answer:(I) Since fleft(0right)=left(a-1right)times0+1=1, we have fleft(fleft(0right)right)=fleft(1right)=1-frac{1}{1}=boxed{0}; (II) When x > 0, fleft(xright)=1-frac{1}{x}, thus fleft(1right)=0, hence fleft(xright) has one zero on the positive half of the x-axis. When xleqslant 0, fleft(xright)=left(a-1right)x+1, if a=1, fleft(xright)=1, thus fleft(xright) does not intersect with the negative half of the x-axis, if a > 1, fleft(xright)=0, thus left(a-1right)x+1=0, thus x=- frac{1}{a-1}, hence fleft(xright) intersects with the negative half of the x-axis at one point. if a < 1, fleft(xright)=0, thus left(a-1right)x+1=0, thus x=- frac{1}{a-1}=frac{1}{1-a} (rejected), hence fleft(xright) does not intersect with the negative half of the x-axis. In summary, when a=1, fleft(xright) has one zero, when a > 1, fleft(xright) has two zeros, when a < 1, fleft(xright) has one zero.

answer:1. Let ( n ) be the total number of questions on the examination. 2. The student answers 15 out of the first 20 questions correctly. 3. For the remaining ( n - 20 ) questions, the student answers one third correctly. Therefore, the number of correctly answered questions in the remaining part is ( frac{1}{3}(n - 20) ). 4. The total number of correctly answered questions is: [ 15 + frac{1}{3}(n - 20) ] 5. The student's mark is 50%, which means the student answered half of the total questions correctly. Therefore, we set up the equation: [ frac{15 + frac{1}{3}(n - 20)}{n} = frac{1}{2} ] 6. To solve for ( n ), we first clear the fraction by multiplying both sides by ( n ): [ 15 + frac{1}{3}(n - 20) = frac{n}{2} ] 7. Simplify the equation: [ 15 + frac{n - 20}{3} = frac{n}{2} ] 8. Multiply through by 6 to clear the denominators: [ 6 cdot 15 + 6 cdot frac{n - 20}{3} = 6 cdot frac{n}{2} ] [ 90 + 2(n - 20) = 3n ] 9. Distribute and simplify: [ 90 + 2n - 40 = 3n ] [ 50 + 2n = 3n ] 10. Subtract ( 2n ) from both sides: [ 50 = n ] 11. Therefore, the total number of questions ( n ) is 50. The final answer is ( boxed{1} )

answer:This is a geometric sequence where the first term (number of cells at the beginning) is 5, and the common ratio is 3 (each cell results in three cells after each three-day period). Over 9 days, cells divide every 3 days, so there are frac{9}{3} = 3 periods of cell division. The nth term of a geometric sequence can be calculated using the formula: [ a_n = a cdot r^{(n-1)} ] where ( a ) is the first term, ( r ) is the common ratio, and ( n ) is the term number. Plugging in the numbers: [ a_3 = 5 cdot 3^{(3-1)} = 5 cdot 3^2 = 5 cdot 9 = 45 ] Therefore, there are boxed{45} cells at the end of the 9^text{th} day.

answer:1. Given the identity involving cosecant and secant: [ (csc^{2n} x - 1)(sec^{2n} x - 1) geq (2^n - 1)^2 ] 2. Define new variables ( a ) and ( b ) such that: [ sin^2 alpha = frac{1}{a} quad text{and} quad cos^2 alpha = frac{1}{b} ] This implies: [ frac{1}{a} + frac{1}{b} = 1 ] From this, we also have: [ a + b = ab ] 3. Consider the expression: [ (a-1)(b-1) = ab - (a+b) + 1 = 1 ] 4. Rewrite the original inequality in terms of ( a ) and ( b ): [ (a^n - 1)(b^n - 1) geq (2^n - 1)^2 ] 5. For ( n = 1 ): [ (a-1)(b-1) = 1 geq 1 ] This base case holds true. 6. Use the induction hypothesis: Assume the inequality is true for ( n ). That is, assume: [ left(a^{n-1} + cdots + a + 1right)left(b^{n-1} + cdots + b + 1right) geq (2^n - 1)^2 ] 7. Consider the expression for ( n+1 ): [ left(a^n + cdots + a + 1right)left(b^n + cdots + b + 1right) ] 8. Expand the product: [ = ableft(a^{n-1} + cdots + a + 1right)left(b^{n-1} + cdots + b + 1right) + a^n + cdots + a + b^n + cdots + b + 1 ] 9. Perform the inequality manipulation: [ geq 4 left(2^n - 1 right)^2 + 2sqrt{a^n b^n} + cdots + 2 sqrt{ab} + 1 ] Since ( ab = 4 ), we have: [ geq 4 left(2^n - 1 right)^2 + 2 cdot 2^n + cdots + 2 cdot 2 + 1 ] 10. Simplify the series: [ = 4 left(2^n - 1 right)^2 + 4 left(2^n - 1 right) + 1 ] [ = left(2^{n+1} - 1right)^2 ] Conclusion: By the principle of mathematical induction, the original inequality ( (csc ^{2 n} x - 1)(sec ^{2 n} x - 1) geqslant (2^n - 1)^2 ) holds for all natural numbers ( n ). Thus, this completes the proof. (blacksquare)