IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

manisha_bhandari

1. sin ax +cos ax and sin x +cos x are periodic of the same fundamental period, then find 'a'

2. the function f:

R

R defined by f(x)=(x-1)(x-2)(x-3) is

a)one-one but not onto b)onto but not one-one

c)both one one and onto d) neither one one nor onto

3. find domain of f(x)=sin^-1(log₂x²/2)

plzzzzzzzzzz answer these

funky

2) for substituting any values like 1 2 and 3
you get the same answer 0 hence it is many to one and not
one-one
The function can take all values of x hence onto

3) sin inv of x lies between minus (pie)/2 and (pie)/2
hence sin^-1(log (x^2/2)<= pie/2 or >= minus pie/2
taking first condition
log(x^2/2)<=1
x^2/2<=2
x^2<=4
modulus of x is<=2
[ -2,2]

taking second condition
log(x^2/2)>= -1
x^2/2>=1/2
x^2>=1
modulus of x is >=1
(negative infinity to -1]U[1 to infinity)

final answer is [-2,-1]U[1,2]

shine

since sin ax +cos ax and sin x +cos x are periodic of the same fundamental period
thus perid(1st fnc) = 2pi/a nd 2pi /a ie 2pi/a (on taking lcm)
nd period(2nd fnc)= 2pi nd 2pi ie 2pi
hence
2pi/a=2pi
=>a=1
hope this is right

Mr.IITIAN007

hey a is 1 .

him26.89

2) the func. is not one-one

it has same value at 3,2,1 i.e.0

the function is onto

as it ranges from [-

,

] b'cos it is a simple polynomial

&can attain any value

joyfrancis

q1)period of sinx+cosx=2pi {the fundamental period is 2pi only, fundamental means the least positive value for which the function repeats...like the sin function repeats after each 4pi,8pi....etc. But the LEAST positive value of the period is 2pi so f.p is 2pi}

period of sin(ax) + cos(ax) = 2pi/a

.:2pi/a=2pi

a=1

joyfrancis

q2) To check if the function is one one or not, put f(x1)=f(x2) if you get x1=x2 then it is one one else it is many one.

so,

(x1-1)(x1-2)(x1-3)=(x2-1)(x2-2)(x2-3)

put x1=1

we get x2=1,2,3

so many one.

range of f(x) is R so it is onto.

joyfrancis

q3)let log₂x²/2 = a

log of any number b can only be determined when b is positive

so x^2/2>0

-root2>x>root2.....(1)

Now, y = sin^-1a

a can lie between [-1,1]

.: log₂x²/2

1........(2)

and log₂x²/2

-1.....(3)

from (2)

x

[-2,2]

from 3

x is less that -1 and greater than 1

so taking intersection of the 3 solutions we get the ans as [-2,-1]U[1,2]

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