# ISRO-2013-49

2.3k views

What is the least value of the function $f(x) = 2x^{2}-8x-3$ in the interval $[0, 5]$?

1. $-15$
2. $7$
3. $-11$
4. $-3$
in Calculus

$f(x) = 2x^{2}-8x-3$

$f'(x) = 4x-8$

For stationary point$:f'(x) = 0\implies 4x - 8 = 0 \implies x = 2$

Therefore, critical points $x = 0,2,5$

Now, $f''(x) = 4>0\:\: \textbf{(Minima)}$

For getting the minimum (or) least value , we should check all the value of critical points (stationary point and closed interval points).

For $x = 0: f(x) = -3$

For $x = 2: f(x) = 8 - 16 - 3 = -11$

For $x = 5: f(x) = 50 - 40 - 3 = 7$

$\therefore x= 2 ,\text{minimum value of}\: f(x) = -11$

So, the correct answer is $(C).$

selected by

f(x) = 2x2 -8x -3

f'(x)=4x-8

f''(x)=4 , then it has minimal value at the point x=2

minimum value is 2⨉(2)2 -8⨉(2) -3= -11

0
YOU have forgot to check the minimum for end points
0
not getting @viv

Substutue all the value of x from 1 to 5

in case of 0 it is -3

in case of 1 you will get -9

in case of 2 you will get -11

in case of 3 you will get -9

incase of 4 you will get -3

incase of 5 you will get 7

so the least value is -11 -->option c

0
how u have taken the value of x from 1 to 5
0
Unreliable way
2
reliable but brute force, in ISRO time will play an important role and you can solve some other easy questions then solving this way. each question carry same mark!

## Related questions

1
578 views
Suppose we have variable logical records of lengths of 55 bytes, 1010 bytes and 2525 bytes while the physical block size in disk is 1515 bytes. What is the maximum and minimum fragmentation seen in bytes? 2525 and 55 1515 and 55 1515 and 00 1010 and 5 what is ... haven't mentioned that we have 2 blocks. All those who is supporting answer D uses 2 blocks. Some please give the concrete solution.