maths

[s]

hey please let me know what answers you get for the following..

1.the set of all x satisfying |x^2-4|>4x
a. x<2root2-2 or x>2root2+2
b. x>2root2+2
c. x<-2root2+2 or x>2root2+2
d. none

2.the number of points at which the function is not differentiable
    f(x)= { min{|x|, x^2} if -infinity<x<1 , min{2x-1, x^2 otherwise }
               
a. 0  
b.1
c.2
d.>2

3. let f(x) = { x^3-x+3 for 0<x<=1, 2x+1 for 1<x<=2, x^2+1 for 2<x<3
then
a. f is differentiable at x=1 and at x=2
b. f is differentiable at x=1 not at x=2
c. f is differentiable at x=2 not at x=1
d. f is differentiable neither at x=1 nor x=2


4. if f(x-y)=f(x)/f(y), f'(0)=p and f'(5)=q then f'(-5) is
a. p^2/q
b. q/p
c. p/q
d. q

[ritu]

optipn d for 3rd

[ritu]

Optipn c for 2nd....nt sure though

[ritu]

Optipn c for 2nd....nt sure though

[ritu]

For 2nd questn its nt differentiable at only one point..sry

[s]

thanks for replying,
i got option d for 3rd but the corrct answer is b.
as for question 2, ur answer is correct however i got not differentiable at 2 points ie x=1 LHD=infinity, RHD=2 and at x=-1 LHD= -1, RHD= -2....please tell me where i went wrong..

[ritu]

At  x=-1 it has a kink but to the left of x=1 u have x^2 applying for min value
 nd LHD is 2...to the right side u have 2x-1 applying for a min nd u have RHD as  2 again...so its differentiable
note that at x=1 line y=2x-1 is tangent to y = x^2 .so it doesnt result in a kink nd wot u get is a smooth differentiable line...i hope u got it:-)

[ritu]

Nd ya option b for 3 is ok..i made a calculation mistake...its nt differentiable at x= 2 coz there LHD=2 but RHD=4...at x=1 both limits are 2....

[Monika]

the ans for the 1st is coming to b
-2-2^2< x< -2+2^2 0r x > 2+2^2
so it will b none of these (D)

[s]

o dear i have made stupid calculation mistakes..thanks a lot ritu i got it :)

[deepak]

for the final question, the answer is Psquared by Q

[s]

deepak and monika unfortunately the correct answers for 1 is A and 4 is D  

[deepak]

s, I'm not sure if q is the answer. great if you could explain how you arrive at the result? thanks!

[s]

deepak the answer provided is q, i have not been able to arrive at the same however.

[s]

1.Given x  y  z, and x + y + z = 9, the maximum value of x + 3y +
5z is
(A) 27,
(B) 42,
(C) 21,
(D) 18.

2.If a2 + b2 + c2 = 1, then ab + bc + ca is,
(A) -0.75,
(B) Belongs to the interval [-1, -0.5],
(C) Belongs to the interval [0.5, 1],
(D) None of the above.

3.What is the maximum value of a(1 - a)b(1 - b)c(1 - c), where a, b, c
vary over all positive fractional values?
A. 1
B . 1/8
C. 1/27
D. 1/64

[seema]

1) ans is 27

[duck]

Hi.. :)

Q) if f(x-y)=f(x)/f(y), f'(0)=p and f'(5)=q then f'(-5)
The correct answer should be p/q
Consider f(x) = e^x
it satisfies f(x-y) = f(x)/f(y)
Now, f'(x) = e^x
=> f'(0) = p =0
and f'(5) = q = e^5
and f'(-5) = e^-5 = 1/e^5 = p/q

Q) If a2 + b2 + c2 = 1, then ab + bc + ca = 0
Consider a=b=0 and c=1

Q)What is the maximum value of a(1 - a)b(1 - b)c(1 - c), where a, b, c
vary over all positive fractional values?
Ans) 1/8
Think on these lines>
If you want to maximise the value of (a)(1-a) then, it will happen at a=1/2
So, it must be the case that a=b=c = 1/2
=> Maximum value = 1/8

[ritu]

Hey duck in that case shouldnt the answer be 1/64?????

[duck]

Sorry.. typo..
1/64.. :)