At what integral value of x will the function (x2 + 3x + 1)/(x2 - 3x + 1) attain its maximum value??
a) 3
b) 4
c) -3
d) none of this
LOD 3, chapter: Function
i hope the answer is c that is (-3)..
i hope the answer is c that is (-3)..
At what integral value of x will the function (x2 + 3x + 1)/(x2 - 3x + 1) attain its maximum value??
a) 3
b) 4
c) -3
d) none of this
LOD 3, chapter: Function
see the denominator=>x(x-3)+1
it should be as low as possible,but positive(as,numerator is positive in all cases except -1 and -2 which give values as -1/5 and -1/11)
it is possible when,x=0 or x=3
x=0=>value is 1
x=3=>value is 19.
option (A)
At what integral value of x will the function (x2 + 3x + 1)/(x2 - 3x + 1) attain its maximum value??
a) 3
b) 4
c) -3
d) none of this
LOD 3, chapter: Function
Another method to solve, worth share i guess.
f(x) = (x2 + 3x + 1)/(x2 - 3x + 1) = 1+ (6x/(x2 - 3x + 1)).
So, f(x) to be max.... 6x/(x2 - 3x + 1) shud be max.
6x/(x2 - 3x + 1) = 6/(x-3 +1/x) ..=> (x+1/x) shud be positive (i.e greater than 3) and min.
for x= 3...exp (x+1/x) be 3+1/3 which gives f(x) be max.
hence x= 3 is ans...
If f(x)=1/g(x), then which of the following is correct?
a) f(f(g(f(x)))) = f(g(g(f(f(x)))))
b) f(g(g(f(f(x))))) = f(f(g(g(g(x)))))
c) g(g(f(f(g(f(x)))))) = f(f(g(g(f(g(x))))))
d) f(g(g(g(f(x))))) = g(g(f(f(f(x)))))
Please just dont post the answer... tell the way this problem has to be solved...
If f(x)=1/g(x), then which of the following is correct?
a) f(f(g(f(x)))) = f(g(g(f(f(x)))))
b) f(g(g(f(f(x))))) = f(f(g(g(g(x)))))
c) g(g(f(f(g(f(x)))))) = f(f(g(g(f(g(x))))))
d) f(g(g(g(f(x))))) = g(g(f(f(f(x)))))
Please just dont post the answer... tell the way this problem has to be solved...
ans is option 3
if u take f(x)=x & g(x)=1/x
in only option 3 , both side g(x) come 3 times
in another option LHS=RHS for g(x)
so ans is 3
If f(x)=1/g(x), then which of the following is correct?
a) f(f(g(f(x)))) = f(g(g(f(f(x)))))
b) f(g(g(f(f(x))))) = f(f(g(g(g(x)))))
c) g(g(f(f(g(f(x)))))) = f(f(g(g(f(g(x))))))
d) f(g(g(g(f(x))))) = g(g(f(f(f(x)))))
Please just dont post the answer... tell the way this problem has to be solved...
I got the answer to be (c)..
the approach..
Lets take f(x)=x and g(x)=1/x
as we have f(x)=1/g(x)
and then checking on the various options....other choices are eliminated..if you still cud nt understand it....tell me again..vill show you how all the options got eliminated..:drinking:
At what integral value of x will the function (x2 + 3x + 1)/(x2 - 3x + 1) attain its maximum value??
a) 3
b) 4
c) -3
d) none of this
LOD 3, chapter: Function
for function max
denominator will be zero or min value
x^2-3x=x(x-3) if u put x=3 function value became 1
so ans is 3
here in the book the answer is given as -3 (i.e option c)... but how?? am nt getting it...
for function max
denominator will be zero or min value
x^2-3x=x(x-3) if u put x=3 function value became 1
so ans is 3
Can we have the final answer from the person who got in this question on this forum??
here in the book the answer is given as -3 (i.e option c)... but how?? am nt getting it...
here in the book the answer is given as -3 (i.e option c)... but how?? am nt getting it...
Hey i solved it two days back...lemme solve it again nd den explain it to u later...my answer came out to be 3 at first but then had a close look and by some manipulations got (c) as the answer......hey others who solved this questions look into this query plz....i am still confused about the answr..lemm solve it once again..
here in the book the answer is given as -3 (i.e option c)... but how?? am nt getting it...
its wrong ans because if u take -3
then numerator=9-9+1=1
denominator=9+9+1=19
so ans is 1/19
and if u take 3 then ans is 19
so definitely ans is 3
here in the book the answer is given as -3 (i.e option c)... but how?? am nt getting it...
its wrong ans because if u take -3
then numerator=9-9+1=1
denominator=9+9+1=19
so ans is 1/19
and if u take 3 then ans is 19
so definitely ans is 3
But then we also have an option (d)none of these..
How will you eliminate this one option if you are going by the options method?
bilas Sayshere in the book the answer is given as -3 (i.e option c)... but how?? am nt getting it...
hey we have seen ypur problem..no need to repaeat it or repost it..
CHEERS..
ya thanx i got the logic... (if u take f(x)=x & g(x)=1/x)........
but is it the easier way to count the number of g(X) and f(x) in both the sides, and go for that option which has the same number of g(x) and f(x) in both the sides???
But then we also have an option (d)none of these..
How will you eliminate this one option if you are going by the options method?
without option u can say function max at 3
because denominator at 3 min, below 3 function go in negative value and above 3 denominator more than 1 so value of the function decrease
so ans is 3
ya thanx i got the logic... (if u take f(x)=x & g(x)=1/x)........
but is it the easier way to count the number of g(X) and f(x) in both the sides, and go for that option which has the same number of g(x) and f(x) in both the sides???
See i can not comment on that approach but then merely counting the number of functions on both side vont help you......Sometimes these numbers not being the same can yield you being LHS=RHS...
So go by the conventional method and these method is not even bulk...moreover at practice stages it gives you an insight in2 functions in the form of a beautifully knitted question..
without option u can say function max at 3
because denominator at 3 min, below 3 function go in negative value and above 3 denominator more than 1 so value of the function decrease
so ans is 3
May be..but i will have to check out once again on my approach because i solved it by the conventional method by taking the whole function equivalent
to Y and then forming a quadratic equation,checking the discriminant and so on......
See i can not comment on that approach but then merely counting the number of functions on both side vont help you......Sometimes these numbers not being the same can yield you being LHS=RHS...
So go by the conventional method and these method is not even bulk...moreover at practice stages it gives you an insight in2 functions in the form of a beautifully knitted question..
yep every time counting didnt work but here it will work because due to F(x) function didnt change
so its better to put value of function both side and check LHS=RHS