Question:
If the function f given by \(f(x)=x^3-3(a-2)x^2+3ax+7\), for some \(a\in R\) is increasing in (0, 1] and decreasing in [1, 5), then a root of the equation, \(\frac{f(x)-14}{(x-1)^2}=0\ (x\ne1)\) is :
If the function f given by \(f(x)=x^3-3(a-2)x^2+3ax+7\), for some \(a\in R\) is increasing in (0, 1] and decreasing in [1, 5), then a root of the equation, \(\frac{f(x)-14}{(x-1)^2}=0\ (x\ne1)\) is :
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f has a local maximum at x = 1, so \(f'(1)=0\); this fixes a, and then \(f(x)-14\) has a double root at 1.
Updated On: Nov 5, 2026
- -7
- 5
- 7
- 6
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The Correct Option is C
Solution and Explanation
\(f'(x)=3x^2-6(a-2)x+3a\). Since f increases up to 1 and decreases after 1, \(f'(1)=0\).
\(3-6(a-2)+3a=0\Rightarrow15-3a=0\Rightarrow a=5\).
Then \(f(x)=x^3-9x^2+15x+7\) and \(f(1)=14\), so \(f(x)-14=x^3-9x^2+15x-7=(x-1)^2(x-7)\).
Hence \(\frac{f(x)-14}{(x-1)^2}=x-7=0\Rightarrow x=7\).
Answer: (3) 7
\(3-6(a-2)+3a=0\Rightarrow15-3a=0\Rightarrow a=5\).
Then \(f(x)=x^3-9x^2+15x+7\) and \(f(1)=14\), so \(f(x)-14=x^3-9x^2+15x-7=(x-1)^2(x-7)\).
Hence \(\frac{f(x)-14}{(x-1)^2}=x-7=0\Rightarrow x=7\).
Answer: (3) 7
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