GUJCET 2021 Mathematics Question Paper with Answer and Solution

(C) The area of an ellipse given by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by the formula $A = \pi ab$.
Given that the area is $\frac{\pi}{6}$,we have $\pi ab = \frac{\pi}{6}$,which implies $ab = \frac{1}{6}$.
Let us check the options:
For option $C$,the equation is $4x^2 + 9y^2 = 1$,which can be written as $\frac{x^2}{1/4} + \frac{y^2}{1/9} = 1$.
Here,$a^2 = \frac{1}{4} \implies a = \frac{1}{2}$ and $b^2 = \frac{1}{9} \implies b = \frac{1}{3}$.
The area is $\pi ab = \pi \times \frac{1}{2} \times \frac{1}{3} = \frac{\pi}{6}$.
Thus,option $C$ is correct.

(C) Since $A$ and $B$ are mutually exclusive events,$P(A \cap B) = 0$.
We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $\frac{3}{5} = \frac{1}{2} + P(B) - 0$.
$P(B) = \frac{3}{5} - \frac{1}{2} = \frac{6 - 5}{10} = \frac{1}{10}$.
We are given $P(B') = p$.
Since $P(B') = 1 - P(B)$,we have $p = 1 - \frac{1}{10} = \frac{9}{10}$.
Thus,the correct option is $C$.

(B) Given $f(x) = \frac{1+x}{1-x}$.
We need to find $f(x) \cdot f(y) = \left(\frac{1+x}{1-x}\right) \cdot \left(\frac{1+y}{1-y}\right)$.
$= \frac{1+y+x+xy}{1-y-x+xy} = \frac{1+xy + (x+y)}{1+xy - (x+y)}$.
Divide the numerator and denominator by $(1+xy)$:
$= \frac{1 + \frac{x+y}{1+xy}}{1 - \frac{x+y}{1+xy}}$.
Comparing this with the form $f(z) = \frac{1+z}{1-z}$,we get $z = \frac{x+y}{1+xy}$.
Therefore,$f(x) \cdot f(y) = f\left(\frac{x+y}{1+xy}\right)$.

(A) To determine if $f(x) = x^3$ is one-one and onto:
$1$. For one-one: Let $f(x_1) = f(x_2)$,then $x_1^3 = x_2^3$. Taking the cube root on both sides,we get $x_1 = x_2$. Since $f(x_1) = f(x_2) \implies x_1 = x_2$,the function is one-one.
$2$. For onto: For any $y \in R$ (codomain),we need to find $x \in R$ (domain) such that $f(x) = y$. Since $x^3 = y$,we have $x = y^{1/3}$. Since $y^{1/3}$ is defined for all real numbers $y$,for every $y \in R$,there exists an $x = y^{1/3} \in R$. Thus,the function is onto.
Therefore,$f$ is one-one and onto.

(D) The set is $A = \{1, 2, 3\}$.
$A$ relation $R$ on set $A$ is reflexive if $(a, a) \in R$ for all $a \in A$. Here,$(1,1), (2,2), (3,3) \in R$,so $R$ is reflexive.
$A$ relation $R$ is symmetric if $(a, b) \in R \implies (b, a) \in R$. Here,all pairs are of the form $(a, a)$,so the condition holds trivially. Thus,$R$ is symmetric.
$A$ relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$. Here,the condition holds trivially for all elements. Thus,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.
Therefore,the correct option is $D$.

(A) The property of the transpose of a product of matrices states that $(AB)^T = B^T A^T$.
Therefore,the statement $(AB)^T = A^T B^T$ is generally incorrect unless $A$ and $B$ commute.
Option $(B)$ is a standard property of transpose: $(A+B)^T = A^T + B^T$.
Option $(C)$ is a fundamental property of the adjoint of a matrix: $A \operatorname{adj} A = |A| I$.
Option $(D)$ is a standard property of the inverse of a product: $(AB)^{-1} = B^{-1} A^{-1}$.
Thus,the incorrect statement is $(A)$.

(C) Given $AB = \begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix}$ and $11B^{-1} = \begin{bmatrix} 5 & -3 \\ 2 & 1 \end{bmatrix}$.
We know that $A = (AB)B^{-1}$.
From the second equation,$B^{-1} = \frac{1}{11} \begin{bmatrix} 5 & -3 \\ 2 & 1 \end{bmatrix}$.
Substituting these into the expression for $A$:
$A = \begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix} \times \frac{1}{11} \begin{bmatrix} 5 & -3 \\ 2 & 1 \end{bmatrix}$
$A = \frac{1}{11} \begin{bmatrix} (-6)(5) + (26)(2) & (-6)(-3) + (26)(1) \\ (-1)(5) + (19)(2) & (-1)(-3) + (19)(1) \end{bmatrix}$
$A = \frac{1}{11} \begin{bmatrix} -30 + 52 & 18 + 26 \\ -5 + 38 & 3 + 19 \end{bmatrix}$
$A = \frac{1}{11} \begin{bmatrix} 22 & 44 \\ 33 & 22 \end{bmatrix}$
$A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$.
Thus,the correct option is $C$.

(B) First,calculate the sum of matrices $A$ and $B$:
$A+B = \begin{bmatrix} 4+1 & 0 & 0 \\ 0 & 3+2 & 0 \\ 0 & 0 & 2+3 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} = 5I_3$.
Now,find the inverse of the resulting matrix:
$(A+B)^{-1} = (5I_3)^{-1}$.
Using the property $(kA)^{-1} = \frac{1}{k} A^{-1}$,we get:
$(A+B)^{-1} = \frac{1}{5} I_3^{-1} = \frac{1}{5} I_3$.

(A) Let the determinant be $\Delta = \left|\begin{array}{ccc}2 & 3 & 5 \\ 1 & 0 & 7 \\ -1 & -2 & 4\end{array}\right|$.
The element $7$ is located at the second row and third column,i.e.,$a_{23} = 7$.
The minor $M_{23}$ is obtained by deleting the second row and third column:
$M_{23} = \left|\begin{array}{cc}2 & 3 \\ -1 & -2\end{array}\right| = (2 \times -2) - (3 \times -1) = -4 + 3 = -1$.
The cofactor $C_{23}$ is given by $(-1)^{2+3} M_{23} = (-1)^5 (-1) = (-1) \times (-1) = 1$.
The sum of the minor and the cofactor is $M_{23} + C_{23} = -1 + 1 = 0$.
Therefore,the correct option is $A$.

(D) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Given vertices are $A(1, 3)$,$B(0, 0)$,and $C(k, 0)$,and the area is $3$ sq. units.
Substituting the values into the formula:
$3 = \frac{1}{2} |1(0 - 0) + 0(0 - 3) + k(3 - 0)|$
$3 = \frac{1}{2} |0 + 0 + 3k|$
$3 = \frac{1}{2} |3k|$
$6 = |3k|$
$|k| = 2$
Therefore,$k = \pm 2$.
Thus,the correct option is $D$.

(B) Given the determinant equation:
$2[\sin(A+B)\sin(A-B) - \cos(A+B)\cos(A-B)] + \sqrt{3} = 0$
Using the trigonometric identity $\cos(x+y) = \cos x \cos y - \sin x \sin y$,we can rewrite the expression inside the brackets as:
$-(\cos(A+B)\cos(A-B) - \sin(A+B)\sin(A-B)) = -\cos((A+B) + (A-B)) = -\cos(2A)$
Substituting this into the equation:
$2[-\cos(2A)] + \sqrt{3} = 0$
$-2\cos(2A) = -\sqrt{3}$
$\cos(2A) = \frac{\sqrt{3}}{2}$
Since $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$,we have:
$2A = \frac{\pi}{6}$
$A = \frac{\pi}{12}$

(B) Let the determinant be $\Delta = \left|\begin{array}{lll}10 & 11 & 12 \\ 11 & 12 & 13 \\ 12 & 13 & 14\end{array}\right|$.
Apply the row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$:
$\Delta = \left|\begin{array}{lll}10 & 11 & 12 \\ 11-10 & 12-11 & 13-12 \\ 12-11 & 13-12 & 14-13\end{array}\right|$
$\Delta = \left|\begin{array}{lll}10 & 11 & 12 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right|$.
Since row $R_2$ and row $R_3$ are identical,the value of the determinant is $0$.

(C) First,we expand the determinant $D$ along the first row:
$D = 1(1 - (-\cos^2 \theta)) - (-\cos \theta)(\cos \theta - (- \cos \theta)) - 1(\cos^2 \theta - 1)$
$D = 1(1 + \cos^2 \theta) + \cos \theta(2 \cos \theta) - (\cos^2 \theta - 1)$
$D = 1 + \cos^2 \theta + 2 \cos^2 \theta - \cos^2 \theta + 1$
$D = 2 + 2 \cos^2 \theta$
Since the range of $\cos \theta$ is $[-1, 1]$,the range of $\cos^2 \theta$ is $[0, 1]$.
Therefore,the range of $D = 2 + 2 \cos^2 \theta$ is $[2 + 2(0), 2 + 2(1)] = [2, 4]$.
Thus,the maximum value $p = 4$ and the minimum value $q = 2$.
Finally,we calculate $2p + 3q = 2(4) + 3(2) = 8 + 6 = 14$.

(B) Given $x+1 = e^{-y}$.
Taking the natural logarithm on both sides,we get $\ln(x+1) = -y$,which implies $y = -\ln(x+1)$.
Differentiating with respect to $x$,we get $\frac{d y}{d x} = -\frac{1}{x+1}$.
Differentiating again with respect to $x$,we get $\frac{d^2 y}{d x^2} = -(-1)(x+1)^{-2} = \frac{1}{(x+1)^2}$.
Since $\frac{d y}{d x} = -\frac{1}{x+1}$,we have $\left(\frac{d y}{d x}\right)^2 = \left(-\frac{1}{x+1}\right)^2 = \frac{1}{(x+1)^2}$.
Thus,$\frac{d^2 y}{d x^2} = \left(\frac{d y}{d x}\right)^2$.

(C) Let $y = \operatorname{cosec}^{-1}(e^x)$.
Using the chain rule,we differentiate with respect to $x$:
$\frac{d y}{d x} = \frac{d}{d x}(\operatorname{cosec}^{-1}(e^x))$
We know that $\frac{d}{d u}(\operatorname{cosec}^{-1} u) = \frac{-1}{|u| \sqrt{u^2 - 1}}$.
Here,$u = e^x$,so $\frac{d u}{d x} = e^x$.
Applying the chain rule:
$\frac{d y}{d x} = \frac{-1}{|e^x| \sqrt{(e^x)^2 - 1}} \cdot \frac{d}{d x}(e^x)$
Since $e^x > 0$ for all real $x$,$|e^x| = e^x$.
$\frac{d y}{d x} = \frac{-1}{e^x \sqrt{e^{2 x} - 1}} \cdot e^x$
$\frac{d y}{d x} = \frac{-1}{\sqrt{e^{2 x} - 1}}$
Therefore,the correct option is $C$.

(A) Let $f(x) = \log \left(\frac{1}{x}\right) + \log \left(\frac{1}{x^2}\right) + \log \left(\frac{1}{x^3}\right)$.
Using the property $\log(a^b) = b \log(a)$ and $\log(\frac{1}{x}) = -\log(x)$,we can simplify the expression:
$f(x) = -\log(x) - 2\log(x) - 3\log(x)$
$f(x) = -6\log(x)$.
Now,differentiate with respect to $x$:
$\frac{d}{dx} f(x) = \frac{d}{dx} (-6\log(x))$
$= -6 \times \frac{1}{x} = -\frac{6}{x}$.
Thus,the correct option is $A$.

(D) Given $f(x) = 4x^3 + 3x^2 + 3x + 4$.
First,find $f\left(\frac{1}{x}\right)$:
$f\left(\frac{1}{x}\right) = 4\left(\frac{1}{x}\right)^3 + 3\left(\frac{1}{x}\right)^2 + 3\left(\frac{1}{x}\right) + 4 = \frac{4}{x^3} + \frac{3}{x^2} + \frac{3}{x} + 4$.
Now,multiply by $x^3$:
$x^3 \cdot f\left(\frac{1}{x}\right) = x^3 \left(\frac{4}{x^3} + \frac{3}{x^2} + \frac{3}{x} + 4\right) = 4 + 3x + 3x^2 + 4x^3$.
Finally,differentiate with respect to $x$:
$\frac{d}{dx}(4 + 3x + 3x^2 + 4x^3) = 0 + 3 + 6x + 12x^2 = 12x^2 + 6x + 3$.
Thus,the correct option is $D$.

(C) Let $I = \int_0^1 \frac{dx}{(3x+2)+\sqrt{3x+2}}$.
Substitute $u = \sqrt{3x+2}$. Then $u^2 = 3x+2$,so $2u du = 3 dx$,which means $dx = \frac{2}{3} u du$.
When $x=0$,$u = \sqrt{2}$. When $x=1$,$u = \sqrt{5}$.
Substituting these into the integral:
$I = \int_{\sqrt{2}}^{\sqrt{5}} \frac{\frac{2}{3} u du}{u^2+u} = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{u du}{u(u+1)} = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{du}{u+1}$.
Integrating,we get $I = \frac{2}{3} [\log |u+1|]_{\sqrt{2}}^{\sqrt{5}}$.
$I = \frac{2}{3} (\log |\sqrt{5}+1| - \log |\sqrt{2}+1|) = \frac{2}{3} \log \left|\frac{\sqrt{5}+1}{\sqrt{2}+1}\right|$.
Thus,the correct option is $C$.

(B) We are given the integral $I = \int \frac{\cos 3x}{\sin x} dx$.
Using the trigonometric identity $\cos 3x = 4 \cos^3 x - 3 \cos x$,we can also write $\cos 3x = \cos(2x + x) = \cos 2x \cos x - \sin 2x \sin x$.
Alternatively,use $\cos 3x = 1 - 4 \sin^2 x$ is incorrect,rather $\cos 3x = \cos(2x+x) = \cos 2x \cos x - \sin 2x \sin x = (1 - 2 \sin^2 x) \cos x - (2 \sin x \cos x) \sin x = \cos x - 2 \sin^2 x \cos x - 2 \sin^2 x \cos x = \cos x - 4 \sin^2 x \cos x$.
Thus,$\frac{\cos 3x}{\sin x} = \frac{\cos x - 4 \sin^2 x \cos x}{\sin x} = \cot x - 4 \sin x \cos x = \cot x - 2 \sin 2x$.
Now,integrate: $\int (\cot x - 2 \sin 2x) dx = \int \cot x dx - 2 \int \sin 2x dx$.
$= \log |\sin x| - 2 (-\frac{\cos 2x}{2}) + C = \log |\sin x| + \cos 2x + C$.
Comparing this with $p \cos 2x + q \log |\sin x| + C$,we get $p = 1$ and $q = 1$.
Therefore,$p + q = 1 + 1 = 2$.

(C) We know that $1 + \tan^2 x = \sec^2 x$.
Substituting this into the integral,we get:
$\int e^x(2021 + \tan x + \tan^2 x) dx = \int e^x(2020 + 1 + \tan^2 x + \tan x) dx$
$= \int e^x(2020 + \sec^2 x + \tan x) dx$
$= \int e^x(2020 + \tan x) dx + \int e^x \sec^2 x dx$.
Let $f(x) = 2020 + \tan x$. Then $f'(x) = \sec^2 x$.
Using the standard integral formula $\int e^x(f(x) + f'(x)) dx = e^x f(x) + C$,we have:
$\int e^x(2020 + \tan x + \sec^2 x) dx = e^x(2020 + \tan x) + C$.
Thus,the correct option is $C$.

(C) Let $I = \int \sqrt{\frac{\cos x(1 - \cos^2 x)}{1 - \cos^3 x}} \, dx = \int \sqrt{\frac{\cos x \sin^2 x}{1 - \cos^3 x}} \, dx$.
Since $\sin x$ is under a square root,we consider the positive root $\sin x$ (assuming the domain allows).
$I = \int \sin x \sqrt{\frac{\cos x}{1 - \cos^3 x}} \, dx$.
Let $u = \cos^{\frac{3}{2}} x$. Then $du = \frac{3}{2} \cos^{\frac{1}{2}} x (-\sin x) \, dx$.
So,$-\frac{2}{3} du = \sqrt{\cos x} \sin x \, dx$.
Substituting into the integral: $I = \int \sqrt{\frac{\cos x}{1 - (\cos^{\frac{3}{2}} x)^2}} \sin x \, dx$.
This simplifies to $I = \int \frac{1}{\sqrt{1 - u^2}} \left(-\frac{2}{3} du\right) = -\frac{2}{3} \sin^{-1}(u) + c$.
Substituting back $u = \cos^{\frac{3}{2}} x$,we get $I = -\frac{2}{3} \sin^{-1}(\cos^{\frac{3}{2}} x) + c$.

(A) The area $A$ of the region bounded by the curve $y^2=x$,the $X$-axis,and the lines $x=1$ and $x=4$ in the first quadrant is given by the definite integral:
$A = \int_{1}^{4} y \, dx$
Since $y^2=x$ and we are in the first quadrant,$y = \sqrt{x} = x^{1/2}$.
$A = \int_{1}^{4} x^{1/2} \, dx$
Using the power rule for integration $\int x^n \, dx = \frac{x^{n+1}}{n+1}$:
$A = \left[ \frac{x^{3/2}}{3/2} \right]_{1}^{4} = \left[ \frac{2}{3} x^{3/2} \right]_{1}^{4}$
$A = \frac{2}{3} (4^{3/2} - 1^{3/2})$
$A = \frac{2}{3} (8 - 1) = \frac{2}{3} (7) = \frac{14}{3}$ sq. units.
Thus,the correct option is $A$.

(B) The given curve is $y^2 = 4x$,which implies $x = \frac{y^2}{4}$.
Since the region is bounded by the $Y$-axis $(x = 0)$,the curve $x = \frac{y^2}{4}$,and the line $y = 3$,the area $A$ is given by the integral with respect to $y$ from $y = 0$ to $y = 3$.
$A = \int_{0}^{3} x \, dy = \int_{0}^{3} \frac{y^2}{4} \, dy$.
$A = \frac{1}{4} \left[ \frac{y^3}{3} \right]_{0}^{3}$.
$A = \frac{1}{4} \left( \frac{3^3}{3} - 0 \right) = \frac{1}{4} \times \frac{27}{3} = \frac{9}{4}$ sq. units.
Therefore,the correct option is $B$.

(A) The given equation of the parabola is $x^2 = 12y$. Comparing this with $x^2 = 4ay$,we get $4a = 12$,which implies $a = 3$.
The focus of the parabola is $(0, a) = (0, 3)$.
The equation of the latus rectum is $y = 3$.
The region is bounded by the parabola $y = \frac{x^2}{12}$ and the line $y = 3$.
The points of intersection are found by substituting $y = 3$ into $x^2 = 12y$,giving $x^2 = 36$,so $x = \pm 6$.
The area $A$ is given by the integral $A = \int_{-6}^{6} (3 - \frac{x^2}{12}) dx$.
Since the function is even,$A = 2 \int_{0}^{6} (3 - \frac{x^2}{12}) dx$.
$A = 2 [3x - \frac{x^3}{36}]_{0}^{6} = 2 [3(6) - \frac{216}{36}] = 2 [18 - 6] = 2(12) = 24$ sq. units.

(A) The given differential equation is $e^{\frac{d^2 y}{d x^2}} = x$.
To find the order,we identify the highest order derivative present in the equation,which is $\frac{d^2 y}{d x^2}$. Thus,the order is $2$.
To find the degree,the differential equation must be a polynomial in terms of its derivatives. Since the term $\frac{d^2 y}{d x^2}$ appears in the exponent of $e$,the equation cannot be expressed as a polynomial in its derivatives.
Therefore,the degree of this differential equation is not defined.

(C) Given the differential equation: $\{1+(\frac{dy}{dx})^2\}^{\frac{3}{2}}=\frac{d^2y}{dx^2}$.
To find the degree,we must eliminate the fractional exponent by squaring both sides:
$\{1+(\frac{dy}{dx})^2\}^3 = (\frac{d^2y}{dx^2})^2$.
The highest order derivative present is $\frac{d^2y}{dx^2}$,so the order $p = 2$.
The power of the highest order derivative after making the equation a polynomial in derivatives is $2$,so the degree $q = 2$.
Therefore,$p+q = 2+2 = 4$.

(A) The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product,i.e.,$|\vec{a} \times \vec{b}|$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix}$
$= \hat{i}((-1)(1) - (3)(-7)) - \hat{j}((1)(1) - (3)(2)) + \hat{k}((1)(-7) - (-1)(2))$
$= \hat{i}(-1 + 21) - \hat{j}(1 - 6) + \hat{k}(-7 + 2)$
$= 20\hat{i} + 5\hat{j} - 5\hat{k}$
Now,find the magnitude of this vector:
$|\vec{a} \times \vec{b}| = \sqrt{20^2 + 5^2 + (-5)^2}$
$= \sqrt{400 + 25 + 25} = \sqrt{450}$
$= \sqrt{225 \times 2} = 15\sqrt{2}$
Thus,the area of the parallelogram is $15\sqrt{2}$ square units.

(C) By definition,a unit vector is a vector with a magnitude of $1$.
Since $\vec{c}$ is defined as a unit vector in the direction of $(\vec{a} + \vec{b})$,its magnitude must be $1$ by definition.
Therefore,$|\vec{c}| = 1$.

(A) To find the minimum value of $Z = 2x + 3y$,we identify the feasible region defined by the constraints:
$1$. $2x + 4y \leq 12 \implies x + 2y \leq 6$
$2$. $x + y \leq 3$
$3$. $x \geq 0, y \geq 0$
The vertices of the feasible region are found by the intersection of the lines:
- Intersection of $x + y = 3$ and $x = 0$ gives $(0, 3)$.
- Intersection of $x + y = 3$ and $y = 0$ gives $(3, 0)$.
- The origin $(0, 0)$ is also a vertex.
Evaluating $Z$ at the vertices:
- At $(0, 0)$: $Z = 2(0) + 3(0) = 0$
- At $(3, 0)$: $Z = 2(3) + 3(0) = 6$
- At $(0, 3)$: $Z = 2(0) + 3(3) = 9$
The minimum value of $Z$ is $0$ at the point $(0, 0)$.

(A) To find the minimum value of the objective function $Z = 6x + 3y$,we evaluate $Z$ at each corner point of the feasible region:
$1$. At point $(2, 72)$: $Z = 6(2) + 3(72) = 12 + 216 = 228$
$2$. At point $(15, 20)$: $Z = 6(15) + 3(20) = 90 + 60 = 150$
$3$. At point $(40, 15)$: $Z = 6(40) + 3(15) = 240 + 45 = 285$
Comparing the values $228$,$150$,and $285$,the minimum value is $150$,which occurs at the point $(15, 20)$.

(D) Given the equation: $P(A) + P(B) - P(A \cap B) = P(A)$.
Subtracting $P(A)$ from both sides,we get: $P(B) - P(A \cap B) = 0$,which implies $P(B) = P(A \cap B)$.
This condition $P(A \cap B) = P(B)$ indicates that the event $B$ is a subset of event $A$ (i.e.,$B \subseteq A$).
By the definition of conditional probability,$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Substituting $P(A \cap B) = P(B)$ into the formula,we get $P(A \mid B) = \frac{P(B)}{P(B)} = 1$ (assuming $P(B) \neq 0$).
Therefore,the correct option is $D$.

(B) Given that $A$ and $B$ are independent events,$P(A) = p$ and $P(B) = 2p$.
The probability of exactly one of the events $A$ or $B$ occurring is given by $P(A \cap B^c) + P(A^c \cap B)$.
Since $A$ and $B$ are independent,$P(A \cap B^c) = P(A)P(B^c) = p(1 - 2p)$ and $P(A^c \cap B) = P(A^c)P(B) = (1 - p)(2p)$.
Thus,$P(\text{exactly one}) = p(1 - 2p) + 2p(1 - p) = \frac{5}{9}$.
Expanding this,we get $p - 2p^2 + 2p - 2p^2 = \frac{5}{9}$.
This simplifies to $3p - 4p^2 = \frac{5}{9}$,or $4p^2 - 3p + \frac{5}{9} = 0$.
Multiplying by $9$,we get $36p^2 - 27p + 5 = 0$.
Factoring the quadratic equation: $36p^2 - 12p - 15p + 5 = 0 \implies 12p(3p - 1) - 5(3p - 1) = 0$.
So,$(12p - 5)(3p - 1) = 0$.
This gives $p = \frac{5}{12}$ or $p = \frac{1}{3}$.

(D) Given that $P(B \mid A) = 1$.
By the definition of conditional probability,$P(B \mid A) = \frac{P(A \cap B)}{P(A)}$.
Since $P(B \mid A) = 1$,we have $\frac{P(A \cap B)}{P(A)} = 1$,which implies $P(A \cap B) = P(A)$.
This equality holds if and only if $A \subseteq B$.
Therefore,the correct option is $D$.

(A) We are given that $P(A) = \frac{6}{11}$,$P(B) = \frac{5}{11}$,and $P(A \cup B) = \frac{7}{11}$.
Using the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $\frac{7}{11} = \frac{6}{11} + \frac{5}{11} - P(A \cap B)$.
$\frac{7}{11} = \frac{11}{11} - P(A \cap B)$.
$P(A \cap B) = 1 - \frac{7}{11} = \frac{4}{11}$.
Now,the conditional probability $P(A \mid B)$ is defined as $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
$P(A \mid B) = \frac{4/11}{5/11} = \frac{4}{5}$.
Therefore,the correct option is $A$.