IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

ashish_banga

if $F(x) = 1 + 2{x}^{2} + {2}^{2}{x}^{4} + .......... + {2}^{10}{x}^{20}$ , then F(x) has

1) more than one minimum

2) exactly one minimum

3) at least one maximum

4) none of the above

allamraju

I think exactly one minimum is the ans because the fun is strictly decresing for x<0 and strictly increasing for x>0 and attains the min.value 1 at x=0

navin_m29

I'll go with exactly one minimum value.
Method:
find first derivative and then second derivative. Second deri. is positive. which means minimum value. Now consider the que. upto 2x^2. double derivative is 4. Therefore only one minimumm value.
Rate if found useful or atleast nudge me for your point of view.

karansingh

exactly 1 min.

heres how:
As terms are additive in nature, derivative is always +ve.
Hence the f(x) always increases. So it has just 1 min.

u need not find the solutions for ' f'(x) = 0 ' as those will be imaginary

sarvpriye

In this

f(x) = 1+ 2x² + 2²x⁴+.................... + 2¹⁰x²⁰

f¹(x) = 4x + 4.2²x³ +................+ 20.2¹⁰x¹⁹

= x(4 +4.2²x²+ ...............+ 20.2¹⁰x¹⁹)

for f¹(x)=0

either x=0

or

4 + 4.2²x² +.................. + 20.2¹⁰x¹⁹=0,

This equation has all terms with even powers of x and also all coff. are positive, thats why it has all imaginary roots.

therefore, there is only one val x=0 for which there may be max or min,

Now, at x=0 F²(x)= 4,

that's why there is only one minima..........

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I think exactly one minimum is the ans because the fun is strictly decresing for x<0 and strictly increasing for x>0 and attains the min.value 1 at x=0