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Solution


Q.

Let (2, 3) be the largest open interval in which the function f(x)=2 loge (x–2)–x2+ax+1 is strictly increasing and (b, c) be the largest open interval, in which the function g(x)=(x–1)3(x+2–a)2 is strictly decreasing. Then 100 (a + b – c) is equal to:          [2025]
1 360  
2 160  
3 280  
4 420  

Ans.

(1)

Given : f(x)=2 loge (x–2)–x2+ax+1 is strictly increasing on (2, 3)

g(x)=(x–1)3(x+2–a)2 is strictly decreasing on (b, c).

Using f(x), 

         f'(x)=2x–2–2x+a

⇒ f''(x)=–2(x–2)2–2<0

SInce f(x) is strictly increasing, so  f'(x)>0.

But we have given that (2, 3) is the largest open interval where f(x) is strictly increasing.

⇒  f'(3)=0 ⇒ 2–6+a=0 ⇒ a=4

Taking, g(x)=(x–1)3(x+2–a)2

                         =(x–1)3(x–2)2           [∵ a = 4]

⇒ g'(x)=(x–1)2(x–2)(2x–2+3x–6)

                  =(x–1)2(x–2)(5x–8)

Here, g'(x)<0          [∵ g(x) is strictly decreasing]

⇒ (x–1)2(x–2)(5x–8)<0

⇒ x∈(85,2)

⇒ b=85 and c=2

Finally, we get 100 (a + b – c) = 100(4+85–2)=360.