f(x)=x3+3x−9‌‌x∈[−2,3] Differentiate with respect to x f′(x)=3x2+3 Hence f(x) is strictly increasing function so its greatest value will be at x=3 ‌f(3)=33+3×3−9=27 ‌‌
a
1−r
=27 ‌⇒‌‌a=27−27r ‌⇒‌‌a+27r=27 ....(i) ‌f′(0)=3 Also given a−ar=f′(0) ‌⇒‌‌a(1−r)=3 ‌⇒‌‌1−r=‌
3
a
....(ii) From equations, (i) and (ii) we get, a+27(1−‌
3
a
)‌=27 a+27−‌
81
a
‌=27 ⇒‌‌a2=81⇒a=±9 ∵ G.P. is decreasing ∴‌‌a=9 Now, ‌‌‌