ISRO-2013-49

Question

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

• A. $-15$
• B. $7$
• C. $-11$
• D. $-3$

Answer 1

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

$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 $(C).$

Answer 2

Answer -11

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

Comments

YOU have forgot to check the minimum for end points

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not getting @viv

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@srestha

After evaluating f'(x) , you got x=2, you just check for critical point x=2, but the start and end of the interval are also critical points i.e 0 and 5

though, in this question function value was minimum at x=2, but it may not be possible everywhere and everytime and we get answer wrong.

that is what @viv696 is also saying.

Answer 3

Answer is -11

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

Comments

how u have taken the value of x from 1 to 5

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Unreliable way

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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!

Answer 4

f(x) = 2x² - 8x - 3

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f(x) = 2(x²-4x-3/2)

f(x) = 2((x-2)² - 11/2)

f(x) = 2(x-2)² - 11

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For the minimum value of function quadratic term must be negative or zero 

As this term can't be negative so will be 0 and

 For 2(x-2)² to be zero  , x will bw 2 

and f(x) will be  = 0 - 11 = -11

So , -11 will be minimum value of function