $RHD = 2$,so it is not differentiable.

(C) Given the curve $f(x) = x^2 + bx - b$. Since the point $(1, 1)$ lies on the curve,we have $1 = 1^2 + b(1) - b$,which is $1 = 1$,confirming the point is on the curve for any $b$.
First,find the derivative: $\frac{dy}{dx} = 2x + b$.
At the point $(1, 1)$,the slope of the tangent is $m = 2(1) + b = 2 + b$.
The equation of the tangent line at $(1, 1)$ is $y - 1 = (2 + b)(x - 1)$.
Simplifying,$y - 1 = (2 + b)x - (2 + b)$,which gives $(2 + b)x - y = 1 + b$.
To find the intercepts,set $y = 0$ to get $x = \frac{1 + b}{2 + b}$ (point $A$),and set $x = 0$ to get $y = -(1 + b)$ (point $B$).
Since the triangle lies in the first quadrant,the intercepts must be positive: $\frac{1 + b}{2 + b} > 0$ and $-(1 + b) > 0$.
This implies $1 + b < 0$ and $2 + b < 0$,so $b < -2$.
The area of the triangle is $\frac{1}{2} \times |OA| \times |OB| = \frac{1}{2} \times \left| \frac{1 + b}{2 + b} \right| \times |-(1 + b)| = 2$.
Since $b < -2$,$1 + b$ is negative and $2 + b$ is negative,so $\frac{1 + b}{2 + b} > 0$ and $-(1 + b) > 0$.
Thus,$\frac{1}{2} \cdot \frac{1 + b}{2 + b} \cdot (-(1 + b)) = 2$.
$-(1 + b)^2 = 4(2 + b) \Rightarrow -(1 + 2b + b^2) = 8 + 4b \Rightarrow b^2 + 6b + 9 = 0$.
$(b + 3)^2 = 0$,so $b = -3$. This satisfies $b < -2$.