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Let a function f:\left [ 0,5 \right ]\rightarrow \textbf{R} be continuous, f\left ( 1 \right )=3 and F be defined as : F(x)=\int_{1}^{x}t^{2}g\left ( t \right )dt, where g(t)=\int_{1}^{t}f\left ( u \right )du. Then for the function F, the point x=1 is :   
Option: 1 a point of infection.
Option: 2  a point of local maxima.
Option: 3 a point of local minima.
Option: 4 not a critical point.   
 

Answers (2)

best_answer

 

 

Integration as Reverse Process of Differentiation -

Integration is the reverse process of differentiation. In integration, we find the function whose differential coefficient is given. 

For example,

\\\mathrm{\frac{d}{dx}(\sin x)=\cos x}\\\\\mathrm{\frac{d}{dx}\left ( x^2 \right )=2x}\\\\\mathrm{\frac{d}{dx}\left ( e^x \right )=e^x}

In the above example,  the function cos x is the derived function of sin x. We say that sin x is an anti derivative (or an integral) of cos x. Similarly, x2 and ex  are the anti derivatives (or integrals) of 2x and ex respectively.  

Also note that the derivative of a constant  (C) is zero. So we can write the above examples as:

\\\mathrm{\frac{d}{dx}(\sin x+c)=\cos x}\\\\\mathrm{\frac{d}{dx}\left ( x^2+c \right )=2x}\\\\\mathrm{\frac{d}{dx}\left ( e^x +c\right )=e^x}

 

Thus, anti derivatives (or integrals) of the above functions are not unique. Actually, there exist infinitely many anti derivatives of each of these functions which can be obtained by selecting C arbitrarily from the set of real numbers. 

For this reason C is referred to as arbitrary constant. In fact, C is the parameter by varying which one gets different anti derivatives (or integrals) of the given function. 

 

If the function F(x) is an antiderivative of f(x), then the expression F(x) + C is the indefinite integral of the function f(x) and is denoted by the symbol ∫ f(x) dx. 

By definition,

\\\mathrm{\int f(x)dx=F(x)+c,\;\;\;where\;\;F'(x)=f(x)\;\;and\;\;'c' \;is\;constant.}

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Maxima and Minima of a Function -

Maxima and Minima of a Function

Let y = f(x) be a real function defined at x = a. Then the function f(x) is said to have a maximum value at x = a, if f(x) ≤ f(a)  ∀ a ≥∈ R.

And also the function f(x) is said to have a minimum value at x = a, if f(x) ≥ f(a)  ∀ a ∈ R

   

Concept of Local Maxima and Local Minima 

The function f(x) is said to have a maximum (or we say that f(x) attains a maximum) at a point ‘a’ if the value of f(x) at ‘a’  is greater than its values for all x in a small neighborhood of ‘a’ .

In other words, f(x) has a maximum at x = ‘a’, if f(a + h) ≤ f(a) and f(a - h) ≤ f(a), where h ≥ 0 (very small quantity).

The function f(x) is said to have a minimum (or we say that f(x) attains a minimum) at a point ‘b’ if the value of f(x) at ‘b’  is less than its values for all x in a small neighborhood of ‘b’ .

In other words, f(x) has a maximum at x = ‘b’, if f(a + h) ≥ f(a) and f(a - h) ≥ f(a), where h ≥ 0 (very small quantity). 

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\\ \begin{aligned} &\mathrm{F}^{\prime}(\mathrm{x})=\mathrm{x}^{2} \mathrm{g}(\mathrm{x})\\ &\Rightarrow \mathrm{F}^{\prime}(1)=1.\;\;g(1)=0\;\;\;\;\;\ldots(1)\;\;\;\;\;\;\;\;(\because\;g(1)=0)\\ &\text { Now }(\mathrm{x})=2 \mathrm{xg}(\mathrm{x})+\mathrm{x}^{2} \mathrm{g}^{\prime}(\mathrm{x}) \end{aligned}

\\ \begin{aligned} &\Rightarrow F^{\prime \prime}(x)=2 x g(x)+x^{2} f(x)\;\;\;\;(\because\;g'(x)=f(x))\\ &\Rightarrow F^{\prime \prime}(1)=0+1 \times 3\\ &\Rightarrow F^{\prime \prime}(1)=3\;\;\;\;\;\;\;\;\;\;\ldots (2) \end{aligned}\\\text{From (1) and (2) F(x) has local minimum at x = 1 }

Correct Option (1)

Posted by

avinash.dongre

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Option 3 currect 

Posted by

Rajesh Yadav

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