32. Let f = {(1,1), (2,3), (0,–1), (–1, –3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b. Determine a, b.
32. Let f = {(1,1), (2,3), (0,–1), (–1, –3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b. Determine a, b.
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1 Answer
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32. Given, f(x) = (ax + b)
= {(1,1),(2,3),(0, – 1),(–1, –3)} .
As (1,1) f.
Then, f(1)=1 [? f(x) = y for (x, y)]
a × 1+b=1
a+b=1…… (1)
and (0, – 1) f .
Then, f(0)= –1
a× 0+b= –1
b= –1…….(2)
Putting value of (2) in (1) we gets
a – 1=1
a=1+1
a=2
So, (a, b)=(2, –1)
Similar Questions for you
Total number of possible relation =
Favourable relations =
Probability =
Circle S? : x² + y² - 10x - 10y + 41 = 0.
Center C? = (5, 5). Radius r? = √ (5² + 5² - 41) = √ (25 + 25 - 41) = √9 = 3.
Circle S? : x² + y² - 16x - 10y + 80 = 0.
Center C? = (8, 5). Radius r? = √ (8² + 5² - 80) = √ (64 + 25 - 80) = √9 = 3.
The solution checks if the center of one circle lies on the other.
Put C? (8, 5) into S? : 8² + 5² - 10 (8) - 10 (5) + 41 = 64 + 25 - 80 - 50 + 41 = 130 - 130 = 0. So C? lies on S?
Put C? (5, 5) into S? : 5² + 5² - 16 (5) - 10 (5) + 80 = 25 + 25 - 80 - 50 + 80 = 130 - 130 = 0. So C? lies on S?
This means bo
Kindly consider then following figure
36. Given, A={9,10,11,12,13}.
f(x)=the highest prime factor of n.
and f: A → N.
Then, f(9)=3 [? prime factor of 9=3]
f (10)=5 [? prime factor of 10=2,5]
f(11)=11 [? prime factor of 11 = 11]
f(12)=3 [? prime factor of 12 = 2, 3]
f(13)=13 [? prime factor of 13 = 13]
?Range of f=set of all image of f(x) = {3,5,11,13}.
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