For any real number x, the maximum value of \[4-6x-{{x}^{2}}\]is [SSC (CGL) 2014] |
A) 4
B) 7
C) 9
D) 13
Correct Answer: D
Solution :
Let the given equation be represented as |
\[\text{f}\,(x)=4-\,\,6x-{{x}^{2}}\] |
Now, differentiating above function w.r.t. x, we get |
\[\text{f}'\,\,(x)=-\,\,6-2x\] |
For value of x put \[\text{f}'(x)=0\] |
\[-\,\,6-2x=0\] |
\[\therefore \] \[x=-\,\,3\] |
For maximum value, we take \[\text{f}'(x)\] |
\[\text{f}'(x)=-\,2\] |
Since, value of \[\text{f}'(x)\] is negative. |
So, \[\text{f}\,(x)\] is maximum at \[x=\,\,-3.\] |
Putting \[x=-\,\,3\]in \[\text{f}\,(x),\] we get |
\[\text{f}\,(-\,\,3)=4\,\,-(6)(-\,\,3)-{{(-\,\,3)}^{2}}\] |
\[=4+18-9=13\] |
The maximum value of \[4-6x-{{x}^{2}}\] is 13. |
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