Let a function f - -0- 5- - R- be continuous- f1 - 3 and F be defined as- Fx-1x t2 gt dt- where gt-1t fu du- Then for the function F- the point x-1 is

F'(x) = x^2 g(x) Put x= 1 ⇒ F'(1) = g(1) = 0 ⋯(1) Now F”(x) = 2xg(x) + g'(x)x^2 F”(1) = 2g(1) + g'(1) nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; ...

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Standard XI

Mathematics

Local Minima

Let a functio...

Question

Let a function $f : [0, 5] \to R$ , be continuous, $f (1) = 3$ and $F$ be defined as:
$F (x) = x \int 1 t^{2} g (t) d t$ , where $g (t) = t \int 1 f (u) d u$ . Then for the function $F$ , the point $x = 1$ is

a point of inflection

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a point of local maxima

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a point of local minima

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not a critical point

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Solution

The correct option is C a point of local minima
$F^{'} (x) = x^{2} g (x)$
Put $x = 1$
$\Rightarrow F^{'} (1) = g (1) = 0 \dots (1)$
Now $F^{''} (x) = 2 x g (x) + g^{'} (x) x^{2}$
$F^{''} (1) = 2 g (1) + g^{'} (1) {∵ g^{'} (x) = f (x)}$
$F^{''} (1) = f (1) = 3 \dots (2)$
From $(1)$ and $(2)$ , $F (x)$ has local minimum at $x = 1$

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Let a function f:[0,5]→R, be continuous, f(1)=3 and F be defined as: F(x)=x∫1t2g(t) dt, where g(t)=t∫1f(u) du. Then for the function F, the point x=1 is

The correct option is C a point of local minimaF′(x)=x2g(x) Put x=1 ⇒F′(1)=g(1)=0⋯(1) Now F′′(x)=2xg(x)+g′(x)x2 F′′(1)=2g(1)+g′(1) {∵ g′(x)=f(x)} F′′(1)=f(1)=3⋯(2) From (1) and (2), F(x) has local minimum at x=1

Let a function $f : [0, 5] \to R$ , be continuous, $f (1) = 3$ and $F$ be defined as:
$F (x) = x \int 1 t^{2} g (t) d t$ , where $g (t) = t \int 1 f (u) d u$ . Then for the function $F$ , the point $x = 1$ is