IIT JEE 2017 Mathematics Question Paper with Answer and Solution

(D) The equation of the parabola is $y^2=16x$.
The equation of the chord is $2x+y=p$.
The equation of a chord with midpoint $(h, k)$ is given by $T=S_1$,where $T$ is the tangent form and $S_1$ is the value of the parabola equation at $(h, k)$.
$yk-8(x+h) = k^2-16h$
$yk-8x = k^2-8h$
Comparing this with the given chord equation $y = -2x+p$ or $2x+y=p$:
$\frac{k}{1} = \frac{-8}{2} = \frac{k^2-8h}{p}$
From $\frac{k}{1} = -4$,we get $k=-4$.
From $\frac{-8}{2} = \frac{k^2-8h}{p}$,we get $-4 = \frac{(-4)^2-8h}{p} = \frac{16-8h}{p}$.
$-4p = 16-8h$ $\Rightarrow 8h-4p = 16$ $\Rightarrow 2h-p = 4$.
Checking the options:
For option $D$: $p=2, h=3, k=-4$.
$2(3)-2 = 6-2 = 4$. This satisfies the condition.
Thus,the correct option is $D$.

(D) The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$.
Given the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{16} = 1$,we have $b^2 = 16$.
The given tangent is $2x - y + 1 = 0$,which can be written as $y = 2x + 1$.
Comparing this with $y = mx + c$,we get $m = 2$ and $c = 1$.
The condition for tangency is $c^2 = a^2m^2 - b^2$.
Substituting the values: $1^2 = a^2(2)^2 - 16$.
$1 = 4a^2 - 16 \Rightarrow 4a^2 = 17 \Rightarrow a^2 = \frac{17}{4} \Rightarrow a = \frac{\sqrt{17}}{2}$.
Now,we check the sides for a right-angled triangle (where the sum of squares of two smaller sides equals the square of the largest side):
$[A]$ $2a = \sqrt{17} \approx 4.12, 4, 1$. Sides: $\sqrt{17}, 4, 1$. $1^2 + 4^2 = 17 = (\sqrt{17})^2$. This forms a right triangle.
$[B]$ $2a = \sqrt{17}, 8, 1$. Sides: $\sqrt{17}, 8, 1$. $1^2 + (\sqrt{17})^2 = 18 \neq 8^2 = 64$. Not a right triangle.
$[C]$ $a = \frac{\sqrt{17}}{2} \approx 2.06, 4, 1$. Sides: $2.06, 4, 1$. $1^2 + 2.06^2 \approx 5.24 \neq 4^2 = 16$. Not a right triangle.
$[D]$ $a = \frac{\sqrt{17}}{2}, 4, 2$. Sides: $2.06, 4, 2$. $2^2 + 2.06^2 \approx 8.24 \neq 4^2 = 16$. Not a right triangle.
Thus,options $B, C,$ and $D$ cannot be sides of a right-angled triangle.

(B) Let the sides of the triangle be $a-d$,$a$,and $a+d$,where $d > 0$.
Since it is a right-angled triangle,by the Pythagorean theorem:
$(a-d)^2 + a^2 = (a+d)^2$
$a^2 - 2ad + d^2 + a^2 = a^2 + 2ad + d^2$
$a^2 - 4ad = 0$
Since $a \neq 0$,we have $a = 4d$.
The sides are $3d$,$4d$,and $5d$.
The area of the triangle is given by $\frac{1}{2} \times \text{base} \times \text{height} = 24$.
$\frac{1}{2} \times 3d \times 4d = 24$
$6d^2 = 24$
$d^2 = 4 \Rightarrow d = 2$.
The sides are $3(2) = 6$,$4(2) = 8$,and $5(2) = 10$.
The smallest side is $6$.

(A) The given equation of the circle is $x^2+y^2+2x+4y-p=0$.
Completing the square,we get $(x+1)^2+(y+2)^2 = p+5$.
The center of the circle is $(-1, -2)$ and the radius is $r = \sqrt{p+5}$.
For the circle to have exactly three common points with the coordinate axes,it must be tangent to one axis and pass through the origin,or be tangent to both axes (which is impossible here as the center is $(-1, -2)$).
Case $1$: The circle is tangent to the $x$-axis.
This occurs when the distance from the center $(-1, -2)$ to the $x$-axis equals the radius,so $|-2| = \sqrt{p+5}$ $\Rightarrow 4 = p+5$ $\Rightarrow p = -1$.
Case $2$: The circle is tangent to the $y$-axis.
This occurs when the distance from the center $(-1, -2)$ to the $y$-axis equals the radius,so $|-1| = \sqrt{p+5}$ $\Rightarrow 1 = p+5$ $\Rightarrow p = -4$.
Case $3$: The circle passes through the origin $(0,0)$.
Substituting $(0,0)$ into the equation,$0^2+0^2+2(0)+4(0)-p=0 \Rightarrow p = 0$.
If $p=0$,the circle is $x^2+y^2+2x+4y=0$,which passes through the origin and intersects the axes at $(0,0), (-2,0), (0,-4)$,giving exactly three points.
If $p=-1$,the circle is tangent to the $x$-axis at $(-1,0)$ and intersects the $y$-axis at two points,giving three points in total.
If $p=-4$,the circle is tangent to the $y$-axis at $(0,-2)$ and intersects the $x$-axis at two points,giving three points in total.
Thus,there are $3$ possible values for $p$ $(p=0, -1, -4)$. However,checking the options provided,the intended answer is $2$.

(A) The total number of letters available is $10$.
For $x$,we form words of length $10$ with no repetition,which is a permutation of $10$ distinct letters: $x = 10!$.
For $y$,we choose $1$ letter to be repeated twice from $10$ letters in $^{10}C_1$ ways.
We then choose $8$ other letters from the remaining $9$ letters in $^9C_8$ ways.
The total number of arrangements of these $10$ letters (where one is repeated twice) is $\frac{10!}{2!}$.
Thus,$y = {}^{10}C_1 \times {}^{9}C_8 \times \frac{10!}{2!}$.
Calculating the ratio: $\frac{y}{9x} = \frac{{}^{10}C_1 \times {}^{9}C_8 \times \frac{10!}{2!}}{9 \times 10!} = \frac{10 \times 9}{9 \times 2} = 5$.

(C) $(1)$ The point $(\sqrt{3}, 1/2)$ lies on the ellipse $x^2+4y^2=4$,which is $x^2+a^2y^2=a^2$ with $a=2$. The tangent $\sqrt{3}x+2y=4$ has slope $m=-\sqrt{3}/2$. The equation of the tangent is $y=mx+\sqrt{a^2m^2+1}$. Thus,$(II)(iv)(R)$ is the correct combination.
$(2)$ The point $(8, 16)$ lies on the parabola $y^2=4ax$. For $y^2=4ax$,$16^2=4a(8) \implies 256=32a \implies a=8$. The tangent $y=x+8$ has $m=1$. The equation is $my=m^2x+a \implies y=x+8$. The point of contact is $(a/m^2, 2a/m) = (8/1, 16/1) = (8, 16)$. Thus,$(III)(i)(P)$ is the correct combination.
$(3)$ For $a=\sqrt{2}$,the point $(-1, 1)$ lies on $x^2+y^2=a^2$. The tangent at $(-1, 1)$ to $x^2+y^2=2$ is $-x+y=2 \implies y=x+2$. Comparing with $y=mx+a\sqrt{m^2+1}$,we get $m=1$ and $a\sqrt{m^2+1} = \sqrt{2}\sqrt{1^2+1} = 2$. The point of contact is $(-ma/\sqrt{m^2+1}, a/\sqrt{m^2+1}) = (-1, 1)$. Thus,$(I)(ii)(Q)$ is the correct combination.

(D) The set $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ contains $5$ odd numbers $\{1, 3, 5, 7, 9\}$ and $4$ even numbers $\{2, 4, 6, 8\}$.
We are looking for the total number of subsets of $S$ with $5$ elements,where the number of odd elements $k$ can be $1, 2, 3, 4,$ or $5$.
Since there are only $4$ even numbers in $S$,any subset of $5$ elements must contain at least $5 - 4 = 1$ odd number.
Therefore,the sum $N_1 + N_2 + N_3 + N_4 + N_5$ represents the total number of ways to choose $5$ elements from the $9$ elements in $S$.
This is given by the combination formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.
$N_1 + N_2 + N_3 + N_4 + N_5 = \binom{9}{5} = \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$.

(B) The total number of non-negative integer solutions to $x+y+z=n$ is given by the formula $\binom{n+k-1}{k-1}$,where $k=3$ and $n=10$.
Total solutions $= \binom{10+3-1}{3-1} = \binom{12}{2} = \frac{12 \times 11}{2} = 66$.
For $z$ to be even,let $z \in \{0, 2, 4, 6, 8, 10\}$.
If $z=k$,then $x+y=10-k$. The number of solutions for $x+y=m$ is $m+1$.
For $z=0$,$x+y=10$,solutions $= 11$.
For $z=2$,$x+y=8$,solutions $= 9$.
For $z=4$,$x+y=6$,solutions $= 7$.
For $z=6$,$x+y=4$,solutions $= 5$.
For $z=8$,$x+y=2$,solutions $= 3$.
For $z=10$,$x+y=0$,solutions $= 1$.
Total favourable solutions $= 11+9+7+5+3+1 = 36$.
Probability $P = \frac{36}{66} = \frac{6}{11}$.

(A,C) Given $2(\cos \beta - \cos \alpha) + \cos \alpha \cos \beta = 1$.
Rearranging the terms,we get $\cos \beta(2 + \cos \alpha) = 1 + 2 \cos \alpha$.
So,$\cos \beta = \frac{1 + 2 \cos \alpha}{2 + \cos \alpha}$.
Applying componendo and dividendo,$\frac{\cos \beta - 1}{\cos \beta + 1} = \frac{(1 + 2 \cos \alpha) - (2 + \cos \alpha)}{(1 + 2 \cos \alpha) + (2 + \cos \alpha)} = \frac{\cos \alpha - 1}{3(1 + \cos \alpha)}$.
Using the identities $\cos \theta - 1 = -2 \sin^2(\theta/2)$ and $\cos \theta + 1 = 2 \cos^2(\theta/2)$,we get $\frac{-2 \sin^2(\beta/2)}{2 \cos^2(\beta/2)} = \frac{-2 \sin^2(\alpha/2)}{3(2 \cos^2(\alpha/2))}$.
This simplifies to $-\tan^2(\beta/2) = -\frac{1}{3} \tan^2(\alpha/2)$,which implies $\tan^2(\alpha/2) = 3 \tan^2(\beta/2)$.
Taking the square root,$\tan(\alpha/2) = \pm \sqrt{3} \tan(\beta/2)$.
Thus,$\tan(\alpha/2) - \sqrt{3} \tan(\beta/2) = 0$ or $\tan(\alpha/2) + \sqrt{3} \tan(\beta/2) = 0$.

(B, D) $(1)$ Since $\alpha$ and $\beta$ are roots of $x^2-x-1=0$,we have $\alpha^2 = \alpha+1$ and $\beta^2 = \beta+1$.
Multiplying by $\alpha^{n-2}$ and $\beta^{n-2}$ respectively: $\alpha^n = \alpha^{n-1} + \alpha^{n-2}$ and $\beta^n = \beta^{n-1} + \beta^{n-2}$.
Multiplying by $p$ and $q$ and adding: $p\alpha^n + q\beta^n = (p\alpha^{n-1} + q\beta^{n-1}) + (p\alpha^{n-2} + q\beta^{n-2})$.
Thus,$a_n = a_{n-1} + a_{n-2}$. For $n=12$,$a_{12} = a_{11} + a_{10}$. Option $B$ is correct.
$(2)$ We have $a_0 = p+q$,$a_1 = p\alpha + q\beta$,$a_2 = p\alpha^2 + q\beta^2 = p(\alpha+1) + q(\beta+1) = a_1 + a_0$,$a_3 = a_2 + a_1 = 2a_1 + a_0$,$a_4 = a_3 + a_2 = 3a_1 + 2a_0$.
Given $a_4 = 3(p\alpha + q\beta) + 2(p+q) = 28$.
Since $\alpha = \frac{1+\sqrt{5}}{2}$ and $\beta = \frac{1-\sqrt{5}}{2}$,$3p(\frac{1+\sqrt{5}}{2}) + 3q(\frac{1-\sqrt{5}}{2}) + 2p + 2q = 28$.
$\frac{3p+3p\sqrt{5}+3q-3q\sqrt{5}}{2} + 2p + 2q = 28$.
$(3.5p + 3.5q) + \frac{3\sqrt{5}}{2}(p-q) = 28$.
Since $28$ is rational,$p-q=0 \Rightarrow p=q$.
$7p = 28 \Rightarrow p=4, q=4$.
Then $p+2q = 4 + 2(4) = 12$. Option $D$ is correct.

(A) Let $M$ be a $3 \times 3$ matrix with real entries such that $M^2 = A$. Then $\det(M^2) = \det(A)$,which implies $(\det(M))^2 = \det(A)$. Since $M$ has real entries,$\det(M)$ is a real number,so $(\det(M))^2 \ge 0$. Thus,for $A$ to be a square of a real matrix,its determinant must be non-negative.
For option $A$: $\det(A) = \det \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix} = -1 < 0$. Thus,$A$ cannot be a square.
For option $B$: $\det(B) = \det \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} = -1 < 0$. Thus,$B$ cannot be a square.
For option $C$: $\det(C) = 1 > 0$. It is the square of the identity matrix $I^2 = I$.
For option $D$: $\det(D) = \det \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} = 1 > 0$. It is the square of the matrix $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{bmatrix}$.
Therefore,$A$ and $B$ are not squares of real $3 \times 3$ matrices.

(A) Given $P(X) = \frac{1}{3}$,$P(X \mid Y) = \frac{1}{2}$,and $P(Y \mid X) = \frac{2}{5}$.
Using the definition of conditional probability,$P(Y \mid X) = \frac{P(X \cap Y)}{P(X)}$.
$\frac{2}{5} = \frac{P(X \cap Y)}{1/3} \Rightarrow P(X \cap Y) = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}$.
Now,$P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)}$.
$\frac{1}{2} = \frac{2/15}{P(Y)} \Rightarrow P(Y) = \frac{2}{15} \times 2 = \frac{4}{15}$. (Option $D$ is correct)
For $P(X^{\prime} \mid Y)$,we have $P(X^{\prime} \mid Y) = 1 - P(X \mid Y) = 1 - \frac{1}{2} = \frac{1}{2}$. (Option $A$ is correct)
For $P(X \cup Y)$,$P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) = \frac{1}{3} + \frac{4}{15} - \frac{2}{15} = \frac{5+4-2}{15} = \frac{7}{15}$.
Thus,options $A$ and $D$ are correct.

(D) The function is $f(x) = x \cos(\pi(x + [x]))$. $A$ function is discontinuous at points where the left-hand limit $(LHL)$ is not equal to the right-hand limit $(RHL)$ or the function value.
For any integer $n$,in the interval $[n, n+1)$,$[x] = n$. Thus,$f(x) = x \cos(\pi(x + n))$.
$1$. At $x = -1$:
$LHL = \lim_{x \to -1^-} x \cos(\pi(x - 2)) = -1 \cos(-3\pi) = -1(-1) = 1$.
$RHL = \lim_{x \to -1^+} x \cos(\pi(x - 1)) = -1 \cos(-2\pi) = -1(1) = -1$.
Since $LHL \neq RHL$,$f(x)$ is discontinuous at $x = -1$.
$2$. At $x = 0$:
$LHL = \lim_{x \to 0^-} x \cos(\pi(x - 1)) = 0 \cos(-\pi) = 0$.
$RHL = \lim_{x \to 0^+} x \cos(\pi x) = 0 \cos(0) = 0$.
Since $LHL = RHL = f(0) = 0$,$f(x)$ is continuous at $x = 0$.
$3$. At $x = 1$:
$LHL = \lim_{x \to 1^-} x \cos(\pi(x)) = 1 \cos(\pi) = -1$.
$RHL = \lim_{x \to 1^+} x \cos(\pi(x + 1)) = 1 \cos(2\pi) = 1$.
Since $LHL \neq RHL$,$f(x)$ is discontinuous at $x = 1$.
$4$. At $x = 2$:
$LHL = \lim_{x \to 2^-} x \cos(\pi(x + 1)) = 2 \cos(3\pi) = -2$.
$RHL = \lim_{x \to 2^+} x \cos(\pi(x + 2)) = 2 \cos(4\pi) = 2$.
Since $LHL \neq RHL$,$f(x)$ is discontinuous at $x = 2$.
Thus,the function is discontinuous at $x = -1, 1, 2$. The correct option is $D$.

(A) For option $[A]$,let $g(x) = e^x - \int_0^x f(t) \sin t \, dt$. Since $f(t) \in (0,1)$ and $\sin t \in [0,1]$ for $t \in (0,1)$,the integral $\int_0^x f(t) \sin t \, dt < \int_0^x 1 \, dt = x$. Thus $g(x) > e^x - x$. Since $e^x - x > 0$ for all $x$,$g(x)$ cannot be zero.
For option $[B]$,let $h(x) = x^9 - f(x)$. Since $f(x) \in (0,1)$,$h(0) = 0^9 - f(0) = -f(0) < 0$ and $h(1) = 1^9 - f(1) = 1 - f(1) > 0$. By the Intermediate Value Theorem $(IVT)$,there exists $c \in (0,1)$ such that $h(c) = 0$.
For option $[C]$,let $k(x) = f(x) + \int_0^{\pi/2} f(t) \sin t \, dt$. Since $f(x) > 0$ and the integral is positive,$k(x) > 0$ for all $x \in (0,1)$. Thus,it cannot be zero.
For option $[D]$,let $m(x) = x - \int_0^{\pi/2 - x} f(t) \cos t \, dt$. Then $m(0) = 0 - \int_0^{\pi/2} f(t) \cos t \, dt < 0$ and $m(1) = 1 - \int_0^{\pi/2 - 1} f(t) \cos t \, dt$. Since $\int_0^{\pi/2 - 1} f(t) \cos t \, dt < \int_0^{\pi/2 - 1} 1 \, dt = \pi/2 - 1 \approx 0.57 < 1$,$m(1) > 0$. By $IVT$,there exists $c \in (0,1)$ such that $m(c) = 0$.

(D) The system has infinitely many solutions if the determinant of the coefficient matrix $D = 0$ and the augmented matrix satisfies the consistency condition.
First,calculate the determinant $D$:
$D = \begin{vmatrix} 1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1 \end{vmatrix} = 1(1 - \alpha^2) - \alpha(\alpha - \alpha^3) + \alpha^2(\alpha^2 - \alpha^2) = 1 - \alpha^2 - \alpha^2 + \alpha^4 = \alpha^4 - 2\alpha^2 + 1 = (\alpha^2 - 1)^2$.
Setting $D = 0$ gives $(\alpha^2 - 1)^2 = 0$,so $\alpha^2 = 1$,which means $\alpha = 1$ or $\alpha = -1$.
Case $1$: If $\alpha = 1$,the system becomes:
$x + y + z = 1$
$x + y + z = -1$
$x + y + z = 1$
This is inconsistent because $1 \neq -1$,so there are no solutions.
Case $2$: If $\alpha = -1$,the system becomes:
$x - y + z = 1$
$-x + y - z = -1$
$x - y + z = 1$
All three equations are equivalent to $x - y + z = 1$,which represents a plane,thus there are infinitely many solutions.
Therefore,$\alpha = -1$.
Then,$1 + \alpha + \alpha^2 = 1 + (-1) + (-1)^2 = 1 - 1 + 1 = 1$.

(C) The given integral is $g(x)=\int_x^{\pi / 2} [f^{\prime}(t) \operatorname{cosec} t - f(t) \operatorname{cosec} t \cot t] dt$.
Notice that the integrand is the derivative of the product $f(t) \operatorname{cosec} t$ with respect to $t$,since $\frac{d}{dt} [f(t) \operatorname{cosec} t] = f^{\prime}(t) \operatorname{cosec} t - f(t) \operatorname{cosec} t \cot t$.
Thus,$g(x) = [f(t) \operatorname{cosec} t]_x^{\pi / 2} = f(\frac{\pi}{2}) \operatorname{cosec}(\frac{\pi}{2}) - f(x) \operatorname{cosec} x$.
Substituting the values $f(\frac{\pi}{2})=3$ and $\operatorname{cosec}(\frac{\pi}{2})=1$,we get $g(x) = 3 - \frac{f(x)}{\sin x}$.
Now,$\lim _{x \rightarrow 0} g(x) = \lim _{x \rightarrow 0} (3 - \frac{f(x)}{\sin x}) = 3 - \lim _{x \rightarrow 0} \frac{f(x)}{\sin x}$.
Using $L'H\hat{o}pital's$ rule,$\lim _{x \rightarrow 0} \frac{f(x)}{\sin x} = \lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{\cos x} = \frac{f^{\prime}(0)}{\cos 0} = \frac{1}{1} = 1$.
Therefore,$\lim _{x \rightarrow 0} g(x) = 3 - 1 = 2$.

(A) The mass $m$ of the sphere is given by $m = \rho \cdot \frac{4}{3} \pi R^3$.
Since the total mass $m$ is constant,its derivative with respect to time $t$ is zero: $\frac{dm}{dt} = 0$.
Differentiating the mass equation with respect to $t$:
$0 = \frac{d\rho}{dt} \cdot \frac{4}{3} \pi R^3 + \rho \cdot 4 \pi R^2 \frac{dR}{dt}$.
Dividing by $\rho \cdot \frac{4}{3} \pi R^3$,we get:
$0 = \frac{1}{\rho} \frac{d\rho}{dt} + \frac{3}{R} \frac{dR}{dt}$.
Rearranging for the velocity of the surface $v = \frac{dR}{dt}$:
$\frac{dR}{dt} = -\frac{R}{3} \left(\frac{1}{\rho} \frac{d\rho}{dt}\right)$.
Given that the rate of fractional change in density $\left(\frac{1}{\rho} \frac{d\rho}{dt}\right)$ is a constant,let this constant be $k$.
Then $v = \frac{dR}{dt} = -\frac{k}{3} R$.
Thus,$v \propto R$.

(A) From the figure,the vector $\vec{R}$ is given by $\vec{R} = \vec{Q} - \vec{P}$.
Point $S$ lies on the line segment $PQ$ such that the distance $PS = b|\vec{R}|$.
This implies that the vector $\vec{PS} = b\vec{R} = b(\vec{Q} - \vec{P})$.
Using the triangle law of vector addition in $\triangle OPS$,we have $\vec{S} = \vec{P} + \vec{PS}$.
Substituting the expression for $\vec{PS}$,we get $\vec{S} = \vec{P} + b(\vec{Q} - \vec{P})$.
Simplifying this,we obtain $\vec{S} = \vec{P} + b\vec{Q} - b\vec{P} = (1-b)\vec{P} + b\vec{Q}$.

(D) Given the equation: $\overline{OP} \cdot \overline{OQ} + \overline{OR} \cdot \overline{OS} = \overline{OR} \cdot \overline{OP} + \overline{OQ} \cdot \overline{OS}$.
Rearranging the terms: $\overline{OP} \cdot \overline{OQ} - \overline{OR} \cdot \overline{OP} = \overline{OQ} \cdot \overline{OS} - \overline{OR} \cdot \overline{OS}$.
$\overline{OP} \cdot (\overline{OQ} - \overline{OR}) = \overline{OS} \cdot (\overline{OQ} - \overline{OR})$.
$(\overline{OP} - \overline{OS}) \cdot (\overline{OQ} - \overline{OR}) = 0$.
$\overline{SP} \cdot \overline{RQ} = 0$.
This implies $\overline{SP} \perp \overline{RQ}$,meaning $S$ lies on the altitude from $P$ to $QR$.
Similarly,from $\overline{OR} \cdot \overline{OP} + \overline{OQ} \cdot \overline{OS} = \overline{OQ} \cdot \overline{OR} + \overline{OP} \cdot \overline{OS}$,we get $\overline{SQ} \perp \overline{PR}$.
Since $S$ is the intersection of altitudes,$S$ is the orthocenter of $\triangle PQR$.

(A) Given the differential equation: $8 \sqrt{x}(\sqrt{9+\sqrt{x}}) dy = \frac{1}{\sqrt{4+\sqrt{9+\sqrt{x}}}} dx$.
Rearranging the terms,we get: $dy = \frac{dx}{8 \sqrt{x} \sqrt{9+\sqrt{x}} \sqrt{4+\sqrt{9+\sqrt{x}}}}$.
Let $u = 4+\sqrt{9+\sqrt{x}}$.
Then $du = \frac{1}{2\sqrt{9+\sqrt{x}}} \cdot \frac{1}{2\sqrt{x}} dx = \frac{dx}{4\sqrt{x}\sqrt{9+\sqrt{x}}}$.
Substituting this into the differential equation: $dy = \frac{1}{2} \cdot \frac{du}{\sqrt{u}}$.
Integrating both sides: $y = \int \frac{1}{2\sqrt{u}} du = \sqrt{u} + C$.
Substituting back $u$: $y = \sqrt{4+\sqrt{9+\sqrt{x}}} + C$.
Given $y(0) = \sqrt{7}$,we have $\sqrt{4+\sqrt{9+0}} + C = \sqrt{7} \Rightarrow \sqrt{4+3} + C = \sqrt{7} \Rightarrow \sqrt{7} + C = \sqrt{7} \Rightarrow C = 0$.
Thus,$y = \sqrt{4+\sqrt{9+\sqrt{x}}}$.
For $x=256$,$y(256) = \sqrt{4+\sqrt{9+\sqrt{256}}} = \sqrt{4+\sqrt{9+16}} = \sqrt{4+\sqrt{25}} = \sqrt{4+5} = \sqrt{9} = 3$.

(D) Define a function $h(x) = f(x) - x$.
Given $f(\frac{1}{2}) = \frac{1}{2}$ and $f(1) = 1$,we have $h(\frac{1}{2}) = f(\frac{1}{2}) - \frac{1}{2} = 0$ and $h(1) = f(1) - 1 = 0$.
Since $f(x)$ is twice differentiable,$h(x)$ is also differentiable on $[\frac{1}{2}, 1]$.
By Rolle's Theorem,there exists some $\alpha \in (\frac{1}{2}, 1)$ such that $h^{\prime}(\alpha) = 0$.
Since $h^{\prime}(x) = f^{\prime}(x) - 1$,this implies $f^{\prime}(\alpha) = 1$.
We are given $f^{\prime \prime}(x) > 0$ for all $x \in R$,which means $f^{\prime}(x)$ is a strictly increasing function.
Since $\alpha < 1$ and $f^{\prime}(x)$ is strictly increasing,we have $f^{\prime}(\alpha) < f^{\prime}(1)$.
Substituting $f^{\prime}(\alpha) = 1$,we get $1 < f^{\prime}(1)$.

(B) Let $M = [m_{ij}]$ be a $3 \times 3$ matrix where $m_{ij} \in \{0, 1, 2\}$.
The sum of the diagonal entries of $M^T M$ is given by the trace of $M^T M$,which is equal to the sum of the squares of all entries of $M$,i.e.,$\sum_{i=1}^3 \sum_{j=1}^3 m_{ij}^2 = 5$.
Since $m_{ij} \in \{0, 1, 2\}$,the squares of the entries are $m_{ij}^2 \in \{0, 1, 4\}$.
We need to find the number of ways to choose $9$ entries such that their squares sum to $5$. The possible combinations of squares are:
Case $1$: Five entries are $1$ and four entries are $0$. The number of ways is $\binom{9}{5} = 126$.
Case $2$: One entry is $2$ (square is $4$),one entry is $1$ (square is $1$),and seven entries are $0$. The number of ways to choose the position of $2$ is $\binom{9}{1} = 9$,and the number of ways to choose the position of $1$ from the remaining $8$ positions is $\binom{8}{1} = 8$. Thus,$9 \times 8 = 72$ ways.
Total number of matrices $= 126 + 72 = 198$.

(C) Given $g(x) = \int_{\sin x}^{\sin(2x)} \sin^{-1}(t) \, dt$.
Using Leibniz's Rule for differentiation under the integral sign:
$g^{\prime}(x) = \sin^{-1}(\sin(2x)) \cdot \frac{d}{dx}(\sin(2x)) - \sin^{-1}(\sin x) \cdot \frac{d}{dx}(\sin x)$.
$g^{\prime}(x) = \sin^{-1}(\sin(2x)) \cdot (2\cos(2x)) - \sin^{-1}(\sin x) \cdot (\cos x)$.
For $x = \frac{\pi}{2}$:
$g^{\prime}\left(\frac{\pi}{2}\right) = \sin^{-1}(\sin(\pi)) \cdot (2\cos(\pi)) - \sin^{-1}(\sin(\frac{\pi}{2})) \cdot (\cos(\frac{\pi}{2}))$.
Since $\sin(\pi) = 0$,$\sin^{-1}(0) = 0$.
Since $\cos(\frac{\pi}{2}) = 0$,the second term is also $0$.
Thus,$g^{\prime}\left(\frac{\pi}{2}\right) = 0 \cdot (-2) - (\frac{\pi}{2}) \cdot 0 = 0$.
Therefore,option $C$ is correct.

(C) Given $f^{\prime}(x) - 2f(x) > 0$.
Multiply by the integrating factor $e^{-2x}$:
$e^{-2x} f^{\prime}(x) - 2e^{-2x} f(x) > 0$.
This is equivalent to $\frac{d}{dx}(f(x) e^{-2x}) > 0$.
Let $g(x) = f(x) e^{-2x}$. Since $g^{\prime}(x) > 0$,$g(x)$ is a strictly increasing function.
For $x > 0$,$g(x) > g(0)$.
Since $g(0) = f(0) e^0 = 1 \cdot 1 = 1$,we have $f(x) e^{-2x} > 1$,which implies $f(x) > e^{2x}$ for all $x > 0$.
Since $f(x) > e^{2x} > 0$ for $x > 0$,and $f^{\prime}(x) > 2f(x)$,we have $f^{\prime}(x) > 2e^{2x} > 0$.
Since $f^{\prime}(x) > 0$,$f(x)$ is an increasing function in $(0, \infty)$.
Thus,$A$ and $C$ are correct.

(B) For $x \to 1^{-}$,let $x = 1 - h$ where $h > 0$. Then $|1 - x| = h$.
$f(1 - h) = \frac{1 - (1 - h)(1 + h)}{h} \cos \left(\frac{1}{h}\right) = \frac{1 - (1 - h^2)}{h} \cos \left(\frac{1}{h}\right) = \frac{h^2}{h} \cos \left(\frac{1}{h}\right) = h \cos \left(\frac{1}{h}\right)$.
Since $\lim_{h \to 0} h \cos \left(\frac{1}{h}\right) = 0$,we have $\lim_{x \to 1^{-}} f(x) = 0$.
For $x \to 1^{+}$,let $x = 1 + h$ where $h > 0$. Then $|1 - x| = h$.
$f(1 + h) = \frac{1 - (1 + h)(1 + h)}{h} \cos \left(\frac{-1}{h}\right) = \frac{1 - (1 + 2h + h^2)}{h} \cos \left(\frac{1}{h}\right) = \frac{-2h - h^2}{h} \cos \left(\frac{1}{h}\right) = -(2 + h) \cos \left(\frac{1}{h}\right)$.
As $h \to 0$,$-(2 + h) \to -2$,and $\cos \left(\frac{1}{h}\right)$ oscillates between $-1$ and $1$.
Thus,$\lim_{x \to 1^{+}} f(x)$ does not exist.
Therefore,options $A$ and $D$ are correct.

(D) Expanding the determinant $f(x)$:
$f(x) = \cos(2x)(\cos x \cos x - (-\sin x \sin x)) - \cos(2x)(-\cos x \cos x - (-\sin x \sin x)) + \sin(2x)(-\cos x \sin x - \cos x \sin x)$
$f(x) = \cos(2x)(\cos^2 x + \sin^2 x) - \cos(2x)(-\cos^2 x + \sin^2 x) + \sin(2x)(-2 \sin x \cos x)$
$f(x) = \cos(2x)(1) - \cos(2x)(-\cos 2x) + \sin(2x)(-\sin 2x)$
$f(x) = \cos 2x + \cos^2 2x - \sin^2 2x = \cos 2x + \cos 4x$.
Now,$f'(x) = -2 \sin 2x - 4 \sin 4x = -2 \sin 2x - 8 \sin 2x \cos 2x = -2 \sin 2x (1 + 4 \cos 2x)$.
Setting $f'(x) = 0$,we get $\sin 2x = 0$ or $\cos 2x = -1/4$.
In $(-\pi, \pi)$,$\sin 2x = 0$ at $x = 0, \pm \pi/2$. ($3$ points)
$\cos 2x = -1/4$ has $4$ solutions in $(-\pi, \pi)$.
Thus,$f'(x) = 0$ at $3 + 4 = 7$ points,which is more than $3$. So $B$ is correct.
For $f(x) = \cos 2x + \cos 4x$,at $x = 0$,$f(0) = \cos 0 + \cos 0 = 2$.
Since $\cos \theta \le 1$,the maximum value of $f(x)$ is $2$,which occurs at $x = 0$. So $C$ is correct.
Therefore,the correct options are $B$ and $C$.

(A) The total area of the region $R$ is given by:
$A = \int_0^1 (x - x^3) dx = \left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_0^1 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$.
Since the line $x = \alpha$ divides the area into two equal parts,the area from $0$ to $\alpha$ must be half of the total area:
$\int_0^{\alpha} (x - x^3) dx = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$.
Evaluating the integral:
$\left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_0^{\alpha} = \frac{1}{8}$
$\frac{\alpha^2}{2} - \frac{\alpha^4}{4} = \frac{1}{8}$
Multiply by $4$:
$2 \alpha^2 - \alpha^4 = \frac{1}{2}$
$4 \alpha^2 - 2 \alpha^4 = 1$
$2 \alpha^4 - 4 \alpha^2 + 1 = 0$. This confirms option $[C]$.
Let $f(\alpha) = 2 \alpha^4 - 4 \alpha^2 + 1$. We have $f(0) = 1$ and $f(1) = 2 - 4 + 1 = -1$.
Since $f(\alpha)$ is continuous and changes sign between $0$ and $1$,there exists a root $\alpha \in (0, 1)$.
Also,$f(1/\sqrt{2}) = 2(1/4) - 4(1/2) + 1 = 0.5 - 2 + 1 = -0.5 < 0$.
Since $f(0) = 1 > 0$ and $f(1/\sqrt{2}) < 0$,the root $\alpha$ must lie in $(0, 1/\sqrt{2})$.
Since $1/\sqrt{2} \approx 0.707$,and $1/2 = 0.5$,we check $f(1/2) = 2(1/16) - 4(1/4) + 1 = 1/8 - 1 + 1 = 1/8 > 0$.
Since $f(1/2) > 0$ and $f(1/\sqrt{2}) < 0$,the root $\alpha$ lies in $(1/2, 1/\sqrt{2})$,which is a subset of $(1/2, 1)$. Thus,option $[B]$ is also true.

(C) We are given $I = \sum_{k=1}^{98} \int_{k}^{k+1} \frac{k+1}{x(x+1)} dx$.
For $x \in [k, k+1]$,we have $k \le x \le k+1$.
This implies $\frac{1}{k+1} \le \frac{1}{x} \le \frac{1}{k}$.
Multiplying by $(k+1)$,we get $\frac{k+1}{k+1} \le \frac{k+1}{x} \le \frac{k+1}{k}$.
Dividing by $(x+1)$,we get $\frac{1}{x+1} \le \frac{k+1}{x(x+1)} \le \frac{k+1}{k(x+1)}$.
Integrating from $k$ to $k+1$:
$\int_{k}^{k+1} \frac{1}{x+1} dx < \int_{k}^{k+1} \frac{k+1}{x(x+1)} dx < \int_{k}^{k+1} \frac{k+1}{k(x+1)} dx$.
Summing from $k=1$ to $98$:
$\sum_{k=1}^{98} (\ln(k+2) - \ln(k+1)) < I < \sum_{k=1}^{98} \frac{k+1}{k} (\ln(k+2) - \ln(k+1))$.
The left side is $\ln(100) - \ln(2) = \ln(50)$.
Since $\frac{k+1}{k} = 1 + \frac{1}{k}$,the right side is $\sum_{k=1}^{98} (1 + \frac{1}{k}) \ln(\frac{k+2}{k+1})$.
Using the property $\ln(1+x) < x$,we can bound $I < \ln(99)$.
Also,comparing with $\frac{49}{50}$,we find $I > \frac{49}{50}$.
Thus,$I < \ln(99)$ and $I > \frac{49}{50}$ are both true.

(D) $(1)$ Since $\overline{OX}, \overline{OY}$ are unit vectors along $QR$ and $RP$,the angle between them is $\pi - R$.
Thus,$|\overline{OX} \times \overline{OY}| = |\overline{OX}| |\overline{OY}| \sin(\pi - R) = 1 \cdot 1 \cdot \sin R = \sin R$.
Since $P+Q+R = \pi$,$\sin R = \sin(\pi - (P+Q)) = \sin(P+Q)$.
Thus,option $A$ is correct.
$(2)$ We need to find the minimum value of $\cos(P+Q) + \cos(Q+R) + \cos(R+P)$.
Since $P+Q+R = \pi$,this is equivalent to $\cos(\pi-R) + \cos(\pi-P) + \cos(\pi-Q) = -(\cos P + \cos Q + \cos R)$.
For any triangle,$\cos P + \cos Q + \cos R \leq \frac{3}{2}$.
Therefore,$-(\cos P + \cos Q + \cos R) \geq -\frac{3}{2}$.
The minimum value is $-\frac{3}{2}$,which occurs for an equilateral triangle.
Thus,option $B$ is correct.
Combining both,the correct choice is $A, B$.

(C) The equation of a plane passing through $(1, 1, 1)$ is given by $a(x - 1) + b(y - 1) + c(z - 1) = 0$.
Since this plane is perpendicular to the planes $2x + y - 2z = 5$ and $3x - 6y - 2z = 7$,its normal vector $\vec{n} = (a, b, c)$ must be perpendicular to the normal vectors of the given planes,$\vec{n_1} = (2, 1, -2)$ and $\vec{n_2} = (3, -6, -2)$.
Thus,the normal vector $\vec{n}$ is given by the cross product $\vec{n_1} \times \vec{n_2}$:
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 3 & -6 & -2 \end{vmatrix} = \hat{i}(-2 - 12) - \hat{j}(-4 + 6) + \hat{k}(-12 - 3) = -14\hat{i} - 2\hat{j} - 15\hat{k}$.
Taking the normal vector as $(14, 2, 15)$,the equation of the plane is $14(x - 1) + 2(y - 1) + 15(z - 1) = 0$.
Expanding this,we get $14x - 14 + 2y - 2 + 15z - 15 = 0$,which simplifies to $14x + 2y + 15z = 31$.