JEE MAIN - Mathematics (2020 - 9th January Evening Slot - No. 8)
Let a function ƒ : [0, 5] $$ \to $$ R be continuous,
ƒ(1) = 3 and F be defined as :
$$F(x) = \int\limits_1^x {{t^2}g(t)dt} $$ , where $$g(t) = \int\limits_1^t {f(u)du} $$
Then for the function F, the point x = 1 is :
$$F(x) = \int\limits_1^x {{t^2}g(t)dt} $$ , where $$g(t) = \int\limits_1^t {f(u)du} $$
Then for the function F, the point x = 1 is :
a point of inflection.
a point of local maxima.
a point of local minima.
not a critical point.
Explanation
$$F(x) = \int\limits_1^x {{t^2}g(t)dt} $$
$$ \Rightarrow $$ F'(x) = x2g(x) = x2$$\int\limits_1^t {f(u)du} $$
$$ \therefore $$ F'(1) = (1)(0) = 0
Now, F''(x) = 2xg(x) + x2g'(x)
F''(1) = 2g(1) + g'(1) = 0 + g'(1) = 3
[ As g'(t) = f(t); g'(1) = f'(1) = 3 ]
So, at x = 1, F'(1) = 0 and F"(1) = 3 > 0
$$ \therefore $$ For the function f(x), x = 1 is a point of local minima.
$$ \Rightarrow $$ F'(x) = x2g(x) = x2$$\int\limits_1^t {f(u)du} $$
$$ \therefore $$ F'(1) = (1)(0) = 0
Now, F''(x) = 2xg(x) + x2g'(x)
F''(1) = 2g(1) + g'(1) = 0 + g'(1) = 3
[ As g'(t) = f(t); g'(1) = f'(1) = 3 ]
So, at x = 1, F'(1) = 0 and F"(1) = 3 > 0
$$ \therefore $$ For the function f(x), x = 1 is a point of local minima.
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