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(b) 2 \(\int_{0}^a\) f(x) dx

Correct option (d) Cannot be determined

Explanation:

The answer to this question cannot be determined because the question is talking about income and asking about expenses. You cannot solve this unless you know the value of the expenditure she incurs over the years. Thus, “Cannot be Determined” is the correct answer. 

Correct option (d) 23

Explanation:

The numbers forming an AP would be:

1234, 1357, 2345, 2468, 3210, 3456, 3579, 4321, 4567, 5432, 5678, 6543, 6420, 6789, 7654, 7531, 8765, 8642, 9876, 9753, 9630. 

A total of 21 numbers. 

If we count the GPs we get: 1248, 8421-a total of 2 numbers.

Hence, we have a total of 23, 4-digit numbers why the digits are either APs or GPs. 

Correct option (c) Both of these

Explanation:

The answer to this question can be seen from the options. Both 3, 6, 12 and 12, 6, 3 satisfy the required conditions— viz, GP with sum of first and third terms as 15.

Correct option (c) 135

Explanation:

From the facts given in the question it is self evident that the common ratio of the GP must be 3 (as the sum of the 2nd and 4th term is thrice the sum of the first and third term). 

a + ar² = 50 or a = 50/(1 + 9) = 5

Largest term = 5.(3)³ = 135.

Let the series be a, ar, ar² , ar³ ,…. 

According to the question a= 3ar/(1 − r) or r = 1/4.

The series would be 4, 4/4, 4/16,….. and so on. 

The product of first three terms of the series would be 4 x 1 x 1/4 =1.

Correct option (d) Both (a) and (b)

Explanation:

The key to this question is what you understand from the statement— ‘for two progressions out of P₁, P₂ and P₃ the average is itself a term of the original progression P.’ For option (a) which tells us that the Progression P has 20 terms, we can see that P₁ would have 7 terms, P₂ would have 7 terms and P₃ would have 6 terms. Since, both P₁ and P₂ have an odd number of terms we can see that for P₁ and P₂ their 4th terms (being the middle terms for an AP with 7 terms) would be equal to their average. Since, all the terms of P₁, P₂ and P₃ have been taken out of the original AP

P, we can see that for P₁ and P₂ their average itself would be a term of the original progression P. This would not occur for P₃ as P₃, has an even number of terms. Thus, 20 is a correct value for n. Similarly, if we go for n = 26 from the second option we get: 

P₁, P₂ and P₃ would have 9, 9 and 8 terms, respectively and the same condition would be met here too. For n = 36 from the third option, the three progressions would have 12 terms each and none of them would have an odd number of terms. Thus, option (d) is correct as both options (a) and (b) satisfy the conditions given in the problem

Correct option (b) 2.5

Explanation: 

Solve this question using options. The average of the sum of the first 5 terms of the AP can be used to get the value of the third term of the AP. If we try to use the options, in option 2, if the sum of the first 5 terms is 2.5, the third term must be 2.5/5 = 0.5. This means our AP is 2,1.25,0.5,−0.25,−1. The corresponding GP with the same 1^st and 3^rd terms is 2,1,0.5,0.25… The condition for the second term is also matched here.

Correct option (a) 16    

Explanation: 

Since the difference between the tenth and the sixth terms is −16, the common difference would be −4. Using a trial and error approach with the options, we can see that if we take the first term as 16, we will get the series 16,12,8,4,0,−4,−8,−12,−16,−20. We can see that both the conditions given in the question are met by this series. Hence, the first term would be 16. 

Correct option (c) 7881

Explanation:

The series would be given by: 13, 17, 21... which essentially means that all the numbers in the series are of the form 4n + 13 or 4k + 1( Where k = n + 3). Only the value in option (c) is a 4k + 1 number and is hence the correct answer. 

Correct option (d) – 66

Explanation: 

A factor search for factor pairs of 456 give us the following possibilities. 1,456; 2,228; 3,152; 4,114; 6,76; 8,57; 12,38 & 19,24. A check of the conditions given in the problem, tells us that if we take 12 as the 4^th term and 38 as the 5^th term, we would get the series till 9 terms as: −66,−40,−14,12,38,64, 90,116,142. In this series we can see that the division of the 9^th term by the 4^th term gives us a quotient of 11 and a remainder of 10. Hence, the required first term is −66.

Correct option (b) 7

Explanation:

Since the sum of 5 numbers in AP is 35, their average would be 7. The average of 5 terms in an AP is also equal to the value of the 3rd term (logic of the middle term of an AP). Hence, the third term’s value would be 7.

Correct option (c) (116P – 39)/90    

Explanation: 

(a + 4d)² + (a + 10d)² = 3 → a² + 14ad + 58d² = 1.5. 

Also, (a + d)(a + 13d) = P → a² + 14ad + 13d² = P. Further, we need to find the value of a² + 14ad (product of the first and fifteenth terms of the AP). From the above two equations, we get that 45d² =3/2-p →13d² =(39-26p)/90

Thus, a² + 14ad = P −(39-26p)/90 = (116P − 39)/90.

Correct option (d) 3200   

Explanation: 

The areas would consecutively get halved. So, the first area being 1600, the next one would be 800, then 400 and so on till infinite terms. Thus, the infinite sum would be 3200. 

Correct option (d) 2048    

Explanation: 

The first term of the given AP is 2 and the common difference is also 2. Thus, the 11^th term of the GP = 2 × 2¹⁰ = 2048.

\(f(x + y) = f(x) f(y)\)     (Given)

\(f( 0 + 0) = f(0) .f(0)\)    (By taking x = 0 = y)

⇒ \(f(0) = (f(0))^2\)

⇒ \((f(0))^2 - f(0) = 0\)

⇒ \(f(0)\, (f(0) -1) = 0\)

⇒ \(f(0) = 1\)      \((\because f(0)\ne 0)\)

Given that \(f(1) = 2\)

Then 

\(f(1 + 1) = f(1) f(1) = 2 \times 2 = 4\)

⇒ \(f(2) = 4\)

\(\therefore k = 2\)

Answer is (C) Domain = R – {4}, Range = {– 1}

The given real function is f (x) = |x – 1|. 

It is clear that |x – 1| is defined for all real numbers. 

∴ Domain of f = R Also, for x ∈ R, |x – 1| assumes all real numbers. 

Hence, the range of f is the set of all non-negative real numbers.  

The given real function is () = √( − 1) It can be seen that √( − 1) is defined for  ≥ 1.

 The given function is  Therefore, the domain of f is the set of all real numbers greater than or equal to 1 i.e.,

 the domain of f = [1, ∞). As x ≥ 1 ⇒ (x – 1) ≥ 0 ⇒ √( − 1) ≥ 0

Therefore, the range of f is the set of all real numbers greater than or equal to 0 i.e., 

the range of f =  [0, ∞).