Home › ISC Class 12 › Mathematics › Application of Derivatives › For $f(x)=3x^4-8x^3+12x^2-48x+25$ on $[0,3]$, th…
For $f(x)=3x^4-8x^3+12x^2-48x+25$ on $[0,3]$, the critical point inside the interval is
A$x=1$
B$x=2$
C$x=3$
D$x=0$
Answer & Solution
Correct answer: B. $x=2$
1. $f'(x)=12x^3-24x^2+24x-48=12(x^3-2x^2+2x-4)$.
2. Factor: $x^3-2x^2+2x-4=x^2(x-2)+2(x-2)=(x-2)(x^2+2)$.
3. $f'(x)=0 \Rightarrow x=2$ (since $x^2+2>0$ has no real root).
4. So the only interior critical point is $x=2$.
_Source: NCERT Class 12 Mathematics Ch 6 "Application of Derivatives", p.28_
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