Solution:

Note that $\frac{1}{100}$ is a very small number. We approximate the given expression: $${(1+\frac{1}{100})}^{200}={\left[{(1+\frac{1}{100})}^{100}\right]}^2$$ By the limit property of exponential functions: $$\underset{x\to 0}{lim}(1+x{)}^{1/x}=e$$ We have: $${(1+\frac{1}{100})}^{100}\approx e$$ Therefore: $${(1+\frac{1}{100})}^{200}\approx e^2$$ Since $e^2\approx 7.389$ , the nearest integer to $e^2$ is $7$ .

The correct answer is: B. 7.

Solution:

To find $\int f(x)dx=\int {\tan }^{-1}(\sqrt{x})dx$ , use integration by parts:

Let: $$u={\tan }^{-1}(\sqrt{x})\text{(1st function)},dv=dx\text{(2nd function)}.$$ Then: $$\int {\tan }^{-1}(\sqrt{x})dx=x{\tan }^{-1}(\sqrt{x})-\int \frac{1}{1+x}\cdot \frac{x}{2\sqrt{x}}dx.$$

Simplify: $$\int {\tan }^{-1}(\sqrt{x})dx=x{\tan }^{-1}(\sqrt{x})-\frac{1}{2}\int \frac{\sqrt{x}}{1+x}dx.$$ Substituting $x=t^2$ and $dx=2tdt$ : $$\int \frac{\sqrt{x}}{1+x}dx=\int \frac{t^2}{1+t^2}dt=\int (1-\frac{1}{1+t^2})dt.$$ Solve: $$\int \frac{\sqrt{x}}{1+x}dx=t-{\tan }^{-1}(t)+C=\sqrt{x}-{\tan }^{-1}(\sqrt{x})+C.$$

Thus: $$\int f(x)dx=(x+1){\tan }^{-1}(\sqrt{x})-\sqrt{x}+C.$$ Given that the graph passes through $(0,2)$ : $$2=(0+1)\cdot {\tan }^{-1}(0)-\sqrt{0}+C⟹C=2.$$

Final answer: $$\int f(x)dx=(x+1){\tan }^{-1}(\sqrt{x})-\sqrt{x}+2.$$

The correct answer is: C. $(x+1){\tan }^{-1}(\sqrt{x})-\sqrt{x}+2$ .

Solution:

Start by using trigonometric identities: $${\cos }^2(x)=\frac{1+\cos (2x)}{2},\cos (A)+\cos (B)=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right).$$

Substitute these into $f(x)$ : $$f(x)=\frac{1+\cos (2x)}{2}+\frac{1+\cos (\frac{2\pi }{3}+2x)}{2}-\frac{1}{2}[\cos (\frac{\pi }{3}+2x)+\cos \left(\frac{\pi }{3}\right)].$$

Simplify: $$f(x)=1+\frac{\cos (2x)}{2}+\frac{\cos (\frac{2\pi }{3}+2x)}{2}-\frac{\cos (\frac{\pi }{3}+2x)}{2}-\frac{1}{4}.$$ Combine terms: $$f(x)=\frac{3}{4}+\frac{1}{2}[\cos (\frac{\pi }{3}+2x)-\cos (\frac{\pi }{3}+2x)].$$

Since all terms involving $x$ cancel out, $f(x)=\frac{3}{4}$ , which is constant. Therefore: $$f(1)=f(0)=f(-1)=\frac{3}{4}.$$

The correct answer is: B. $f(1)=f(0)=f(-1)$ .

Solution:

Split the integral at 0 since $|x|$ changes sign: $${\int }_{-1}^1{x}^3|x|dx={\int }_{-1}^0-x^{4}dx+{\int }_0^1{x}^{4}dx.$$

Compute each integral: $${\int }_{-1}^0-x^{4}dx=-{\left[\frac{x^5}{5}\right]}_{-1}^0=-(0-\frac{(-1{)}^5}{5})=\frac{1}{5}.$$ $${\int }_0^1{x}^{4}dx={\left[\frac{x^5}{5}\right]}_0^1=\frac{1}{5}.$$

Combine the results: $${\int }_{-1}^1{x}^3|x|dx=\frac{1}{5}-\frac{1}{5}=0.$$

The correct answer is: B. 0.

Solution:

The given limit is: $$\underset{x\to 0}{lim}\frac{(1+x{)}^{1/6}-(1-x{)}^{1/6}}{x}.$$

Substituting $x=0$ , we get a $\frac{0}{0}$ indeterminate form. Hence, we use L'Hopital's Rule by differentiating the numerator and denominator: $$\underset{x\to 0}{lim}\frac{(1+x{)}^{1/6}-(1-x{)}^{1/6}}{x}=\underset{x\to 0}{lim}[\frac{1}{6}(1+x{)}^{-5/6}+\frac{1}{6}(1-x{)}^{-5/6}].$$

Evaluating at $x=0$ : $$\frac{1}{6}(1+0{)}^{-5/6}+\frac{1}{6}(1-0{)}^{-5/6}=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}.$$

The correct answer is: A. $\frac{1}{3}$ .

Solution:

The given region is symmetric about both axes. To simplify the calculation, we calculate the area of the region in the first quadrant and multiply by 4.

The area of the shaded region in the first quadrant is given by: $${\int }_0^1{x}^{2}dx={\left[\frac{x^3}{3}\right]}_0^1=\frac{1}{3}.$$

The total area of the enclosed region is: $$4\times \frac{1}{3}=\frac{4}{3}.$$

Therefore, the correct answer is: D. $\frac{4}{3}$ .

Solution:

Total letters are $12$ . Letters $O,I,N$ appear twice.

Consider $C----S$ as one block. The remaining 6 positions are to be filled, and this block can be placed in 7 different positions between the 6 other letters. Since $C$ and $S$ are fixed in this block, the remaining 10 positions can be filled in: $$\frac{10!}{2!2!2!}$$

Total number of arrangements: $$7\times \frac{10!}{2!2!2!}=7\times \frac{10\times 9\times 8\times 7!}{8}=630\times 7!$$

The correct answer is: C. $630\times 7!$ .

Solution:

Total number of players = 9 (2 batsmen, 3 bowlers, 4 fielders).
Number of ways to select 3 players from 9: $${}^9{C}_3=\frac{9\times 8\times 7}{6}=84$$

Number of ways to select 3 players such that no bowlers are included (i.e., only batsmen and fielders are selected): $${}^6{C}_3=\frac{6\times 5\times 4}{6}=20$$

Number of ways to select 3 players with at least one bowler: $${}^9{C}_3-{}^6{C}_3=84-20=64$$

The correct answer is: B. 64.

Solution:

Given circles: $$(x+1{)}^2+(y+4{)}^2=40,(x-2{)}^2+(y-5{)}^2=10.$$

The center and radius of the circles are: $$c_1=(-1,-4),r_1=\sqrt{40},c_2=(2,5),r_2=\sqrt{10}.$$

The distance between the centers is: $$c_1{c}_2=\sqrt{(2+1{)}^2+(5+4{)}^2}=\sqrt{9+81}=3\sqrt{10}.$$

The sum of the radii is: $$r_1+r_2=\sqrt{40}+\sqrt{10}=3\sqrt{10}.$$

Since $c_1{c}_2=r_1+r_2$ , the circles are just touching, and there are exactly 3 common tangents.

Hence, the correct answer is: B. 3.

Solution:

Given points $A(0,4)$ and $B(3,0)$ , we need to find the coordinates of $C$ for the square $ABCD$ .

It can be shown that $\mathrm{△}AOB$ is congruent to $\mathrm{△}BEC$ because:

• $\mathrm{\angle }AOB=\mathrm{\angle }ABC={90}^{\circ }$
• $AB=BC$ , as they are sides of the square.
• Other angles and distances are equal by the properties of congruent triangles.

The coordinates of $E$ (the reflection of $O(0,0)$ across $AB$ ) are $(7,0)$ , and the coordinates of $C$ can be found as $C(7,3)$ .

Hence, the correct answer is: D. (7,3).

Solution:

We are given: $$(1+x^2)f(x)={\int }_1^x[2t^2-f^{\prime }(t)]dt$$ Differentiate both sides using the Leibniz rule: $$2xf(x)+(1+x^2)f^{\prime }(x)=(2x^2-f^{\prime }(x))\cdot \frac{dx}{dx}-(2-f^{\prime }(1))\cdot \frac{d1}{dx}$$ At $x=1$ , note that $\frac{d1}{dx}=0$ . Substituting $x=1$ : $$2f(1)+2f^{\prime }(1)=2-f^{\prime }(1)$$ From the integral definition, $f(1)=0$ . Substituting $f(1)=0$ : $$3f^{\prime }(1)=2⟹f^{\prime }(1)=\frac{2}{3}$$ Hence, the correct answer is: B. $\frac{2}{3}$ .

Solution:

We are given: $$f(x)=\frac{\tan (\pi [x-\frac{\pi }{2}])}{2+[x{]}^2}$$ - The denominator $2+[x{]}^2>0$ , as $[x{]}^2\ge 0$ . - From trigonometric properties, $\tan (n\pi )=0$ for all integers $n$ . - Substituting, $\tan (\pi [x-\frac{\pi }{2}])=0$ for integers, making $f(x)=0$ . Since $f(x)=0$ (a constant function), it is differentiable everywhere.
$\therefore$ Option A. Differentiable everywhere is correct.

Solution:
Given the equation of the parabola: $$x^2=4ay$$ Differentiating with respect to $x$ , we get: $$2x=4a\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{x}{2a}$$ The tangents from point $P$ are perpendicular, so if the slope of one tangent is $m$ , the slope of the other tangent will be $-\frac{1}{m}$ . Using this, we find that the coordinates of the point $P$ satisfy the equation: $$m^{2}y+mx+a=0$$ After simplifying, we obtain: $$(m^2+1)k+(m^2+1)a=0\Rightarrow k=-a$$ This shows that the locus of $P$ is a straight line. Therefore, the correct answer is: $\boxed{{\displaystyle B.\text{A straight line}}}$

Solution:
We are given that: $$AE=1\text{km},AB=AE\cos ({30}^{\circ })=\frac{\sqrt{3}}{2}\text{km}$$ $$BC=x\text{and}FC=BE=AE\sin ({30}^{\circ })=\frac{1}{2}\text{km}$$ The distance DF is related to the tangent angle at ${60}^{\circ }$ , and we get: $$DF+FC=AB+BC$$ Solving the equation, we find: $$x(\sqrt{3}-1)=\frac{\sqrt{3}-1}{2}\Rightarrow x=\frac{1}{2}$$ Thus, the height of the rock is: $$h=DF+FC=\frac{\sqrt{3}}{2}+\frac{1}{2}=\frac{\sqrt{3}+1}{2}\text{km}$$ Therefore, the correct answer is: $\boxed{{\displaystyle D.\frac{\sqrt{3}+1}{2}\text{km}}}$

Solution:
The sides of the right-angled triangle are in Geometric Progression. Let the smallest side be $\frac{a}{r}$ , the middle one be $a$ , and the hypotenuse be $ar$ . The angle opposite to the middle side is the greater acute angle. Using the Pythagorean theorem, we have: $$(ar{)}^2={\left(\frac{a}{r}\right)}^2+a^2$$ Solving for $r^2$ , we find: $$r^2=\frac{1+\sqrt{5}}{2}$$ Therefore, the value of $\frac{1}{r^2}$ is: $$\frac{1}{r^2}=\frac{2}{1+\sqrt{5}}$$ Hence, the correct answer is: $\boxed{{\displaystyle A.\frac{2}{1+\sqrt{5}}}}$

Solution:
Let: - $A$ be the set of Mastercard users - $B$ be the set of Visa Card users - $C$ be the set of American Express users Given information: - $A=60$ - Only $A=30$ - $A\cap B-(A\cap B\cap C)=20$ - $A\cap C-(A\cap B\cap C)=6$ Using the formula for total number of users in each category: $$60=30+20+(A\cap B\cap C)+6$$ This gives: $$A\cap B\cap C=4$$ The total number of users across all 3 categories is 100: $$100=30+20+10+20+6+4+x$$ Solving for $x$ : $$x=100-30-20-10-20-6-4=10$$ Therefore, the number of executives who use both Visa and American Express cards is: $$B\cap C=10+4=14$$ Hence, the correct answer is: $\boxed{{\displaystyle D.14}}$

Solution:
The given equation is: $$x-\frac{1}{x}=ax+b$$ Rearranging this equation: $$x^2-1=a{x}^2+bx$$ Simplifying: $$(a-1)x^2+bx+1=0$$ To have distinct real roots, the discriminant must be greater than zero: $$D=b^2-4(a-1)$$ For $a<1$ , we have $a-1<0$ , so: $$b^2-4(a-1)>0$$ Thus, for any $b$ , the equation will have two distinct roots as long as $a<1$ . Hence, the correct answer is: $\boxed{{\displaystyle C.\text{Will have two distinct real roots if }a<1\text{ and for any }b.}}$

Solution:
Let $\alpha +\beta$ be the roots of the equation $a{x}^2+bx+c=0$ . The roots $\alpha$ and $\beta$ are real and of opposite signs: $$\alpha +\beta =\frac{-b}{a}\text{and}\alpha \beta =\frac{c}{a}$$ Since the sum of the roots is negative: $$\alpha +\beta <0\Rightarrow -\frac{b}{a}<0\Rightarrow \frac{b}{a}>0$$ This means $a$ and $b$ are of the same sign. Also, since $\alpha \beta <0$ (as the roots have opposite signs), we have: $$\frac{c}{a}<0\Rightarrow a\text{ and }c\text{ are of opposite signs.}$$ Thus, the correct answer is: $\boxed{{\displaystyle D.\text{the signs of }a\text{ and }b\text{ are the same and opposite to the sign of }c.}}$

Solution:
$$(5+\sqrt{17}{)}^{2021}=I+F\text{where}I=\text{integer part},F=\text{fraction part}$$ Consider the equation for $(5-\sqrt{17}{)}^{2021}=F^{\prime }$ , where $F^{\prime }$ is a fraction, and note that $(5-\sqrt{17})$ is a fraction with value approximately $0.1$ , and the power of a fraction remains a fraction. Adding both equations: $$(5+\sqrt{17}{)}^{2021}+(5-\sqrt{17}{)}^{2021}=I+F+F^{\prime }$$ After cancellation of positive and negative terms, we get: $$2[5^{2021}+(\genfrac{}{}{0ex}{}{2021}{1})5^{2020}\sqrt{17}+\dots ]=I+F+F^{\prime }$$ The left-hand side (LHS) is even, and the right-hand side is a natural number. Therefore: $$F+F^{\prime }=1\text{implies}I=\text{odd}.$$ Therefore, the integral part is odd, and the correct answer is: $\boxed{{\displaystyle D.\text{odd}}}$

Solution:
Note: Any prime number greater than 3 can be written as $$6k+1\text{or}6k-1$$ **Proof**: Any natural number can be written as one of $6k,6k+1,6k+2,6k+3,6k+4,6k+5$ . However, only the forms $6k+1$ and $6k-1$ are prime. Therefore, a prime number $p$ can be written as either $6k+1$ or $6k-1$ . The equation is: $$y^2=1+2x^2$$ Let $x=6k\pm 1$ , then: $$y^2=1+2(6k\pm 1{)}^2=1+2(36k^2+1\pm 12k)=1+72k^2+2\pm 24k$$ For $y$ to be in the form $6m\pm 1$ , we have: $$72k^2\pm 24k+2=6m\text{or}72k^2\pm 24k+4=6m$$ In both cases, the left-hand side is not divisible by 6, implying that $x\ne 6k\pm 1$ . Hence, $x$ must be 2 or 3. Taking $x=3$ gives: $$y^2=1+2(3^2)=19$$ Since 19 is not a perfect square, $x=3$ is not valid. Taking $x=2$ , we get: $$y^2=1+2(2^2)=9\Rightarrow y=3$$ Therefore, the only solution is $x=2$ and $y=3$ . The equation has a unique solution: $x=2$ and $y=3$ . Hence, the correct answer is: $\boxed{{\displaystyle B.\text{the equation has a unique solution}}}$

Solution:
Let $P$ be a prime number and $n$ be a natural number. The maximum power of $P$ that divides $n!$ is given by: $$\lfloor \frac{n}{p}\rfloor +\lfloor \frac{n}{p^2}\rfloor +\lfloor \frac{n}{p^3}\rfloor +\cdots$$ where $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$ . For this problem, $P=5$ and $n=518$ . We need to calculate: $$\lfloor \frac{518}{5}\rfloor +\lfloor \frac{518}{25}\rfloor +\lfloor \frac{518}{125}\rfloor +\lfloor \frac{518}{625}\rfloor +\cdots$$ Let's calculate these terms: $$\lfloor \frac{518}{5}\rfloor =103,\lfloor \frac{518}{25}\rfloor =20,\lfloor \frac{518}{125}\rfloor =4,\lfloor \frac{518}{625}\rfloor =0$$ Adding these terms gives: $$103+20+4+0=127$$ Therefore, the highest power of 5 that divides $518!$ is $\boxed{{\displaystyle 127}}$ . Hence, the correct answer is: $\boxed{{\displaystyle C.\text{127}}}$

Solution:
We are given the matrix $P$ and need to compute $P^{20}(I-P{)}^{21}$ .
Step 1: Compute the higher powers of $P$ and $(I-P)$ : $$P=\left[\begin{array}{ccc}0& 1& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right],I=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right],I-P=\left[\begin{array}{ccc}1& -1& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$$ Now, observe that: $$P^2=P\Rightarrow P^{10}=P^5=P^2=P$$ and similarly: $$(I-P{)}^2=I-P\Rightarrow (I-P{)}^{20}=I-P$$ Step 2: Now, calculate the desired product: $$P^{20}(I-P{)}^{21}=P^{20}(I-P{)}^{20}(I-P)=P(I-P)(I-P)=P(I-P)$$ Finally: $$P(I-P)=\left[\begin{array}{ccc}0& 1& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& -1& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$$ Thus, $P^{20}(I-P{)}^{21}=0$ , which is the $3\times 3$ zero matrix. Hence, the correct answer is: $\boxed{{\displaystyle C.0}}$

Solution:
The probability for each roll of a fair die is: $$P(1)=P(2)=P(3)=P(4)=P(5)=P(6)=\frac{1}{6}$$ Let's break down the cases: **Case 1**: Obtaining two faces in 2 throws: The second number is different from the first one. The probability is: $$P(\text{2 different faces in 2 throws})=1\times \frac{5}{6}$$ **Case 2**: Obtaining two faces in 3 throws: The third number is different from the first two. The probability is: $$P(\text{2 different faces in 3 throws})=1\times \frac{1}{6}\times \frac{5}{6}$$ **Case 3**: Obtaining two faces in 4 throws: The fourth number is different from the first three. The probability is: $$P(\text{2 different faces in 4 throws})=1\times \frac{1}{6}\times \frac{1}{6}\times \frac{5}{6}$$ Simplifying: $$P(\text{two different faces in 4 throws})=\frac{5}{216}$$ The correct probability for the desired event is: $$\boxed{{\displaystyle D.\frac{5}{216}}}$$

Solution:
To find information about maxima and minima, we need to find the critical points, i.e., solve $f^{\prime }(x)=0$ . The given polynomial is: $$f(x)=\underset{i=0}{\overset{n}{\sum }}(a_0+i)x^{2i},\text{where }{a}_0>0$$ The first derivative of $f(x)$ is: $$f^{\prime }(x)=\underset{i=1}{\overset{n}{\sum }}2i(a_0+i)x^{2i-1}$$ Expanding this expression: $$f^{\prime }(x)=2(a_0+1)x+4(a_0+2)x^3+\cdots +2n{x}^{2n-1}(a_0+n)$$ Factor out $x$ : $$f^{\prime }(x)=x[2(a_0+1)+4(a_0+2)x^2+\cdots +2n{x}^{2n-2}(a_0+n)]$$ For $f^{\prime }(x)=0$ , the only critical point is $x=0$ . To determine whether $x=0$ is a minimum or maximum, we calculate the second derivative: $$f^{″}(x)=2(a_0+1)+x[12(a_0+2)x^2+\cdots +2n(2n-1)x^{2n-3}]$$ Evaluating at $x=0$ : $$f^{″}(0)=2(a_0+1)>0\text{(since }{a}_0>0\text{)}$$ Therefore, $x=0$ is a point of minima. Since there are no other critical points, the function has only one minimum and no maxima. The correct answer is: $$\boxed{{\displaystyle C.\text{only one minima but no maxima}}}$$