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Let $f(1)=-2$ and $f^{\prime}(x) \geq 4.2$ for $1 \leq x \leq 6$. The possible value of $f(6)$ lies in the interval :
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The correct answer is:
$[19, \infty)$
$[19, \infty)$
Given $f(1)=-2$ and $f^{\prime}(x) \geq 4.2$ for $1 \leq x \leq 6$
Consider $f^{\prime}(x)=\frac{f(x+h)-f(x)}{h}$ $\Rightarrow f(x+h)-f(x)=f^{\prime}(x) \cdot h \geq(4.2) h$
So, $f(x+h) \geq f(x)+(4.2) h$
put $x=1$ and $h=5$, we get
$f(6) \geq f(1)+5(4.2) \Rightarrow f(6) \geq 19$
Hence $f(6)$ lies in $[19, \infty)$
Consider $f^{\prime}(x)=\frac{f(x+h)-f(x)}{h}$ $\Rightarrow f(x+h)-f(x)=f^{\prime}(x) \cdot h \geq(4.2) h$
So, $f(x+h) \geq f(x)+(4.2) h$
put $x=1$ and $h=5$, we get
$f(6) \geq f(1)+5(4.2) \Rightarrow f(6) \geq 19$
Hence $f(6)$ lies in $[19, \infty)$
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