π/2
1
-1
– π/2
D.
– π/2
A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched?
Interval | Function |
(-∞, ∞) | x^{3} – 3x^{2} + 3x + 3 |
Interval | Function |
[2, ∞) | 2x^{3} – 3x^{2} – 12x + 6 |
Interval | Function |
(-∞, 1/3] | 3x^{2} – 2x + 1 |
Interval | Function |
(- ∞, -4] | x^{3} + 6x^{2} + 6 |
C.
Interval | Function |
(-∞, 1/3] | 3x^{2} – 2x + 1 |
Clearly function f(x) = 3x2 – 2x + 1 is increasing when
f′(x) = 6x – 2 ≥ 0 ⇒ x∈[1/3, ∞)
A real valued function f(x) satisfies the functional equation f(x – y) = f(x) f(y) – f(a – x) f(a + y) where a is a given constant and f(0) = 1, f(2a – x) is equal to
–f(x)
f(x)
f(a) + f(a – x)
f(-x)
A.
–f(x)
f(a – (x – a)) = f(a) f(x – a) – f(0) f(x)
= - f(x) [ ∵ x = 0, y= 0, f(0) = f^{2} (0)-f^{2}(a) = 0 ⇒ f(a) = 0]
Consider the following relations:
R = {(x, y)| x, y are real numbers and x = wy for some rational number w}; S = {(m/p, p/q)| m, n, p and q are integers such that n, q ≠ 0 and qm = pn}. Then
R is an equivalence relation but S is not an equivalence relation
neither R nor S is an equivalence relation
S is an equivalence relation but R is not an equivalence relation
R and S both are equivalence relations
D.
R and S both are equivalence relations
For real x, let f(x) = x^{3}+ 5x + 1, then
f is one–one but not onto R
f is onto R but not one–one
f is one–one and onto R
f is neither one–one nor onto R
C.
f is one–one and onto R
f(x) = x^{3}+ 5x + 1
f'(x )3x^{2} +5>0
⇒ f is one–one
therefore, f is cubic
⇒ f is onto
‘f’ is one–one and onto.