Solution:
Since the consonants must be distinct, the number of ways to choose and arrange 4 consonants out of 11 is: $${}^{11}{P}_4=11\times 10\times 9\times 8=7920.$$ The vowels can repeat, so for each of the middle two positions, there are 5 choices. The total arrangements of vowels are: $$5\times 5=25.$$ The total number of six-letter words is the product of the arrangements for consonants and vowels: $$\text{Total words}={}^{11}{P}_4\times 25=7920\times 25=198,000.$$ **Final Answer:** $\boxed{{\displaystyle D}}$

Solution:
For an 8-digit palindrome of the form $\text{ABCDDCBA}$ to be divisible by 5, the last digit $A$ must be 5 (since $A=0$ would make it a 7-digit number).
This makes the structure $5BCDDBC5$ .
Here, $B,C,$ and $D$ can each take any digit from 0 to 9.
Therefore, the total number of palindromic numbers is: $$1\times 10\times 10\times 10=1000.$$ **Final Answer:** $\boxed{{\displaystyle A}}$

Solution:
The given equation is: $$x^2+4|x|-4=0$$ **Case 1:** $x>0$
Substituting $|x|=x$ : $$x^2+4x-4=0$$ Solving using the quadratic formula: $$x=-2+2\sqrt{2},x=-2-2\sqrt{2}$$ Since $x>0$ , we take $x=-2+2\sqrt{2}$ . **Case 2:** $x<0$
Substituting $|x|=-x$ : $$x^2-4x-4=0$$ Solving using the quadratic formula: $$x=2+2\sqrt{2},x=2-2\sqrt{2}$$ Since $x<0$ , we take $x=2-2\sqrt{2}$ . The product of the solutions is: $$(-2+2\sqrt{2})\cdot (2-2\sqrt{2})=-4(\sqrt{2}-1{)}^2$$ **Final Answer:** $\boxed{{\displaystyle C}}$

Solution:
In the expansion of $(x^3+1{)}^m$ , the general term is: $$T_k=(\genfrac{}{}{0ex}{}{m}{k})\cdot x^{3k}$$ For $x^{12}$ , set $3k=12⟹k=4$ .
The coefficient of $x^{12}$ is: $$(\genfrac{}{}{0ex}{}{m}{4})=210$$ Using the formula for combinations: $$(\genfrac{}{}{0ex}{}{m}{4})=\frac{m!}{4!(m-4)!}=210$$ Solving, we find $m=10$ . For $x^{15}$ , set $3k=15⟹k=5$ .
The coefficient of $x^{15}$ is: $$(\genfrac{}{}{0ex}{}{10}{5})=\frac{10\cdot 9\cdot 8\cdot 7\cdot 6}{5\cdot 4\cdot 3\cdot 2\cdot 1}=252$$ **Final Answer:** $\boxed{{\displaystyle A}}$

Solution:
Given: $$\frac{S_m}{S_n}=\frac{m^2}{n^2}$$ Substituting the formula for the sum of an arithmetic progression: $$S_n=\frac{n}{2}(2\alpha +(n-1)d)$$ For the given ratio: $$\frac{\frac{m}{2}(2\alpha +(m-1)d)}{\frac{n}{2}(2\alpha +(n-1)d)}=\frac{m^2}{n^2}$$ Simplifying: $$\frac{m}{n}=\frac{2\alpha +(m-1)d}{2\alpha +(n-1)d}$$ Cross-multiplying: $$m(2\alpha +(n-1)d)=n(2\alpha +(m-1)d)$$ Expanding and rearranging: $$(n-m)d=(n-m)2\alpha ⟹d=2\alpha$$ The ${50}^{th}$ term is: $$T_{50}=\alpha +49d=\alpha +49\cdot 2\alpha =99\alpha$$ **Final Answer:** $\boxed{{\displaystyle B}}$

Solution:
The series alternates between odd and even terms: $$1,3\times 1,7\times 3\times 1,3\times 7\times 3\times 1,\dots$$ **Odd terms:** Form a geometric progression with the first term $1$ and common ratio $7\times 3=21$ .
**Even terms:** Form a geometric progression with the first term $3$ and the same common ratio $21$ .

The sum of 50 odd terms: $$S_{\text{odd}}=\frac{1\cdot ({21}^{50}-1)}{21-1}=\frac{{21}^{50}-1}{20}.$$ The sum of 50 even terms: $$S_{\text{even}}=\frac{3\cdot ({21}^{50}-1)}{21-1}=\frac{3({21}^{50}-1)}{20}.$$ Adding both: $$S_{100}=S_{\text{odd}}+S_{\text{even}}=\frac{1}{5}({21}^{50}-1).$$ **Final Answer:** $\boxed{{\displaystyle A}}$

Solution:
Start with the equation: $$4{\sin }^{2}x-4\cos x=1$$ Substituting ${\sin }^{2}x=1-{\cos }^{2}x$ , we get: $$4(1-{\cos }^{2}x)-4\cos x=1$$ Simplifying: $$4{\cos }^{2}x+4\cos x-3=0$$ Let $\cos x=y$ . The equation becomes: $$4y^2+4y-3=0$$ Solving the quadratic equation: $$y=\frac{-4\pm \sqrt{16+48}}{8}=\frac{-4\pm 8}{8}$$ Thus, $y=\frac{1}{2}$ (valid) or $y=-\frac{3}{2}$ (not valid).
For $\cos x=\frac{1}{2}$ , the solutions in $[0,2\pi ]$ are: $$x=\frac{\pi }{3},\frac{5\pi }{3}.$$ The sum of solutions: $$\frac{\pi }{3}+\frac{5\pi }{3}=2\pi .$$ **Final Answer:** $\boxed{{\displaystyle D}}$

Solution:
Given that: $$\frac{h}{SQ}=\frac{8}{25},\frac{h}{RQ}=\frac{17}{25}.$$ From the Pythagoras theorem: $$S{Q}^2=R{Q}^2+R{S}^2.$$ Substituting $RS=375$ , we find $SQ$ and $RQ$ . Let $h=PQ$ , then solving the above equations yields: $$h=136.$$ **Final Answer:** $\boxed{{\displaystyle C}}$

Solution:
Using the AM-GM inequality: $$4^{\sin x}+4^{\cos x}\ge 2\sqrt{4^{\sin x+\cos x}}.$$ The minimum value of $\sin x+\cos x$ for $x\in \mathbb{R}$ is $-\sqrt{2}$ .
Substituting, the minimum value of the given expression becomes: $$2^{1-\sqrt{2}}.$$ **Final Answer:** $\boxed{{\displaystyle B}}$

Solution:
The numerator of the given expression is: $$n{C}_0+n{C}_1+\dots +n{C}_n=2^n.$$ Thus, the given expression becomes: $$\underset{n=0}{\overset{\mathrm{\infty }}{\sum }}\frac{2^n}{n!}.$$ This is the Taylor series expansion of $e^x$ evaluated at $x=2$ :
$$\underset{n=0}{\overset{\mathrm{\infty }}{\sum }}\frac{2^n}{n!}=e^2.$$ **Final Answer:** $\boxed{{\displaystyle D}}$

Solution:
Calculate the determinant of $f(x)$ : $$f(x)=\cos x(x^2\cdot 2-x\cdot 2x)-x(2\sin x\cdot 2-2x\cdot \tan x)+1(2\sin x\cdot x-\tan x\cdot x^2)$$ Simplifying: $$f(x)=x^2\tan x-2x\sin x$$ Now, evaluate the limit: $$\underset{x\to 0}{lim}\frac{f(x)}{x^2}=\underset{x\to 0}{lim}(\tan x-\frac{2\sin x}{x})$$ Using standard limits: $$\tan x\to 0\text{ and }\frac{\sin x}{x}\to 1\text{ as }x\to 0,$$ so: $$\underset{x\to 0}{lim}\frac{f(x)}{x^2}=0-2=-2.$$ **Final Answer:** $\boxed{{\displaystyle B}}$

Solution:
Compute $det(P)$ : $$det(P)=2(2\alpha )-\alpha (-{\alpha }^2)+3(2\alpha -6)$$ Simplifying: $$det(P)={\alpha }^3+10\alpha -18.$$ The cofactor of $a_{22}$ is $2\alpha -9$ . Equating: $${\alpha }^3+10\alpha -18=2\alpha -9⟹{\alpha }^3+8\alpha -9=0.$$ Solving, we find $\alpha =1$ . Thus: $$det(P)=2\alpha -9=-7.$$ Now calculate $det(\text{adj}(P^{-1}))$ : $$det(\text{adj}(P^{-1}))={[det(P^{-1})]}^{3-1}={\left[\frac{1}{det(P)}\right]}^2=\frac{1}{(-7{)}^2}=\frac{1}{49}.$$ **Final Answer:** $\boxed{{\displaystyle B}}$

Solution:
The integral is: $$I={\int }_0^{1.7}\left[x^2\right]dx.$$ Break the integral into intervals based on the values of $\left[x^2\right]$ : $$I={\int }_0^1[x^2]dx+{\int }_1^{\sqrt{2}}[x^2]dx+{\int }_{\sqrt{2}}^{1.7}[x^2]dx.$$ - For $0\le x<1$ , $[x^2]=0$ . - For $1\le x<\sqrt{2}$ , $[x^2]=1$ . - For $\sqrt{2}\le x<\sqrt{3}\approx 1.732$ , $[x^2]=2$ . Therefore: $$I={\int }_1^{\sqrt{2}}1dx+2{\int }_{\sqrt{2}}^{1.7}1dx.$$ Solve each term: $${\int }_1^{\sqrt{2}}1dx=\sqrt{2}-1,$$ $${\int }_{\sqrt{2}}^{1.7}1dx=1.7-\sqrt{2}.$$ Combine: $$I=(\sqrt{2}-1)+2(1.7-\sqrt{2}).$$ Simplify: $$I=2.4-\sqrt{2}.$$ **Final Answer:** $\boxed{{\displaystyle B}}$

Solution:
Let's analyze each option for differentiability at $x=0$ :

**Option A:** $f(x)=e^{-|x|}-|x|$
- At $x>0$ : $f^{\prime }(x)=-e^{-x}-1$
- At $x<0$ : $f^{\prime }(x)=e^x+1$
- At $x=0$ : LHD $=2$ , RHD $=-2$ .
**Not differentiable.**

**Option B:** $f(x)=e^{|x|}+|x|$
- At $x>0$ : $f^{\prime }(x)=e^x+1$
- At $x<0$ : $f^{\prime }(x)=-e^{-x}-1$
- At $x=0$ : LHD $=-2$ , RHD $=2$ .
**Not differentiable.**

**Option C:** $f(x)=-e^{|x|}+|x|$
- At $x>0$ : $f^{\prime }(x)=-e^x+1$
- At $x<0$ : $f^{\prime }(x)=e^{-x}-1$
- At $x=0$ : LHD $=0$ , RHD $=0$ .
**Differentiable.**

**Option D:** $f(x)=-e^{-|x|}+|x|$
- At $x>0$ : $f^{\prime }(x)=e^{-x}+1$
- At $x<0$ : $f^{\prime }(x)=-e^x-1$
- At $x=0$ : LHD $=-2$ , RHD $=2$ .
**Not differentiable.**

**Correct Answer:** $\boxed{{\displaystyle C}}$

Solution:
Differentiate $f(x)$ for $x\ne 0$ :
$$f^{\prime }(x)=1+2x\sin \left(\frac{\pi }{x}\right)-\pi \cos \left(\frac{\pi }{x}\right).$$ Substitute $x=1$ :
$$f^{\prime }(1)=1+2\sin (\pi )-\pi \cos (\pi ).$$ Since $\sin (\pi )=0$ and $\cos (\pi )=-1$ :
$$f^{\prime }(1)=1+\pi .$$ For $x=0$ :
$$f^{\prime }(0)=\underset{x\to 0}{lim}\frac{x+x^2\sin \left(\frac{\pi }{x}\right)}{x}.$$ Simplify:
$$f^{\prime }(0)=\underset{x\to 0}{lim}(1+x\sin \left(\frac{\pi }{x}\right)).$$ As $x\to 0$ , $x\sin \left(\frac{\pi }{x}\right)\to 0$ :
$$f^{\prime }(0)=1.$$ Therefore:
$$f^{\prime }(1)-f^{\prime }(0)=(1+\pi )-1=\pi .$$ **Correct Answer:** $\boxed{{\displaystyle \text{C}}}$

Solution:

The given equation is: \[ \sqrt{(x-1)^2 + (y+2)^2} = \sqrt{(x+3)^2 + (y-2)^2} \] Squaring both sides, we get: \[ (x-1)^2 + (y+2)^2 = (x+3)^2 + (y-2)^2 \] Expanding and simplifying: \[ x^2 - 2x + 1 + y^2 + 4y + 4 = x^2 + 6x + 9 + y^2 - 4y + 4 \] Canceling terms and rearranging: \[ -2x + 4y + 5 = 6x - 4y + 13 \] \[ 8y - 8x = 8 \implies y - x = 1 \]

This is the equation of a straight line. Hence, the correct answer is: A. a straight line

Solution:

The lines \( x^2 - 7x + 6 = 0 \) factorize as \( x = 1 \) and \( x = 6 \). Similarly, \( y^2 - 14y + 40 = 0 \) factorize as \( y = 4 \) and \( y = 10 \).

The diagonals of the rectangle intersect at: \[ W = \left(\frac{1+6}{2}, \frac{4+10}{2}\right) = \left(\frac{7}{2}, 7\right). \]

The length of one side (QR) is: \[ QR = 6 - 1 = 5. \] The radius of the circle is: \[ r = \frac{10 - 4}{2} = 3. \]

The perpendicular distance \( PM \) is: \[ PM = \sqrt{PW^2 - MW^2} = \sqrt{3^2 - \left(\frac{5}{2}\right)^2} = \frac{\sqrt{11}}{2}. \] The length \( PQ \) is: \[ PQ = 2 \times PM = \sqrt{11}. \]

The area of the rectangle is: \[ \text{Area of } PQRS = QR \times PQ = 5 \times \sqrt{11} = 5\sqrt{11}. \] Hence, the correct answer is: B. \( 5\sqrt{11} \).

Solution:

Given the parabola \( y^2 = 4x \), we have \( a = 1 \). The equation of the normal is: \[ y = mx - 2m - m^3, \] where \( m \) is the slope of the normal.

The normal passes through \( R(9, -6) \): \[ -6 = 9m - 2m - m^3. \] Simplifying: \[ (m + 1)(m - 3)(m + 2) = 0. \] Thus, \( m = -1, 3, -2 \).

For each \( m \): - When \( m = -1 \), \( y = 2 \), \( x = 1 \) (point \( P(1, 2) \)). - When \( m = 3 \), \( y = -6 \), \( x = 9 \) (point \( R(9, -6) \)). - When \( m = -2 \), \( y = 4 \), \( x = 4 \) (point \( Q(4, 4) \)).

The distance between \( P(1, 2) \) and \( Q(4, 4) \) is: \[ PQ = \sqrt{(4 - 1)^2 + (4 - 2)^2} = \sqrt{9 + 4} = \sqrt{13}. \]

Hence, the correct answer is: A. \( \sqrt{13} \).

Solution:

Given \( f(x) + f(3 - x) = 4 \), we can use the definite integral property: \[ \int_a^b f(x) \, dx = \int_a^b f(b + a - x) \, dx. \]

Let \( I = \int_0^3 f(x) \, dx \). Using the property: \[ I = \int_0^3 f(3 - x) \, dx. \] Adding the two integrals: \[ I + I = \int_0^3 \left[f(x) + f(3 - x)\right] \, dx = \int_0^3 4 \, dx. \]

Simplifying: \[ 2I = \int_0^3 4 \, dx = 4 \times [x]_0^3 = 4 \times (3 - 0) = 12. \] Hence: \[ I = \frac{12}{2} = 6. \]

Therefore, the value of \( \int_{0}^{3} f(x) \, dx \) is: C. 6.

Solution:

- \( \angle PRQ = \angle POQ \) is incorrect. In a cyclic quadrilateral, the angles subtended at the circumference (like \( \angle PRQ \)) are half of the angles subtended at the center (like \( \angle POQ \)). Hence, \( \angle POQ = 2 \angle PRQ \), not equal to it.
- \( \angle POQ = 2 \angle PSQ \) is correct.
- \( \angle OPS = \angle OSP \) is correct because \( OP = OS \) (the radius of the circumcircle), making the triangle \( OSP \) isosceles.
- \( \angle PRQ = \angle PSQ \) is correct because angles in the same segment of a circle are equal.
Therefore, the statement \( \angle PRQ = \angle POQ \) is NOT true.

The correct answer is: A. \( \angle PRQ = \angle POQ \).

Solution:

We are given: \[ (P \cup Q)^c \cup \left( P^c \cap Q \right) \] Simplifying the expression: \[ (P \cup Q)^c = P^c \cap Q^c \] So the original expression becomes: \[ (P^c \cap Q^c) \cup (P^c \cap Q) = P^c \cap (Q^c \cup Q) \] Since \( Q^c \cup Q = U \) (the universal set), we get: \[ P^c \cap U = P^c \] Therefore, the correct answer is \( P^c \).

The correct answer is: A. \( P^c \).