જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : False
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : True
જવાબ : 1
જવાબ : 1
જવાબ : P(41)=412−41+41=1681 (not prime)
જવાબ : P(3) =32−3+41=9−3+41=47 (prime)
જવાબ : P(2)=22−2+41 =4−2+41 =43 (prime)
જવાબ : P(1) =12−1+41=41 (prime)
જવાબ : 8
જવાબ : 7
જવાબ : 6
જવાબ : 5
જવાબ : 4
જવાબ : 4
જવાબ : 3
જવાબ : 3
જવાબ : 3
જવાબ : 4
જવાબ : Assume that P(n): -6^{n} < n
If n =1, -6^{1}>1. Hence P(1) is true If n =2, -6^{2} = 36 > 2. Hence P(2) isn’t true Therefore, P(n) isn’t true for every positive integer n is proved using the principle of mathematical induction.જવાબ : Assume that P(n): -11^{n} < n
If n =1, -11^{1}>1. Hence P(1) is true If n =2, -11^{2} = 121>2 . Hence P(2) isn’t true Therefore, P(n) isn’t true for every positive integer n is proved using the principle of mathematical induction.જવાબ : Assume that P(n): -√2^{n} < n
If n =1, -√2^{1}>1. Hence P(1) is true If n =2, -√2^{2} = 2. Hence P(2) isn’t true Therefore, P(n) isn’t true for every positive integer n is proved using the principle of mathematical induction.જવાબ : Assume that P(n): -2^{n} < n
If n =1, -2^{1}>1. Hence P(1) is true If n =2, -2^{2}>1 = 4 > 1. Hence P(2) isn’t true Therefore, P(n) isn’t true for every positive integer n is proved using the principle of mathematical induction.જવાબ : Assume that P(n): -23^{n} < n
If n =1, -23^{1}>1. Hence P(1) is true If n =2, -23^{2}>1 = 529 > 1. Hence P(2) isn’t true Therefore, P(n) isn’t true for every positive integer n is proved using the principle of mathematical induction.જવાબ : Assume that P(n): 23^{n} > n
If n =1, 23^{1}>1. Hence P(1) is true Let us assume that P(k) is true for any positive integer k, It means that, i.e., 23^{k} > k …(1) We shall now prove that P(k +1) is true whenever P(k) is true. Now, multiplying both sides of the equation (1) by 23, we get 23 x 23^{k} > 23k Now by using the property, i.e., 23^{k+1}> 23k = k + k > k + 1 Hence, P(k + 1) is true when P(k) is true. Therefore, P(n) is true for every positive integer n is proved using the principle of mathematical induction.જવાબ : Assume that P(n): 13^{n} > n
If n =1, 13^{1}>1. Hence P(1) is true Let us assume that P(k) is true for any positive integer k, It means that, i.e., 13^{k} > k …(1) We shall now prove that P(k +1) is true whenever P(k) is true. Now, multiplying both sides of the equation (1) by 13, we get 13 x 13^{k} > 13k Now by using the property, i.e., 13^{k+1}> 13k = k + k > k + 1 Hence, P(k + 1) is true when P(k) is true. Therefore, P(n) is true for every positive integer n is proved using the principle of mathematical induction.જવાબ : Assume that P(n): 2n > n
If n =1, 2>1. Hence P(1) is true Let us assume that P(k) is true for any positive integer k, It means that, i.e., 2k > k …(1) We shall now prove that P(k +1) is true whenever P(k) is true. Now, addinging both sides of the equation (1) by 1, we get 2k + 1> k+1 Hence, P(k + 1) is true when P(k) is true. Therefore, P(n) is true for every positive integer n is proved using the principle of mathematical induction.જવાબ : Assume that P(n): √3^{n} > 2n
If n =1, √3^{1}>2. Hence P(1) isn’t true Therefore, P(n) isn’t true for every positive integer n is proved using the principle of mathematical induction.જવાબ : Assume that P(n): 3^{n} > n
If n =1, 3^{1}>1. Hence P(1) is true Let us assume that P(k) is true for any positive integer k, It means that, i.e., 3^{k} > k …(1) We shall now prove that P(k +1) is true whenever P(k) is true. Now, multiplying both sides of the equation (1) by 3, we get 3 x 3^{k} > 3k Now by using the property, i.e., 3^{k+1}> 3k = k + k > k + 1 Hence, P(k + 1) is true when P(k) is true. Therefore, P(n) is true for every positive integer n is proved using the principle of mathematical induction.જવાબ : Given, 1 + 3 + 5 + … + (2n – 1) = n^{2}
Let, P(n) : 1 + 3 + 5 +…+ (2n – 1) = n^{2} , for n ∈ N P(1) is true, as P(1) : 1 = 1^{2} Let as assume P(k) is true for some k ∈ N, It means that, P(k) : 1 + 3 + 5 + … + (2k – 1) = k^{2} To prove that P(k + 1) is true, we have 1 + 3 + 5 + … + (2k – 1) + (2k + 1) = k^{2} + (2k + 1) = k^{2} + 2k + 1 = (k + 1)^{2} Hence, P(k + 1) is true, whenever P(k) is true. Therefore, P(n) is true for all n ∈ N is proved by the principle of Mathematical induction.જવાબ : P(n) be the given statement, that is
P(n) : 1 × 1! + 2 × 2! + 3 × 3! + … + n × n! = (n + 1)! – 1 for all natural numbers n. P (1) is true, as P (1) : 1 × 1! = 1 = 2 – 1 = 2! – 1. Let P(n) is true for natural number k, Hence, P(k) : 1 × 1! + 2 × 2! + 3 × 3! + … + k × k! = (k + 1)! – 1 to prove P (k + 1) is true, we have P (k + 1) : 1 × 1! + 2 × 2! + 3 × 3! + … + k × k! + (k + 1) × (k + 1)! = (k + 1)! – 1 + (k + 1)! × (k + 1) = (k + 1 + 1) (k + 1)! – 1 = (k + 2) (k + 1)! – 1 = ((k + 2)! – 1 Therefore, P (k + 1) is true, whenever P (k) is true. Hence, P(n) is true for all natural number n is proved using the Principle of Mathematical Inductionજવાબ : Assume that P(n): 5^{n} > n
If n =1, 5^{1}>1. Hence P(1) is true Let us assume that P(k) is true for any positive integer k, It means that, i.e., 5^{k} > k …(1) We shall now prove that P(k +1) is true whenever P(k) is true. Now, multiplying both sides of the equation (1) by 5, we get 5 x 5^{k} > 5k Now by using the property, i.e., 5^{k+1 }> 5k = k + k > k + 1 Hence, P(k + 1) is true when P(k) is true. Therefore, P(n) is true for every positive integer n is proved using the principle of mathematical induction.જવાબ : Given, 2 + 4 + 6 + … + (2n) = n(n+1)
Let, P(n) : 2 + 4 + 6 + … + (2n) = n(n+1) , for n ∈ N P(1) is true, as P(1) : 2 = 1(1+1) Let as assume P(k) is true for some k ∈ N, It means that, P(k) : 2 + 4 + 6 + … + (2k) = k(k+1) To prove that P(k + 1) is true, we have 2 + 4 + 6 + … + (2n) + 2(n + 1) = k(k+1) + 2(k + 1) = k^{2} + k + 2k + 2 = (k+1)(k+2) Hence, P(k + 1) is true, whenever P(k) is true. Therefore, P(n) is true for all n ∈ N is proved by the principle of Mathematical induction.જવાબ : Assume that P(n): 7^{n} > n
If n =1, 7^{1}>1. Hence P(1) is true Let us assume that P(k) is true for any positive integer k, It means that, i.e., 7^{k} > k …(1) We shall now prove that P(k +1) is true whenever P(k) is true. Now, multiplying both sides of the equation (1) by 7, we get 7 x 7^{k} > 7k Now by using the property, i.e., 7^{k+1}> 7k = k + k > k + 1 Hence, P(k + 1) is true when P(k) is true. Therefore, P(n) is true for every positive integer n is proved using the principle of mathematical induction.જવાબ : Assume that P(n): -1^{n} > n
If n =1, -1^{1}<1. Hence P(1) is False Therefore, P(n) isn’t true for every positive integer n is proved using the principle of mathematical induction.જવાબ : Assume that P(n): -1^{n} < n
If n =1, -1^{1}<1. Hence P(1) is true If n =2, -1^{2}=1. Hence P(2) isn’t true Therefore, P(n) isn’t true for every positive integer n is proved using the principle of mathematical induction.જવાબ : Given, 1 + 2 + 3 + … + n = n(n+1)/2
Let, P(n) : 1 + 2 + 3 + … + n = n(n+1)/2 , for n ∈ N P(1) is true, as P(1) : 1 = 1(1+1)/2 Let as assume P(k) is true for some k ∈ N, It means that, P(k) : k(k+1)/2 To prove that P(k + 1) is true, we have 1 + 2 + 3 + … + k + (k + 1) = k(k+1)/2 + (2k + 1) = (k^{2} + k + 4k + 2)/2 = (k + 1)(k + 2)/2 Hence, P(k + 1) is true, whenever P(k) is true. Therefore, P(n) is true for all n ∈ N is proved by the principle of Mathematical induction.જવાબ : Given, 1^{3} + 2^{3} + 3^{3} + … + n^{3} = [n(n+1)/2]^{2}
Let, P(n) : 1^{3} + 2^{3} + 3^{3} + … + n^{3} = [n(n+1)/2]^{2} , for n ∈ N P(1) is true, as P(1) : 1 = [1(1+1)/2]^{2} Let as assume P(k) is true for some k ∈ N, It means that, P(k) : [k(k+1)/2]^{2} To prove that P(k + 1) is true, we have 1^{3} + 2^{3} + 3^{3} + … + k^{3} + (k+1)^{3} = [k(k+1)/2]2 + (k+1)^{3} = [(k+1)(k+2)/2]^{2} Hence, P(k + 1) is true, whenever P(k) is true. Therefore, P(n) is true for all n ∈ N is proved by the principle of Mathematical induction.જવાબ : Assume that P(n): √3^{n} > n
If n =1, √3^{1}>1. Hence P(1) is true Let us assume that P(k) is true for any positive integer k, It means that, i.e., √3^{k} > k …(1) We shall now prove that P(k +1) is true whenever P(k) is true. Now, multiplying both sides of the equation (1) by √3, we get √3 x √3^{k} > √3k Now by using the property, i.e., √3^{k+1}> √3k = k + k > k + 1 Hence, P(k + 1) is true when P(k) is true. Therefore, P(n) is true for every positive integer n is proved using the principle of mathematical induction.
1 |
1.2 + 2.3 + 3.4 + ... + n.(n + 1) |
A |
n(4n^{2}−1)/3 |
2 |
½ + ¼ + 1/8 +...+ 1/2n |
B |
n(n+1)(n+2)/3 |
3 |
1^{2} + 3^{2} + 5^{2} + ... + (2n − 1)^{2} |
C |
1 – 1/2n |
જવાબ :
1-B,2-C,3-A
1 |
1.2 + 2.2^{2} + 3.2^{3} + ... + n.2^{n} |
A |
n(n+1)(2n+7)/6 |
2 |
2 + 5 + 8 + 11 + ... + (3n − 1) |
B |
½ n(3n+1) |
3 |
1.3 + 2.4 + 3.5 + ... + n. (n + 2) |
C |
(n − 1) 2^{n}^{+1}+2 |
જવાબ :
1-C, 2-B, 3-A
1 |
12.5+15.8+18.11+...+1(3n−1)(3n+2) |
A |
n/3n+1 |
2 |
1/1.4 + 1/4.7 + 1/7.10 +...+ 1/(3n−2)(3n+1) |
B |
n/3(2n+3) |
3 |
1/3.5 + 1/5.7 + 1/7.9 +...+ 1/(2n+1)(2n+3) |
C |
n/3(4n+3) |
4 |
1/3.7 + 1/7.11 + 1/11.5 +...+ 1/(4n−1)(4n+3) |
D |
n/(6n+4) |
જવાબ :
1-D, 2-A, 3-B, 4-C
1 |
7 + 77 + 777 + ... + 777 ..........._{n−digits}7 |
A |
x + y |
2 |
For all n ∈ N, x^{2n−1} + y^{2n−1} is divisible by |
B |
n^{2} |
3 |
For all n ∈ N, the sum of first n odd natural numbers is |
C |
7(10^{n+1}−9n−10)/81 |
જવાબ :
1-C, 2-A, 3-B
1 |
1^{2} + 3^{2} + 5^{2} + ... + (2n − 1)^{2} |
A |
2^{n}^{+1} – 1 |
2 |
2^{2} + 4^{2} + 6^{2} + …+ 2n^{2} |
B |
3 |
3 |
For all n ∈ N, n^{3} −- 7n + 3 is divisible by |
C |
[n(2n+1)(2n-1)]/3 |
4 |
For all n ∈ N, 1 + 2 + 2^{2} + ... + 2^{n} |
D |
n(4n^{2}−1)/3 |
જવાબ :
1-D, 2-C, 3-B, 4-A
1 |
For all n ∈ N, n(n + 1) (n + 5) is a multiple of |
A |
133 |
2 |
For all n ∈ N, 7^{2n} + 2^{3n−3}. 3^{n}^{−1} is divisible by |
B |
24 |
3 |
For all n ∈ N, 2.7^{n} + 3.5^{n} − 5 is divisible by |
C |
3 |
4 |
For all n ∈ N, 11^{n}^{+2} + 12^{2n+1} is divisible by |
D |
25 |
જવાબ :
1-C, 2-D, 3-B, 4-A
1 |
The sum of the series 1² + 2² + 3² + ………..n² is |
A |
n(n+1)/2 |
2 |
The sum of the series 1 + 2 + 3 + ………..n is |
B |
(n(n+1)/2)^{2} |
3 |
The sum of the series 1^{3} + 2^{3} + 3^{3} + ………..n^{3} is |
C |
n(n+1)(2n+1)/6 |
જવાબ :
1-C, 2-A, 3-B
1 |
For all n ∈ N, 5^{2n+2} −24n −25 is divisible by |
A |
2 |
2 |
For all n ∈ N, 3^{2n+2} −8n − 9 is divisible by |
B |
576 |
3 |
For all n ∈ N, n^{3} – n is divisible by |
C |
8 |
4 |
For all n ∈ N, n^{2} – n is divisible by |
D |
6 |
જવાબ :
1-B, 2-C, 3-D, 4-A
1 |
If P (n) is the statement "n^{2} − n + 41, then P(1) is |
A |
24 |
2 |
If P (n) is the statement "n^{2} − n + 41, then P(41) is |
B |
8 |
3 |
For all n ∈ N, 5^{2n} −1 is divisible by |
C |
Prime |
4 |
For all n ∈ N, 3^{2n}+7 is divisible by |
D |
Not Prime |
જવાબ :
1-C, 2-D, 3-A, 4-B
1 |
P(n) : 2n < n! |
A |
n ≥ 8 |
2 |
P(n) : 3n < n! |
B |
n ≥ 5 |
3 |
P(n) : 4^{n} < n^{4} |
C |
n ≥ 3 |
4 |
P(n) : 7^{n} < n^{7} |
D |
n ≥ 4 |
જવાબ :
1-D, 2-C, 3-B, 4-A
Math
The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.
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