functions TMH

Q1. Let the function f(x) = x2+x+sinx-cosx+log(1+|x|) be defined on the interval [0,1]. The function g(x) on [-1,1] satisfying g(-x) = -f(x) is

ans given: -x2+x+sinx-cosx+log(1+|x|) (wrong naa?)

Q2. If f:R-->R is defined by f(x) = x2+1, teh value of f-1(17) and f-1(3) are respectively

ans given: {-4,4}, φ (how can inverse even be defined????)

Q3. Domain of f(x) = log2(log0.5(x2+4x+3)) + sin-1[2x2-3] is [.] is GIF

Q4. Consider the following sets
R = \begin{Bmatrix} (x,y) |x,y\epsilon R,\; x^2+y^2\leq 25 \end{Bmatrix}
R' = \begin{Bmatrix} (x,y) |x,y\epsilon R,\; y\geq 4x^2/9 \end{Bmatrix}

Is R ∩ R' a function? If so, find its domain and range

11 Answers

106
Asish Mahapatra ·

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 ·

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 ·

Q5. R1 => sinx ≥ 0
This implies x → [0, pi]

R2 => 1 + 3sinx ≠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 ·

Q5. why not option B ? this also satisfies naa?

1
utd4ever ·

ans for the fifth one is a b and c ... sinx>0 is the only condition ..

29
govind ·

In 5th option B is also possible...

1
utd4ever ·

for the sixth question also the answer is a b and c same concept as the 5th question

1
utd4ever ·

is the answer for the third question (-52,-1]

1
Arka Halder ·

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 ·

Q3.

Final Domain : x → (-√5/2, -√2] U [√2, √5/2)

1
Arka Halder ·

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.

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