logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
+1 vote

It is given that at $x = 1$, the function $x^4 – 62x^2 + ax + 9$ attains its maximum value, on the interval $[0, 2].$ Find the value of $a.$

$\begin{array}{1 1} a=0 \\ a=130 \\ a=120 \\ a=150 \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
  • For Maxima and Minima $f'(x)=0$
Step 1:
Let $f(x)=x^4-62x^2+ax+9$
Differentiating with respect to $x$
$f'(x)=4x^3-124x+a$
Now $f'(x)=0$ at $x=1$
$f'(1)=4(1)^3-124(1)+a$
$\qquad=4-124+a$
$f'(x)=0$
$\Rightarrow 4-124+a=0$
$a=120$
Step 2:
Now differenting again with respect to x
Now $f''(x)=12x^2-124$
Substitute the value of x=1
$f''(1)=12(1)-124$
$\qquad\;=-112 < 0$
$\Rightarrow f(x)$ has a maximum at $x=1$ when $a=120$
Hence $a=120$
answered Aug 7, 2013 by sreemathi.v
edited Aug 19, 2013 by sharmaaparna1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...