106
Asish Mahapatra
·2010-04-05 04:51:21
adding more questions...
MULTI ANSWERS (ALL)
Q5. Let f(x) = \frac{\sqrt{sinx}}{1+\sqrt[3]{sinx}}. If D is the domain of f, then D contains
(a) (0,pi) (b) (-2pi,-pi) (c) (2pi,3pi) (d) (4pi, 6pi)
ans given: ac
Q6. Let f(x) = sin(\frac{\pi}{x}) and D+ = {x|f(x) > 0}. Then D+ contains
(a) (1/3, 1/2) (b) (1/5, 1/4) (c) (-1, -1/2) (d) (-pi -1/2)
ans given: abc
Q7. Let f(x) = x2, g(x) = √x, tehn
(a) gof(-2) = 2
(b) gof(4) = 4
(c) gof(3) = 6
(d) gof(2) = 4
ans given: b
Q8. SOLVED (VERIFIED BY WOLFRAM ANS: 1/3)
\lim_{x\rightarrow 0}\frac{tanx-x}{x^2tanx}
ans shud be 1/3 naa?
1
utd4ever
·2010-04-05 05:07:54
ans for the 8th question must be 1/3 .. and in the sevenh question g(x) is not mentioned if we take the second function as g(x) then ul get gof=x so answer is a and b .. since f(x) is an even function f(2)=f(-2)=4 ...
39
Pritish Chakraborty
·2010-04-05 05:14:20
Q5. R1 => sinx ≥ 0
This implies x → [0, pi]
R2 => 1 + 3√sinx ≠0
This implies sinx ≠3√-1
Assuming real domain, sinx ≠-1
So x ≠(2n + 1)pi/2, n = 1,2,3..
So [0, pi] and [2pi, 3pi] possible.
106
Asish Mahapatra
·2010-04-05 05:15:41
Q5. why not option B ? this also satisfies naa?
1
utd4ever
·2010-04-05 05:17:16
ans for the fifth one is a b and c ... sinx>0 is the only condition ..
29
govind
·2010-04-05 05:19:02
In 5th option B is also possible...
1
utd4ever
·2010-04-05 05:25:23
for the sixth question also the answer is a b and c same concept as the 5th question
1
utd4ever
·2010-04-05 05:39:51
is the answer for the third question (-52,-1]
1
Arka Halder
·2010-04-05 05:41:56
7.gof=√(x2)=|x|
from the given options,a and b are thus correct.
dont know why only b has been given.
39
Pritish Chakraborty
·2010-04-05 05:44:05
Q3.
Final Domain : x → (-√5/2, -√2] U [√2, √5/2)
1
Arka Halder
·2010-04-05 05:52:08
for q 2,the function
i)is not onto as it takes values in [1,∞)
ii)is not one one as any value a and -a have the same y(a2+1)
the function clearly is not invertible.