BITSAT 2006 Solved Question Paper

p(t)=1000+

1000t

100+t2

...(i)
On differentiating both side w.r.t.t.
p′(t)=0+

(100+t2)(1000)−1000t(2t)

(100+t2)2

=1000

(100−t2)

(100+t2)2

...(ii)
Put p′(t)=0 for maxima or minima
⇒100−t2=0
⇒t=±10
Now, again differentiating Eq. (ii), w.r.t. x
p′′(t)=1000
×[

(100+t2)2(−2t)−(100−t2)2(100+t2)2t

(100+t2)4

]
=1000t

[(100+t2)(−2)−(100−t2)(4)]

(100+t2)3

=−1000t

[600−2t2]

(100+t2)3

At t=10
p′′(t)<0
∴ The maximum value is
p(10)=1000+

10000

100+100

=1000+

10000

200

=1000+50
= 1050