Q.1. Which of the following defines a oneone function?
(a)
(b)
(c)
(d)
Correct Answer is options (a, c, d)
If f(x) = e^{x} then f(x_{1}) = f(x_{2})
⇒
⇒
Thus e^{x} is oneone. If f(x) = cos x then, f(x_{1}) = f(x_{2})
⇒ cos x_{1}  cos x_{2}= 0
⇒ x_{1} = x_{2}
Thus cos x is oneone. If f(x) = ax + b then f(x_{1}) = f(x_{2})
⇒ ax_{1} + b = ax_{2} + b
⇒ a(x_{1} – x_{2}) = 0
⇒ x_{1} = x_{2}, as a ≠ 0 .
Thus ax + b is oneone.
Hence f is not 11.
Q.2. Which of the following functions f : R → R are bijective
(a) f ( x) = x sin x
(b) f ( x) = x sin^{2} x
(c) f ( x) = x+ √x^{2}
(d) f ( x) =  x + cos^{2} x
Correct Answer is options (b, d)
(A) f is not oneone as f cuts x axis twice in (0,2π] If fact f is continuous and achieves every real number infinite times.
(B) f is monotone as f '( x) = 1  sin 2 x ≥ 0 ∀ x
and f is continuous
Hence f is bijective.
(C) is not oneone as f is a constant function for x ≤ 0
(D) f is monotone as f '(x) ≤ 0 ∀x and as in (B) f is bijective.
Q.3. Let the function f : R → ( 0,π) be defined by
Then
(a) f is odd
(b) f is even
(c) range of f is
(d) range of f is
Correct Answer is options (a, c)
Thus f is odd.
Further 2^{x} ranges from 0 to ∞. The range of cot^{1}(2^{x}) is .
Hence range of f is
Q.4. Let S = {1, 2, 3, .....,n} and f n be the no. of those subsets of S which do not contain consecutive elements of S, then
(a)
(b)
(c)
(d) f_{4} = 8
Correct Answer is options (c, d)
Let n = 4 , then the subsets of {1, 2, 3, 4} which do not contain consecutive elements of this set and
Ф ,{1},{2},{3},{4},{1, 3},{1, 4},{2, 4}
f_{4} = 8 similarly f = 3,f_{3} = 5
Q.5. Let f : R  {0,1} → R satisfying then
(a)
(b) the graph of y = f ( x ) crosses xaxis at x = 1
(c) f ( 2) +f (3)= 5
(d) f ( 2) +f (3)= 6
Correct Answer is options (a, b, c)
Replacing x by 1/1x and by (1  1/x) in equation (1) one by one and on solving, we get
Q.6. If a function satisfies (x  y)f (x + y)(x + y)f (x  y) = 2(x^{2}y  y^{3})∀ x, y ∈ R andf (1) = 2 , then
(a) f(x) must be polynomial function
(b) f(3) = 12
(c) f(0) = 0
(d) f(x) may not be differentiable
Correct Answer is options (a, b, c)
(xy)f (x+ y)(x + y)f (x y) = 2y(x y(x + y))
Let x  t = u; x + y = v
uf ( v )  vf (u ) = 2uv ( v  u )
Let
f (1) = 2
λ+1 = 2
⇒ λ = 1
Q.7. If
then which of the following statement(s) is/are true?
(a)
(b) f ( 2) + f (1)= 0
(c)
(d)
Correct Answer is options (a, b, c)
Replace x by 2,
Replace x by 1, f (1) = 1      (2)
Replace x by
Solve (1) & (3)
⇒
Q.8. If f(x) = sin{[x + 5] + {x – {x –{x}}} for x ∈ , is invertible, where {.} and [.] represent fractional part and greatest integer functions respectively, then f^{–1}(x) is
(a) sin^{–1}x
(b)
(c) sin^{–1}{x}
(d) cos^{–1}{x}
Correct Answer is options (a, b, c)
Θ {[x + 5] + {x – {x – {x}}}}
= {[x + 5] + {x – {[x]}}
= {[x + 5] + {x}} = {{x}} = {x}
So, f(x) = sin{x} for x ∈
so f(x) = sinx
so f^{–1}(x) = sin^{–1}(x)
= sin^{–1}{x}
=
Q.9. The integer 'n' for which n^{2} + 19n + 92 is a perfect square
(a) –8
(b) 8
(c) –11
(d) 11
Correct Answer is options (a, c)
n^{2} + 19n + 92 = x^{2}
⇒ (n^{2} + 19n + 92) = 4x^{2}
⇒ (2n + 19)2 + 7 = 4x^{2}
⇒ 4x^{2} – 7 must be square of integer
Let 4x^{2} – 7 = k^{2}
So 4x^{2} – k^{2} = 7
⇒ (2x – k)(2x + k) = 7
from above ⇒ x = 2
(2n + 19)^{2} + 7 = 16
⇒ (2n + 19)^{2} = 9
⇒ n = –8 or –11
Q.10. Let A = R – {3}, B = R –{1}. Let f : A → B be defined by f(x) = x  2 / x  3 . Then
(a) f is bijective
(b) f is oneone
(c) f is onto
(d) onetoone but not onto
Correct Answer is options (a, b, c)
Let x, y ∈ A be such that f (x) = f(y)
⇔
⇔ xy –3x – 2y + 6 = xy –3y –2x + 6
⇔ –2y + 3y = 3x –2x Û x = y
Hence f is oneone
Let y ∈ B, we want to get x ∈ A such that f(x) = y, i.e., x  2 / x  3 = y. Cross multiplying and Solving, we obtain .
Hence f is onto and also bijective.
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