AP EAPCET 23-Aug-2021 Shift 2 Paper

f(x)=ax3+bx2+26x−24... (i) on [2,4]
∵f(x) satisfies the Role's theorem
⇒f(2)=f(4)
⇒a(2)3+b(22)+26(2)−24 =a(4)3+b(4)2+26(4)−24
⇒8a+4b+28=64a+16b+80
⇒56a+12b+52=0
⇒14a+3b+13=0 ...(ii)
On differentiating Eq. (i) w.r.t. x
f′(x)=

d

dx

(ax3+bx2+26x−24)
f′(x)=3ax2+2bx+26
At, x=3+

1

√3

f′(3+

1

√3

)=3a(3+

1

√3

)2+2b(3+

1

√3

)+26
0=3a(9+

1

3

+

6

√3

)+6b+

2b

√3

+26
0=28a+6b+6√3a+

2b

√3

+26
⇒(28+6√3)a+(

6√3+2

√3

)b+26=0 ...(iii)
(28a+6b+26)+(18a+2b)

1

√3

=0+0

1

√3

On comparing rational and irrational part,
28a+6b+26=0 and 18a+2b=0 and 54a+6b=0
∴28a−54a+26=0 and 6b=−54a
⇒−26a+26=0
⇒a=1
b=

−54×1

6

⇒b=−9
∴ab=−9