IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

sankydreams

1) The function y = (ax + b) / [(x-1)(x-4)] has a critical point at P(2, - 1), find a and b so that y has local maximum at P.

2)Let f(x) = x - sinx and g(x) = x - tanx where x

(0,

/2). then for these values of x

A) f(x).g(x) > 0

B) f(x)g(x) < 0

C) f(x)/g)x) > 0

D) none of these

Plz post detailed solns for assured rates....

sankydreams

no replies, weird....

spideyunlimited

Ans 2) (b)
for b use product rule and plug in a value in given domain.. turns out negative so f(x).g(x) is a decreasing function.
then we find maxima..( it's at x=0 ) and also find the value at x = pi/2 which comes out as negative too
and as maxima is at 0 and function is decreasing.. so the function is < 0.
f(x).g(x) < 0

*****
for f(x) / g(x) use quotient rule.. the function is negative in given domain and increasing. and becomes 0 at x = pi/2

rini

ans. b]

for all x between the given range..

x> sinx and tanx> x

so.f(x) > 0 and g(x) <0

f(x) g(x) <0 ...............(b)

priyesh

f(x) = x - sinx

so

f ' (x) = 1 - cosx

now for x b/w (0,pi/2)

f ' (x) is >0

=> f(x) is increasing

so minimum value of f(x) is lim x>>>>0 f(x) = 0 - 0 = 0

so f(x) is > 0 in the given interval

similarly g(x) = x - tanx

g ' (x) = 1 - sec^2 x = 1 - 1/cos^2 x

now when x lies b/w (0,pi/2)

cos^2 x lies b/w (0,1)

so g ' (x) < 0 in the given interval

=>so maximum value of g(x) is limx>>>>0 g(x) = 0 - 0 = 0

so g(x) is < 0 in the given interval

hence f(x) g(x) < 0 in the given interval

ramyadiamond

Given that the function has a maxima at P, hence f^'(2)=0

Now, f^'(x) =( -ax²-2bx+4a+5b) / (x-1)²(x-4)²

f^'(2) = -4a -4b+4a+5b / 4 =0

therefore, b=0

Now, P satisfies the function

f(2) = -1

a(2) +0 / (4-10+4) = -1

hence, a =1

So f(x) = x/(x-1)(x-4)

priyesh

Ramya is absolutely correct!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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