Page 7 - mathsvol1ch1to3ans

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Hence atleast two elements in A have one image. Then for remaining two elements we cannot
have three images.
(c) one–to–one but not onto. This is also not possible and similar explanation as above could be
given.
(d) one–to–one and onto. f = {(1, a), (2, b), (3, c), (4, d)}
1
6. Find the domain of .
1 − 2sinx
Solution:
1
The function is defined for all x ∈ R except for 1 − 2sinx = 0. That is, except sinx = . That is,
π n π o 2
except x = 2nπ ± , n ∈ Z. Hence the domain is R − C 2nπ ± , n ∈ Z.
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6 6
√
4 − x 2
7. Find the largest possible domain of the real valued function f(x) = √ .
2
x − 9
Solution:
2
2
If x ≥ −3 or x ≤ 3, then x will be less than 9 and hence x − 9 will become negative or zero.
If it is negative then it has no square root in R. If it is zero, f is not defined. So x must lie outside
the interval [−3, 3].That is, x must lie on (−∞, −3) ∪ (3, ∞).
2
2
Also if x ≤ −2 and x ≥ 2, then 4 − x will become negative . Then, 4 − x has no square root in
R. So x must lie on the interval [−2, 2].
Combining these two conditions, the largest possible domain forf is
[−2, 2] ∩ ((−∞, −3) ∪ (3, ∞))

That is, ∅.
1
8. Find the range of the function .
2cosx − 1
Solution:
Clearly,
−1 ≤ cos x ≤ 1
⇒ 2 ≥ 2 cos x ≥ −2
⇒ −2 ≤ 2 cos x ≤ 2
⇒ −3 ≤ 2 cos x − 1 ≤ 1
1 1 1
By taking reciprocals, we get ≤ − and ≥ 1. Hence the range of f is
2 cos x − 1 3 2 cos x − 1

1
−∞, − ∪ [1, ∞).
3
9. Show that the relation xy = −2 is a function for a suitable domain. Find the domain and the range
of the function.
Solution:
−2
y =
x
Hence the domain is R − {0}
The range is also R − {0}
10. If f, g : R → R are defined by f(x) = |x| + x and g(x) = |x| − x, find g ◦ f and f ◦ g.
Solution:
When x > 0, we have f(x) = x + x = 2x and g(x) = x − x = 0. Since g(x) is a constant
function, (g ◦ f)(x) = 0 when x > 0.
When x < 0, we have f(x) = x − x = 0 which is a constant function. Here g(constant) = 0. In
both cases (g ◦ f)(x) = 0
When x > 0, we have g(x) = x − x = 0 and f(x) = x + x = 2x. Since g(x) is a constant
function, (f ◦ g)(x) = 0 when x > 0.