| Assertion [A]: The local maximum value of |
| \[f\left( x \right)={{x}^{3}}-6{{x}^{2}}+9x+15\] at \[x=1\]is 19. |
| Reason [R]: For local maximum, \[f''\left( x \right)<0.\] |
A) Both A and R are individually true and R is the correct explanation of A.
B) Both A and R are individually true and R is not the correct explanation of A.
C) 'A' is true but 'R' is false
D) 'A' is false but 'R' is true
E) Both A and R are false.
Correct Answer: A
Solution :
Given \[f(x)={{x}^{3}}-6{{x}^{2}}+9x+15\] \[f'\left( x \right)=3{{x}^{2}}-12x+9=3\left( {{x}^{2}}-4x+3 \right)\] \[=3\left( x-1 \right)\left( x-3 \right)\] Also \[f''(x)=3(2x-4)\] For critical points, \[f'\left( x \right)=0\Rightarrow x=1,\text{ x}=3\] At \[x\text{ }=\text{ 1},f''\left( x \right)=3\left( 2-4 \right)=-6<0\] \[\Rightarrow \,\,\,f\left( x \right)\]is maximum at \[x=1\] \[\therefore \] Maximum value \[={{\left( 1 \right)}^{3}}-6{{\left( 1 \right)}^{2}}+9\times 1+15\] \[=1-6+9+15=19\] \[\therefore \] Assertion [A] is true Since for local maxima, \[f''\left( x \right)<0\] \[\therefore \] Reason 'R' is true and is correct explanation of Assertion 'A' Hence option [A] is the correct answer.
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