11th mathematics exercise questions with answers, Mathematics

Linear Inequalities – Class XI – Exercise 6.1 [1 – 10]

Solve 24x < 100, when (i) x is a natural number (ii) x is an integer Solution: We have the given inequality 24x < 100. then, 24x/24 < 100/24 Thus, x < 25/6 (i) We know that 1, 2, 3, and 4 are the only natural numbers less than 25/6. Thus, when x is a… Continue reading Linear Inequalities – Class XI – Exercise 6.1 [1 – 10]

11th mathematics exercise questions with answers, Mathematics

Principle of Mathematical Induction – Class XI – Exercise 4.1[Part 2]

11: Prove the following by using the principle of mathematical induction for all n€ N: 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + … +1/n(n+1)(n+2) = n(n+3)/4(n+1)(n+2) Solution: Let the given statement be P(n), i.e., P(n): 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + … +1/n(n+1)(n+2) = n(n+3)/4(n+1)(n+2) For n = 1, we have P(1): 1/1.2.3 = 1.(1+3)/4(1+1)(1+2) =… Continue reading Principle of Mathematical Induction – Class XI – Exercise 4.1[Part 2]

11th mathematics exercise questions with answers, Mathematics

Principle of Mathematical Induction – Class XI – Exercise 4.1[Part 1]

Prove the following by using the principle of mathematical induction for all n € N: 1 + 3 + 32+ ………..+3n-1 = (3n – 1)/2 Solution: Let the given statement be P(n) , i.e., P(n): 1 + 3 + 32+ ………..+3n-1 = (3n – 1)/2 For n = 1, we have P(1): (31 – 1)/2… Continue reading Principle of Mathematical Induction – Class XI – Exercise 4.1[Part 1]

11th mathematics exercise questions with answers, Mathematics

Trigonometry Functions – Class XI – Exercise 3.4

Find the principal and general solutions of the equation tanx = √3 Solution: tanx = √3 We know that, tan π/3 = √3 and tan4π/3 = tan(π + π/3) = tan π/3 = √3 therefore, the principal solutions are x = π/3 and 4π/3 Now, tanx = tan π/3 x =n π+ π/3, where n… Continue reading Trigonometry Functions – Class XI – Exercise 3.4

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Trigonometry Functions – Class XI – Exercise 3.3

sin2 π/6 + cos2 π/3 – tan2 π/4 = -1/2 Solution: L.H.S = sin2 π/6 + cos2 π/3 – tan2 π/4 = (1/2)2 + (1/2)2 – (1)2 = 1/4 + 1/4 – 1 = -1/2 = R.H.S Prove that 2sin2 π/6 + coses2 7π/6 cos2 π/3 = 3/2 Solution: L.H.S = 2sin2 π/6 + coses2… Continue reading Trigonometry Functions – Class XI – Exercise 3.3

11th mathematics exercise questions with answers, Mathematics

Trigonometry Functions – Class XI – Exercise 3.2

Find the values of other five trigonometric functions if cos x = -1/2, x lies in third quadrant. Solution: cos x = ‑1/2 sec x = 1/cosx = 1/(-1/2) = -2 We know, sin2x + cos2x = 1 ⇒sin2x = 1 - cos2x ⇒sin2x = 1 – (‑1/2)2 ⇒sin2x = 1 – 1/4 = 3/4… Continue reading Trigonometry Functions – Class XI – Exercise 3.2

11th mathematics exercise questions with answers, Mathematics

Trigonometry Functions – Class XI -Exercise 3.1

Find the radian measures corresponding to the following degree (i) 25˚ (ii) -47˚30’ (iii) 240˚ (iv) 520˚ Solution: (i) 25˚ We know that 180˚ = π radian 25˚ = π/180 x 25radian = 5π/36 radians (ii) -47˚30’ -47˚30’ = -471/2 = -95/2 degree -95/2 degree = π/180 x -95/2 radians = (-19/36x2) π radian =… Continue reading Trigonometry Functions – Class XI -Exercise 3.1