MCQ
The maximum value of $f(x) = (7-x)^4 (2+x)^5$ is
- A$(4 × 5)^{4+5}$
- B$4^55^4$
- ✓$4^45^5$
- DNone of these
✓
Answer
Correct option: C.
$4^45^5$
c
Points where derivative of the function is equal to zero are either maxima or minima since it is a
Points where derivative of the function is equal to zero are either maxima or minima since it is a
continuous function.
putting derivative $=0,$ we get
$\left(4 \times-1 \times\{7-x\}^{3} \times\{2+x\}^{5}\right)+\left(5 \times\{7-x\}^{4}\{2+x\}^{4}\right)=0$
or $, 5\{7-x\}=4\{2+x\}$ since $x \in(-2,7)$ so $\{x+2\} \neq 0$ \& $\{7-x\} \neq 0$
Therefore $9 x=27$ or, $x=3$
Hence the maximum value will be $(7-3)^{4}(2+3)^{5}$ or $8 \times 10^{5}$ or $4^45^5$
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