{"id":40594,"date":"2024-02-17T07:42:36","date_gmt":"2024-02-17T02:12:36","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=40594"},"modified":"2024-02-17T16:04:22","modified_gmt":"2024-02-17T10:34:22","slug":"plus-one-maths-chapter-wise-questions-answers-chapter-2","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/plus-one-maths-chapter-wise-questions-answers-chapter-2\/","title":{"rendered":"Plus One Maths Chapter Wise Questions and Answers Chapter 2 Relations and Functions"},"content":{"rendered":"
Question 1.
\nLet A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by R = {(a, b): a, b \u2208 A, b is exactly divisible by a}<\/p>\n
Answer:<\/p>\n
Question 2.
\nDetermine the domain and range of the relation R defined by R = {(x, x + 5) : x \u2208 {0, 1, 2, 3, 4, 5}}
\nAnswer:
\nR = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9),(5, 10)}
\nDomain of R = {0, 1, 2, 3, 4, 5}
\nRange of R = {5, 6, 7, 8, 9, 10}<\/p>\n
Question 3.
\nA function f is defined as f(x) = 2x – 5, Write down the values of f(0), f(7), f(-3).
\nAnswer:
\nGiven; f(x) = 2x – 5
\nf(0) = -5;
\nf(7) = 2(7) – 5 = 14 – 5 = 9
\nf(-3) = 2(-3) – 5 = -6 – 5 = -11<\/p>\n
Question 4.
\nFind the range of the following functions.<\/p>\n
Answer:<\/p>\n
Question 1.
\nLet A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that<\/p>\n
Answer:
\n1. A \u00d7 (B \u2229 C) ={1, 2} \u00d7 \u03a6 = \u03a6
\nA \u00d7 B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
\nA \u00d7 C = {(1, 5), (1, 6), (2, 5), (2, 6)}
\n(A \u00d7 B) \u2229 (A \u00d7 C) = \u03a6
\nHence; A \u00d7 (B \u2229 C) = (A \u00d7 B) \u2229 (A \u00d7 C)<\/p>\n
2. A \u00d7 C = {(1, 5), (1, 6), (2, 5), (2, 6)}
\nB \u00d7 D = {1, 2, 3, 4} \u00d7 {5, 6, 7, 8}
\n= {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3,6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}
\nHence A \u00d7 C is a subset of B \u00d7 D.<\/p>\n
Question 2. Question 3. Answer: 2. Given; f(x) = \\(\\frac{x^{2}+3 x+5}{x^{2}-5 x+4}\\) Question 4. Question 5. Question 6. Answer: 2. Any Subset of A \u00d7 B (say R={(1, 3),(2, 4)})<\/p>\n 3. Question 1. Answer: 2. (A \u00d7 B) \u2229 (A \u00d7 C) 3. A \u00d7 (B \u222a C) = {1, 2, 3} \u00d7 {3, 4, 5, 6} 4. (A \u00d7 B) \u222a (A \u00d7 C) Question 2. ii) Given; f(x) = \\(\\sqrt{9-x^{2}}\\) iii) Given; f(x) = |x – 1| iv) Given; f(x) = \\(\\sqrt{x-1}\\) Question 1. Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. Answer: 2. f(x) = x2<\/sup> <\/p>\n 3. f(x) = x3<\/sup> <\/p>\n 4. f(x) = \\(\\frac{1}{x}\\) <\/p>\n 5. f(x) = (x – 1)2<\/sup> <\/p>\n 6. f(x) = 3x2<\/sup> – 1 <\/p>\n 7. f(x) = |x| – 2 <\/p>\n Question 9. Answer: 2. Range of R = {3, -3, 5}<\/p>\n Question 10. Answer:<\/p>\n Question 11. Answer: 2. Find the value for which denominator is zero. Question 12. Question 13. Answer:<\/p>\n Question 14.
\nThe arrow diagram given below shows a relation R from P to Q. Write the relation in roster form, set-builder form. Find its domain and range.
\n
\nAnswer:
\nR – {(9, -3), (9, 3), (4, -2), (4, 2), (25, -5), (25, 5)}
\nR = {{x, y) : y2<\/sup> = x}
\nDomain of R = {9, 4, 25}
\nRange of R = {5, 3, 2, -2, -3, -5}<\/p>\n
\nFind the domain of the following.<\/p>\n\n
\n1. Given; f(x) = \\(\\frac{x^{2}+2 x+1}{x^{2}-8 x+12}\\)
\nThe function is not defined at points where the denominator becomes zero.
\nx2<\/sup> – 8x +12 = 0 \u21d2 (x – 6)(x – 2) = 0 \u21d2 x = 2, 6
\nTherefore domain of fis R – {2, 6}.<\/p>\n
\nThe function is not defined at points where the denominator becomes zero.
\nx2<\/sup> – 5x + 4 = 0 \u21d2 (x – 4)(x -1) = 0 \u21d2 x = 1, 4
\nTherefore domain of f is R – {1, 4}.<\/p>\n
\nLet f(x) = \\(=\\sqrt{x}\\) and g(x) = x be two functions defined over the set of nonnegative real numbers. Find (f + g)(x), (f – g)(x), (fg)(x) and \\(\\left(\\frac{f}{g}\\right)(x)\\).
\nAnswer:
\n(f + g)(x) = f(x) + g(x) = \\(=\\sqrt{x}\\) + x
\n(f – g)(x) = f(x) – g(x) = \\(=\\sqrt{x}\\) – x
\n(fg)(x) = f(x) \u00d7 g(x) = \\(=\\sqrt{x}\\) \u00d7 x = \\(x^{\\frac{3}{2}}\\)
\n<\/p>\n
\nLet f(x) = x2<\/sup> and g(x) = 2x + 1 be two functions defined over the set of nonnegative real numbers. Find (f + g)(x), (f – g)(x), (fg)(x) and \\(\\left(\\frac{f}{g}\\right)(x)\\).
\nAnswer:
\n(f + g)(x) = f(x) + g(x) = x2<\/sup> + 2x + 1
\n(f – g)(x) = f(x) – g(x) = x2<\/sup> – 2x – 1
\nf(fg)(x) = f(x) \u00d7 g(x)
\n= x2<\/sup>(2x +1) = 2x3<\/sup> + x2<\/sup>
\n<\/p>\n
\nA = {1, 2}, B = {3, 4}<\/p>\n\n
\n1. A \u00d7 B = {(1, 3), (1, 4), (2, 3), (2, 4)}<\/p>\n
\n<\/p>\nPlus One Maths Relations and Functions Six Mark Questions and Answers<\/h3>\n
\nLet A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}. Find<\/p>\n\n
\n1. A \u00d7 (B \u2229 C) = {1, 2, 3} \u00d7 {4}
\n= {(1, 4), (2, 4), (3, 4)}<\/p>\n
\n= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} \u2229 {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5) , (2, 6), (3, 4), (3, 5), (3, 6)}
\n= {(1, 4), (2, 4), (3, 4)}<\/p>\n
\n= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}<\/p>\n
\n= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} \u222a {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
\n= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}<\/p>\n
\nFind the domain and range of the following
\n
\nAnswer:
\ni) Given; f(x) = -|x|
\nD(f) = R, R(f) = (-\u221e, 0]<\/p>\n
\nx can take values where 9 – x2<\/sup> > 0
\n\u21d2 x2<\/sup> \u2264 9 \u21d2 -3 \u2264 x \u2264 3 \u21d2 x \u2208 [-3, 3]
\nTherefore domain of f is [-3, 3]
\nPut \\(\\sqrt{9-x^{2}}\\) = y, where y \u2265 0
\n\u21d2 9 – x2<\/sup> = y2<\/sup>\u21d2 x2<\/sup> = 9 – y2<\/sup>
\n\u21d2 x = \\(\\sqrt{9-x^{2}}\\)
\n\u21d2 9 – y2<\/sup> \u2265 0 \u21d2 y2<\/sup> \u2264 9 \u21d2 -3 \u2264 y \u2264 3
\nTherefore range of fis [0, 3].<\/p>\n
\nDomain of f is R
\nThe range of |x| is [0, \u221e) , then the range of
\nf(x) = |x -1| is [0, \u221e)<\/p>\n
\nx can take values where x – 1 \u2265 0
\n\u21d2 x \u2265 1 \u21d2 x \u2208 [1, \u221e]
\nTherefore domain of fis [1, \u221e]
\nThe range of \\(\\sqrt{x}\\) is [0, \u221e), then the range of
\nf(x) = \\(\\sqrt{x-1}\\) is [0, \u221e).<\/p>\nPlus One Maths Relations and Functions Practice Problems Questions and Answers<\/h3>\n
\nIf (x + 1, y – 2) = (3, 1), find the values of x and y.
\nAnswer:
\n(x + 1, y – 2) = (3, 1) \u21d2 x + 1 = 3, y – 2 = 1 \u21d2 x = 2, y = 3.<\/p>\n
\nIf \\(\\left(\\frac{x}{3}+1, y-\\frac{2}{3}\\right)=\\left(\\frac{5}{3}, \\frac{1}{3}\\right)\\), find the values of x and y.
\nAnswer:
\n<\/p>\n
\nIf G = {7, 8}; H = {2, 4, 5}, find G \u00d7 H and H \u00d7 G.
\nAnswer:<\/p>\n\n
\nif A = {-1, 1} find A \u00d7 A \u00d7 A
\nAnswer:
\nA \u00d7 A ={-1, 1} \u00d7 {-1, 1}
\n= {(-1, -1), (-1, 1), (1, -1), (1, -1)}
\nA \u00d7 A \u00d7 A
\n= {(-1, -1), (-1, -1), (1,-1), (1, -1)} \u00d7 {-1, 1}
\n= {(-1, -1, -1), (-1, 1, -1), (1, -1, -1), (-1, 1, -1), (-1, -1, 1), (-1, 1, 1), (1, -1, 1), (-1, 1, 1)}.<\/p>\n
\nWrite the relation R = {(x, x3<\/sup>): x is a prime number less than 10} in roster form.
\nAnswer:
\n2, 3, 5, 7 are the prime number less than 10.
\nR = {(2, 8),(3, 27),(5, 125),(7, 343)}<\/p>\n
\nIf f(x) = x2<\/sup>, find \\(\\frac{f(1.1)-f(1)}{(1.1-1)}\\)?
\nAnswer:
\n<\/p>\n
\nLet \\(\\left\\{\\left(x, \\frac{x^{2}}{1+x^{2}}\\right), x \\in R\\right\\}\\) be a real function from R to R. Determine the domain and range of f.
\nAnswer:
\nDomain of f is R.
\nLet \\(\\frac{x^{2}}{1+x^{2}}\\) = y \u21d2 x2<\/sup> = y(1 + x2<\/sup>)
\n\u21d2 x2<\/sup> = y + yx2<\/sup> \u21d2 x2<\/sup> – yx2<\/sup> = y
\n\u21d2 x2<\/sup>(1 – y) = y
\n
\n\u21d2 y \u2265 0, 1 – y > 0
\n\u21d2 y \u2265 0, y < 1 \u21d2 0 \u2264 y \u2264 1
\nTherefore range of f is [0, 1).<\/p>\n
\nGraph the following real functions. (each carries 2 scores)<\/p>\n\n
\n1. f(x) = |x – 2| = \\(\\left\\{\\begin{aligned}x-2, & x \\geq 2 \\\\-x+2, & x<2 \\end{aligned}\\right.\\)
\n
\n<\/p>\n
\n<\/p>\n
\n<\/p>\n
\n<\/p>\n
\n<\/p>\n
\n<\/p>\n
\n<\/p>\n
\nConsider the relation, R = {(x, 2x – 1)\/x \u2208 A) where A = (2, -1, 3}<\/p>\n\n
\n1. x = 2 \u21d2 2x – 1 = 2(2) – 1 = 3
\nx = -1 \u21d2 2x – 1 = 2(-1) – 1 = -3
\nx = 3 \u21d2 2x – 1 = 2(3) – 1 = 5
\nR = {(2, 3), (-1, -3), (3, 5)}<\/p>\n
\nLet A = {1, 2, 3, 4, 6} and R be a relation on A defined by R = {(a, b): a, b \u2208 A, b is exactly divisible by a}<\/p>\n\n
\n
\nConsider the real function<\/p>\n\n
\n1. Given; f(x) = 1 \u21d2 1 = \\(\\frac{x^{2}+2 x+3}{x^{2}-8 x+12}\\)
\n\u21d2 x2<\/sup> – 8x + 12 = x2<\/sup> + 2x + 3
\n\u21d2 10x = 9 \u21d2 x = \\(\\frac{9}{10}\\)<\/p>\n
\n\u21d2 x2<\/sup>\u00a0– 8x + 12 = 0
\n\u21d2 (x – 6)(x – 2) = 0 \u21d2 x = 6, 2
\nTherefore domain of f is R – {2, 6).<\/p>\n
\nIf f(x) = x3<\/sup> + 5x and g(x) = 2x +1, find (f + g)(2) and {fg)(1).
\nAnswer:
\n(f + g)(2) = f(2) + g(2) = (2)3<\/sup> + 5(2) + 2(2) + 1
\n= 8 + 10 + 4 + 1 = 23
\n(fg)(1) = f(1)g(1) = (1 + 5)(2 + 1) = 6 \u00d7 3 = 18.<\/p>\n
\nLet A = {1, 2, 3, 4, 5} and R be a relation on A defined by R = {(a, b):b = a2<\/sup>}<\/p>\n\n
\n
\nDraw the graph of the function
\nf(x) – |x| + 1, x \u2208 R
\nAnswer:
\n<\/p>\n