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Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a function such that
$\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+\mathrm{x}^{2} \mathrm{f}^{\prime}(1)+\mathrm{xf}^{\prime}(2)+\mathrm{f}^{\prime \prime}(3)$
for $\mathrm{x} \in \mathbf{R}$
Consider the following:
$1 \quad f(2)=f(1)-f(0)$
$2 \quad \mathrm{f}^{\prime \prime}(2)-2 \mathrm{f}^{\prime}(1)=12$
Which of the above is/are correct?
Options:
$\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+\mathrm{x}^{2} \mathrm{f}^{\prime}(1)+\mathrm{xf}^{\prime}(2)+\mathrm{f}^{\prime \prime}(3)$
for $\mathrm{x} \in \mathbf{R}$
Consider the following:
$1 \quad f(2)=f(1)-f(0)$
$2 \quad \mathrm{f}^{\prime \prime}(2)-2 \mathrm{f}^{\prime}(1)=12$
Which of the above is/are correct?
Solution:
2146 Upvotes
Verified Answer
The correct answer is:
Both 1 and 2
1 $f(1)-f(0)=4-6$
$=-2$
$f(2)=8-20+4+6=-2$
$\quad$ Hence $f(2)=f(1)-f(0)$
$\therefore$ Statement $(1)$ is correct.
2 $f^{\prime \prime}(2)-2 f^{\prime}(1)=2-2(-5)$
$f^{\prime \prime}(2)-2 \mathrm{f}^{\prime}(1)=12$
$\therefore$ Statement $(2)$ is correct.
$=-2$
$f(2)=8-20+4+6=-2$
$\quad$ Hence $f(2)=f(1)-f(0)$
$\therefore$ Statement $(1)$ is correct.
2 $f^{\prime \prime}(2)-2 f^{\prime}(1)=2-2(-5)$
$f^{\prime \prime}(2)-2 \mathrm{f}^{\prime}(1)=12$
$\therefore$ Statement $(2)$ is correct.
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