1
Che
·2010-02-14 00:45:43
1) c ?
since orthogonal
d(f(x))dxat P*d(g(x))dxat P=-1
now
two cases may be der
d(f(x))dxat P can be negative and d(g(x))dxat P must be +ve
or
vice versa
so they hav opposite monotonocity
2) statement 2 false
1
Che
·2010-02-14 00:55:48
i guess it shud be must hav instead of may hav :P
so NOT
106
Asish Mahapatra
·2010-02-14 00:57:17
1. what if one of the derivatives was 0 at the pt of intersection?
I think may comes into picture there.
1
Che
·2010-02-14 01:02:13
firstly one of deravative cant be 0
Two curves intersect orthogonally when their tangent lines at each point of intersection are perpendicular.
so m1* m2 need to be -1
so any of m1 m2 cant be 0
actully c option has may have opposite monotonity
if it had ...must hav then C wud hav been correct
so NON of these
for second
substitute n=3
u can see statement 2 is false
1
Che
·2010-02-14 01:34:20
r u sure?
infact
curve 1 is decreasing
and
curve 2 is increasing at point of intersection
106
Asish Mahapatra
·2010-02-14 01:36:21
at the point of intersection .... you cant say it is decreasing (strictly) isnt it?
Or else let curve (1) = x2
1
Che
·2010-02-14 01:43:14
ok ya i agree if function was x^2
but then ans shud be C
106
Asish Mahapatra
·2010-02-14 01:44:39
this IS fiitjee isnt it.. (last yrs full test question)
so mistakes can b der na?
1
Che
·2010-02-14 01:48:20
well i dun hav any soln for 2nd
but u can alwys substitute :P
substitute n=3
so u can see statement second is false
may be i m rong 