Solveeit Logo
Login

Question

Question: Let \( I=\int_{a}^{b}{\left( {{x}^{4}}-2{{x}^{2}} \right)}dx \) . If \( I \) is minimum, then the or...

Let I=∫ab(x4−2x2)dxI=\int_{a}^{b}{\left( {{x}^{4}}-2{{x}^{2}} \right)}dx . If II is minimum, then the ordered pair (a,b)\left( a,b \right) is
A. (−2,0)\left( -\sqrt{2},0 \right)
B. (0,2)\left( 0,\sqrt{2} \right)
C. (2,−2)\left( \sqrt{2},-\sqrt{2} \right)
D. (−2,2)\left( -\sqrt{2},\sqrt{2} \right)

Explanation

Solution

To find the ordered pair of minima in the given function, we will use the differentiation concept to find it. Which says that to find out the minima of maxima first find f′(x)f'\left( x \right) and find its roots. And put these roots in f′′(x)f''(x) . Which will give the point which will give minima.

Complete step by step answer:
Moving ahead with the question in step-wise manner;
As we know that to find the point of minima and maxima using differentiation of function, first find out the first derivative of function and put it equal to zero to find out its roots i.e. f′(x)=0f'(x)=0 , from here you will get the roots, put them in double derivative of and find for which root the double gives you greater than zero, that value will give minima in that function.
So going with the same concept in our question, we have function I=∫ab(x4−2x2)dxI=\int_{a}^{b}{\left( {{x}^{4}}-2{{x}^{2}} \right)}dx . So put its first derivative equal to zero. i.e.
dIdx=0 d∫ab(x4−2x2)dxdx=0 \begin{aligned} & \dfrac{dI}{dx}=0 \\\ & \dfrac{d\int_{a}^{b}{\left( {{x}^{4}}-2{{x}^{2}} \right)}dx}{dx}=0 \\\ \end{aligned}
As we know that differentiation of integration is the function itself, i.e.d∫f(x)dx=f(x)\dfrac{d\int{f(x)}}{dx}=f(x)
So by using above statement we can say that;
f′(x)=d∫ab(x4−2x2)dxdx=0 f′(x)=(x4−2x2) \begin{aligned} & f'(x)=\dfrac{d\int_{a}^{b}{\left( {{x}^{4}}-2{{x}^{2}} \right)}dx}{dx}=0 \\\ & f'(x)=\left( {{x}^{4}}-2{{x}^{2}} \right) \\\ \end{aligned}
So now we had to find out the roots of f′(x)=0f'(x)=0 so for roots let us solve it;
x4−2x2=0 x2(x2−2)=0 \begin{aligned} & {{x}^{4}}-2{{x}^{2}}=0 \\\ & {{x}^{2}}({{x}^{2}}-2)=0 \\\ \end{aligned}
By using the identity a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=(a+b)(a-b) in x2−2{{x}^{2}}-2 we will get;
x2(x−2)(x+2)=0{{x}^{2}}(x-\sqrt{2})(x+\sqrt{2})=0
From here we can say that x2=0{{x}^{2}}=0 or (x−2)=0(x-\sqrt{2})=0 or (x+2)=0(x+\sqrt{2})=0 . So from here we got four roots i.e. 0,0,2,−20,0,\sqrt{2},-\sqrt{2} .
Now according to the concept, put them in the second derivative. For that first let us find out second derivative i.e.
df′(x)dx=d(x4−2x2)dx f′′(x)=4x3−4x \begin{aligned} & \dfrac{df'(x)}{dx}=\dfrac{d\left( {{x}^{4}}-2{{x}^{2}} \right)}{dx} \\\ & f''(x)=4{{x}^{3}}-4x \\\ \end{aligned}
Now in this f′′(x)f''(x) put the roots and find the for which value of roots double derivative of function will give value greater than nor equal to zero, that value will give minimum for the function. So let us check in the order 0,0,2,−20,0,\sqrt{2},-\sqrt{2}
For 0;
f′′(x)=4x3−4x f′′(0)=4(0)3−4(0) f′′(0)=0 \begin{aligned} & f''(x)=4{{x}^{3}}-4x \\\ & f''(0)=4{{(0)}^{3}}-4(0) \\\ & f''(0)=0 \\\ \end{aligned}
So at zero function will be minimum. Now check for other roots also;
For 2\sqrt{2} ;
f′′(x)=4x3−4x f′′(2)=4(2)3−4(2) f′′(2)=82−42 f′′(2)=42 \begin{aligned} & f''(x)=4{{x}^{3}}-4x \\\ & f''\left( \sqrt{2} \right)=4{{\left( \sqrt{2} \right)}^{3}}-4\left( \sqrt{2} \right) \\\ & f''\left( \sqrt{2} \right)=8\sqrt{2}-4\sqrt{2} \\\ & f''\left( \sqrt{2} \right)=4\sqrt{2} \\\ \end{aligned}
For 2\sqrt{2} double derivative is greater than zero, so the function will give a minimum at 2\sqrt{2} also.
For −2-\sqrt{2} ;
f′′(x)=4x3−4x f′′(−2)=4(−2)3−4(−2) f′′(−2)=−82+42 f′′(2)=−42 \begin{aligned} & f''(x)=4{{x}^{3}}-4x \\\ & f''\left( -\sqrt{2} \right)=4{{\left( -\sqrt{2} \right)}^{3}}-4\left( -\sqrt{2} \right) \\\ & f''\left( -\sqrt{2} \right)=-8\sqrt{2}+4\sqrt{2} \\\ & f''\left( \sqrt{2} \right)=-4\sqrt{2} \\\ \end{aligned}
So for −2-\sqrt{2} double derivative is giving less than zero, so at this point the function will not give a minimum.
So we can say that the function will give a minimum at 00 and at 2\sqrt{2} .
As the function is continuous, we can say that the function will give a minimum in the interval of 00 and 2\sqrt{2} .

So, the correct answer is “Option B”.

Note: First derivative and second derivative are represented asf′(x)f'\left( x \right)andf′′(x)f''\left( x \right)respectively. Moreover if the roots we got from the first derivative, when we put them in the second derivative and got a value less than zero then at that point the function will give a maxima.