1ya right
somebody paint me!!
COMPREHENSION
let f(x) and g(x) are two distinct functions such that f(x) is an odd function and g(x) is an even function for all X ε R . Let a function h(x)=f(x) +g(x) is an odd function and φ(x)=f(g(x))+g(f(x)).
NOW ANSWER THE FOLLOWING QUESTIONS (1-%):
1. Function g(x) is
(a) a trigonometrical function (b) a polynomial function
(c) an absolute function (d) a constant function
2. The number of solutions of f(x)= g(x) is
(a) 1 (b) 2
(c) 0 (d) 3
3. The behaviour of h(x) for all x ε R IS
(a) An increasing function (b)a decreasing function
(c) a constant function (d) nothing can be said
4. The number of solutions of φ(x) = h(x) is
(a) 2 (b) 1
(c) 0 (d) 3
5. For the functions h(x) and φ (x)
(a) h' (x) > φ'(x) (b) h' (x) < φ'(x)
(c) h'(x) ≤ φ'(x) (d) nothing can be said
MULTIPLE CHOICE
7. if f(x) is a polynomial of degree 5 with real coefficients defined for real x such that f(|x|) = 0 has 8 non-zero real roots then f(x) =0 has
(a) 4 positive roots (b) 1 negative root
(c) 5 real roots (d) at least two imaginary roots
11. Let : R→R be defined by f(x) = x2 +ax+1/x2+x+1. The set of exhaustive values of a such that f(x) is onto is
(a) (-∞,∞) (b) (-∞,0)
(c) (1,∞) (d) φ
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31 Answers
24A couple of conditions missed in Q1 to 5..thats why some problem..
ques will be::
let f(x) and g(x) are two distinct functions such that f(x) is an odd function and g(x) is an even function for all X ε R and f'(x)>g'(x) for all x ε R . Let a function h(x)=f(x) +g(x) is an odd function and φ(x)=f(g(x))+g(f(x)).
Now anyone trying ??
622, 4, 6, 8 or 10.. (sorry my mistake) (*Typing and mind not in sync)
4Nishant Bhaiya the qs says f(mod x) = 0 has 8 non zero real roots
but u posted that it has 2 or 6 or 10 roots???
4@Nishant : Qs and Ans is correct Bhaiya
f (x) is never onto so solution set is phi
1sorry
0 is not neccesarily a root
it maybe a root though
was this a part of comprehension??
then 0 will be a root definitely
in that equation will have repeated roots
1f(modx) has 5 real roots which is symmetric about Y axis,f(modx) is formed when we make graph of f(x) for x>0 and take it;s image for x<0
so f(x)=0 has 4 positive roots
the equation cannot have only 4 roots and imaguinary roots occur in pairs,so it has 5 roots,of which 4 are positive and one negative
hope u r satisfied
62It is given that f(x) is odd, while g(x) is even
also, f+g is odd.. hence3, we can conclude that g is both even and odd
g(x)=-g(-x)
and g(x)=+g(-x)
Thus, g(x) = 0 for all x.. hence g is a constant valued function which is equal to zero
Seeing the options and the questions, I think you have misseed soem information!
4@Akshay : We don't know whether the function is strictly increasing or not ! right!!?
1as f(x) is strictly increasing,implies will cut X axis max once,and definitely it will once as function is odd
1so wat if it is symmetric about origin,it dosen't mean it will have neccesarily one solution
it may not even have on,or even have many
4SORRY!!!!!!!!
THE INFO I MISSED IS THAT f ' (x) > g ' (x) for all x belongs to R
1but den u urself proved dat g(x) is a constant function...and dat is acordin to the question!!!
62NO archana..
look at f(x) = x
and g(x) = |x|
Both one is even one is odd.. but they have infiniite solutions for f(x)=g(x) ...
12. The number of solutions of f(x)= g(x) is
ans..A)1....f(x) an odd function is symmetric bout origin
6211. Let : R→R be defined by f(x) = (x2 +ax+1)/(x2+x+1). The set of exhaustive values of a such that f(x) is onto is
(a) (-∞,∞) (b) (-∞,0)
(c) (1,∞) (d) φ
You can simplify the given function as 1+ (a-1)xx2+x+1
Which has to be onto... so (a-1)xx2+x+1 is onto R.
hence, (a-1)x+1/x+1 is onto R
Which will never be onto R!!
Have I written your question correctly?