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Correct answer is (a) ∞∑n=0 (X^n/n!)

To EXPLAIN I would say: The exponential function e^x can DESCRIBED as ∞∑n=0 (x^n/n!) which is an example of a Maclaurin series. This series converges for all x.

The CORRECT CHOICE is (c) ln(7)∞∑n=0 x^n+1/7^n+1

The best EXPLANATION: We KNOW that ∫1/7−x dx=−ln(7−x) and there is a power series representation for 1/7−x. So, ln(7−x)=−∫1/7−xdx

=−∫ ∞∑n=0 x^n/7^n+1dx=C

⇒ ∞∑n=0 x^n+1/7^n+1

So, the answer is, ln(7−x)=ln(7)∞∑n=0 x^n+1/7^n+1.

Correct choice is (c) 6.8%

BEST explanation: The events are not independent. There will be a \(\FRAC{10}{18} = \frac{5}{9}\) CHANCE that any of the mangoes in the bag is even. The probability that the first one is even is \(\frac{1}{2}\), for the second mango, given that the first one was even, there are only 9 even numbered balls that could be drawn from a total of 17 balls, so the probability is \(\frac{9}{17}\). For the third mango, since the first two are both ODD, there are 8 even numbered mangoes that could be drawn from a total of 16 remaining balls and so the probability is \(\frac{8}{16}\) and for fourth mango, the probability is = \(\frac{7}{15}\). So the probability that all 4 mangoes are even numbered is \(\frac{10}{18}*\frac{9}{17}*\frac{8}{16}*\frac{7}{16}\) = 0.068 or 6.8%.

The CORRECT choice is (b) ∞, −∞<X<∞

Explanation: So, L=limn→∞∣(x−7)^n+1/n^n∣

L=limn→∞∣x−7/n∣

L=|x−7|limn→∞1/n=0

So, since L=0<1 any of the value of x, this power SERIES will CONVERGE for every x. In these cases, the radius of CONVERGENCE is R=∞ and interval of convergence is −∞

Right CHOICE is (d) 0, 1/2

Explanation: SUPPOSE, L=limn→∞ |(m+1)!(2x+1)^m+1/m!(2x+1)^m|

= limm→∞∣(m+1)m!(2x-1)/m!|

= |2x-1|limm→∞(m+1)

So, this power series will only CONVERGE if x=1/2. We know that every power series will converge for x=a and in this case a=1/2. Remember that we GET a from (x−a)^n. In this case, the radius of convergence is R=0 and the INTERVAL of convergence is x=1/2.

Correct option is (b) 7|3x−1|

Easiest EXPLANATION: Okay, LET’s start off with the Ratio TEST to get our HANDS on L = limn→ ∞∣7^N+1(3x−1)^n/(n+1)^n7^n(3x−1)^n-1∣=limn→∞∣7^n(3x−1)^n+1∣=|3x−1|limn→∞7^n/(3n-1)=7|3x−1|.

The CORRECT choice is (a) ∞∑n=0 ((-1)^nE2n / (2n)!)*x^2n

Easy EXPLANATION: A trigonometric SERIES is an example of a Maclaurin series. Here, sec(x) can be REPRESENTED as ∞∑n=0 ((-1)^nE2n / (2n)!)*x^2n.

Correct option is (a) b0+b1x+b2x^2+…+bnx^n

Easiest explanation: A formal power series is also CALLED a “formal series”, of a field F is an infinite SEQUENCE b0, b1, B2, … over F. It is a function from the set of nonnegative integers to F i.e., 0, 1, 2, 3, … → F. A formal power series can also be WRITTEN as b0+b1x+b2x^2+…+bnx^n.

Right option is (c) 512

Explanation: By using exponential FORM: a^-5/9 = 2/8. Now, raise both SIDES of the above equation to the POWER -9/5: (x^-5/9)^-9/5 = (1/32)^-9/5. By SIMPLIFYING we get, a = 32^9/5 = 2^9 = 512.

The correct answer is (d) log54 (n+1)

To explain I WOULD say: By USING the EQUIVALENT expression: a^y = X ⇔ y = loga (x) to write 3^x = m as a LOGARITHM: y = log54 (n+1).

Correct OPTION is (c) ln [( 1 + \(\SQRT{37}\)) / 2] / ln 4

To elaborate: GIVEN: 16^x – 4^x – 9 = 0. Since 16^x = (4^x)^2, the equation may be written as: (4^x)^2 – 4^x – 9 = 0. Let t = 3^x and so t: t^2 – t – 9 = 0 which gives t: t = (1 + \(\sqrt{37}\)) / 2 and (1 – \(\sqrt{37}\)) / 2

Since t = 4x, the ACCEPTABLE solution is y = (1 + \(\sqrt{37}\)) / 2 ⇒ 4x = (1 + \(\sqrt{37}\))/2. By using ln on both sides: ln 4^x = ln [ (1 + \(\sqrt{37}\)) / 2]⇒ x = ln [ ( 1 + \(\sqrt{37}\))/2] / ln 3.

Right option is (d) 1/(2*LN(3))

To explain: By using change of BASE formula we can have ln (X) / ln(4) = B ln(x) / ln(2/3) ⇒

B = 1/(2*ln(3)).

The CORRECT answer is (d) 3, 5

To ELABORATE: By using the property if logax = logay then x=y, GIVES 2x^2-3x=10-6x. Now, to solve the equation x^2-3x-5x+15=0 ⇒ x^2-8x+15 ⇒ x=3, x=5

For x=3:log2(3^2-3*3) = log2(5*3-15) ⇒ true

For x=5:log2(5^2-3*5) = log2(5*5-15) ⇒ true

The solutions to the equation are : x=3 and x=5.

Correct OPTION is (d) 4.8

The EXPLANATION is: Since, a^logax = x. The given EQUATION MAY be WRITTEN as: 3x^2 x = 348 ⇒ x = (116)^1/3 = 4.8.

The correct OPTION is (a) 2x – 8x^2 + \(\frac{152x^3}{3}\) – …

To explain: To SOLVE the logarithmic function LN(\(\frac{1+5x}{1+3X}\)) = ln(1+5x) – ln(1+3x) = (5x – \(\frac{(5x)^2}{2} + \frac{(5x)^3}{3}\) – …) – (3x – \(\frac{(3x)^2}{2} + \frac{(3x)^3}{3}\) – …) = 2x – 8x^2 + \(\frac{152x^3}{3}\) – …

The correct OPTION is (c) Diffie-Hellman key exchange

Explanation: A discrete logarithm modulo of an integer to the base is an integer such that a^x ≡ b (mod g). The PROBLEM of computing the discrete logarithm is a well-known challenge in the FIELD of cryptography and is the basis of the cryptographic SYSTEM i.e., the Diffie-Hellman key exchange.

Right option is (d) -25X + \(\frac{125x^2}{2} – \frac{625x^3}{3} + \frac{3125x^4}{4}\) …

To explain: APPLY the logarithmic law, that is logax = xlog(a). Now the function is ln(1+5x)^-5 = -5log(1+5x). By taking the series = -5(5x – \(\frac{(5x)^2}{2} + \frac{(5x)^3}{3} – \frac{(5x)^4}{4}\) + …) = -25x + \(\frac{125x^2}{2} – \frac{625x^3}{3} + \frac{3125x^4}{4}\) …

The correct choice is (B) 42.5%

To elaborate: Number of MULTIPLES of 7=45 and number of multiples of 3=10^6 and number of numbers which are multiples of both 7 and 3 = 15 Thus, P (selecting either a multiple of 7 or a multiple of 3) = \(\FRAC{45}{320} + \frac{106}{320} – \frac{15}{320} = \frac{136}{320} = \frac{2}{5}\) = 0.425 or 42.5%.

The correct answer is (a) 260100

To elaborate: Let N be the total number of ways we can distribute the letters. Each letter can be placed into any one of the 4 BOXES, so |N| = 4^9. Let T be the SET of ways such that the tetrahedron box has no letters, C be the set of ways such that the cube box has no letters, P be the set of ways such that the cube box has no letters, and D be the set of ways such that the dodecahedron box has no letters. Now, to find |N| – |T U C U P U D|. We have |T|=|C|=|P|=|D|=2^7 and SINCE the letters can be placed into one of the two other boxes, and |TUC| = |C U P| = |P U D| = |D U T| = 1^7, since all the letters must be placed in the remaining box, andT ⋂ C ⋂ P ⋂ D| = 0. Hence, PIE implies |N| – |T U C U P U D| = 4^9 – 4 x 2^9 + 4 x 1^9 – 0 = 260100.

The correct answer is (a) \(\frac{171}{900}\)

The best explanation: The number of numbers that don’t have one ANYWHERE 9^3 = 729 is (9 possibilities for each individual digit), and there are 9*10^2 = 900 numbers overall (9 possibilities for hundreds, 10 for the tens and units), so there are 900 – 729 = 171 numbers with at LEAST a one and thus \(\frac{171}{900}\) PROBABILITY.

Right option is (B) 208

Best explanation: PIE is USED to count the elements of a set and stated as the sum of elements in A or B is equal to the sum of elements in A PLUS the sum of elements in B minus the sum of elements in A and B. Let A be the set of multiples of 4 and B be the set of multiples of 5, then A ⋂ B is the set of multiples of 20, and hence

\(\frac{520}{4} + \frac{520}{5} – \frac{520}{20}\) = 208.

The correct CHOICE is (d) 30

The EXPLANATION: There are 15 PEOPLE who wish to join football, but 9 of those people also join BASKETBALL. By using the principle of inclusion and exclusion, we have: 15 people joining football + 24 people joining basketball – 9 people who will join both = 30 people TOTAL.

Right option is (d) 303

The BEST I can explain: We ADD the number of numbers that are DIVISIBLE by 2 and 6 and subtract the numbers which are divisible by 12. Hence, the required probability is

\(\frac{520}{2} + \frac{520}{6} – \frac{520}{12}\) = 303.3 = 303(APPROXIMATELY).

Right option is (b) 14

The best explanation: SINCE 18 are skilled in all 3. Subtract 18 from all three to get a total with single skilled and double skilled workers including the duplicates. SOFTWARE engineers = 48 – 18 = 30, Hardware engineers = 35 – 18 = 17, Network engineers = 42 – 18 = 24 making a total of 71 and this is a total set of single and double skilled workers including duplicates. Out of 75 candidates, 18 were skilled in three areas. So, 75 – 18 = 57 (actual no of workers skilled with single and both skills) Now the difference between the number without duplicates (57) and with duplicates (71), 71 – 57 = 14. So, 14 are skilled in exactly two JOBS.

The correct choice is (a) 0, 0, 1, 1, 2, 3, 5, 8,…

To EXPLAIN: Based on the GIVEN generating function, the SEQUENCE will be 0, 0, 1, 1, 2, 3, 5, 8,… which is generated by 1/1−x^2−x^4.

Right answer is (C) 0, 4, 8, 12, 16, 20,…

Easy explanation: The sequence should be 0, 4, 8, 12, 16, 20,…for the generating function 4x/(1-x)^2, when basic generating function: 1/(1-x).

The correct ANSWER is (a) (1−x)G−4/x

The explanation: (1−x)G = 4 + 3X + 6x^2 + 9x^3 +⋯ which can be accepted. We can COMPUTE it like this:

3 + 6x + 9x^2 + ⋯ = (1−x)G−4/x.

Correct CHOICE is (C) (4/1−7x)+(6/1+2x)

EXPLANATION: For the given sequence after EVALUATING the FORMULA the generating formula will be (4/1−7x)+(6/1+2x).

Right answer is (B) an = 3an-1−2an-2

Best explanation: The recurrence relation for the sequence 1, 3, 7, 15, 31, 63,… should bean = 3an-1−2an-2. The solution for A: A=1/1 − 3X + 2x^2.

Right OPTION is (B) (1-A)-1/x

The EXPLANATION: The generating function for the sequence A. Using differencing:

A = 1 + 9X + 25x^2 + 49x^3 + ⋯(1)

−xA = 0 + x + 9x^2 + 25x^3 + 49x^4 + ⋯(2)

(1−x)A = 1 + 8x + 16x^2 + 24x^3 +⋯. Since 8x + 16x^2 + 24x^3 + ⋯ = (1-x)A-1 ⇒8 + 16x + 24x^2 +…= (1-A)-1/x.

The correct choice is (c) \(\frac{1}{(1−X)^2}\)

To explain: Basic GENERATING function is \(\frac{1}{1-x}\). If we differentiate term by term in the power SERIES, we GET (1 + x + x^2 + x^3 +⋯)′ = 1 + 2x + 3x^2 + 4x^3 +⋯ which is the generating series for 1, 2, 3, 4,….

The correct option is (b) \(\frac{1}{(1-6x)}\)

For explanation I would say: For the SEQUENCE 1, 6, 36, 216,… the GENERATING FUNCTION must be \(\frac{1}{(1-6x}\), when basic generating function: \(\frac{1}{1-x}\).

Correct answer is (C) 4, 0, 15, 10, 25, 16,…

For explanation I would say: Consider the coefficients of each x^n term. So a0=4, since the COEFFICIENT of x0 is 4 (x0=1 so this is the constant term). Since 15 is the coefficient of x^2, so 15 is the term a2 of the sequence. To find a1 check the coefficient of X1 which in this CASE is 0. So a1=0. Continuing with these we have a2=15, a3=10, a4=25, and a5=16. So we have the sequence 4, 0, 15, 10, 25, 16,…

Right option is (B) \(\frac{1}{5}\)

Best explanation: TOTAL no of balls = 10. Number of ways drawing 3 balls at random out of 10 = ^10C3 = 120. Probability of drawing 3 balls of same colour = ^4C3 + ^6C3 = 24. HENCE, the required probability is \(\frac{24}{120} = \frac{1}{5}\).

The correct option is (a) \(\frac{13}{27}\)

The explanation is: We let Y be the event that the family has one CHILD who is a girl born on Tuesday and X be the event that both children are boys, and apply Bayes’ Theorem. Given that there are 7 days of the week and there are 49 possible combinations for the days of the week the two GIRLS were born on and 13 of these have a girl who was born on a Monday, so P(Y|X) = \(\frac{13}{49}\). P(X) remains unchanged at \(\frac{1}{4}\). To calculate P(Y), there are 142 = 196 possible ways to select the gender and the day of the week the child was born on. There are 132 = 169 ways which do not have a girl born on Monday and which 196 – 169 = 27 which do, so P(Y) = \(\frac{27}{196}\). This GIVES is that P(X|Y) = \(\frac{\frac{13}{19}*\frac{1}{4}}{\frac{27}{196}} = \frac{13}{27}\).

The correct ANSWER is (c) \(\frac{2}{9}\)

To explain I would say: Assume that the probability of an EMPLOYEE being a man or WOMAN is (\(\frac{1}{2}\)). By using Bayes’ theorem: LET B be the event that the meeting has 3 employees who is a woman and let A be the event that all employees are WOMEN. We want to find P(A|B) = \(\frac{P(B|A)*P(A)}{P(B)}\). P(B|A) = 1, P(A) = \(\frac{1}{12}\) and P(B) = \(\frac{8}{12}\). So, P(A|B) = \(\frac{1*\frac{1}{12}}{\frac{8}{12}} = \frac{1}{8}\).

Correct choice is (a) 50%

Best explanation: 18% email are SPAM and 82% email are not spam. Now, 18% of mail MARKED as spam is spam and 82% mail marked as spam are not spam. By Bayes theorem the probability that a mail marked spam is really a spam = (Probability of being spam and being detected as spam)/(Probability of being detected as spam)= (0.18 * 0.82)/(0.18 * 0.82) + (0.18 * 0.82) = 0.5 or 50%.

Right answer is (b) \(\frac{1}{3}\)

BEST explanation: The probability that the card drawn is a queen = \(\frac{4}{52}\), since there are 4 QUEENS in a standard deck of 52 cards. If the event is “this card is a queen” the prior probability P(queen) = \(\frac{4}{52} = \frac{1}{13}\). The POSTERIOR probability P(queen|face) can be calculated USING Bayes theorem: P(king|face) = P(face|king)/P(face)*P(king). Since every queen is ALSO a face card, P(face|queen) = 1. The probability of a face card is P(face) = (\(\frac{3}{13}\)). [since there are 3 face cards in each suit (Jack, Queen, King)]. Using Bayes theorem gives P(queen|face) = \(\frac{13}{3}*\frac{1}{13} = \frac{1}{3}\).