How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?

# Understand the problem

[/et_pb_text][et_pb_text _builder_version="4.1" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Let f : [0, 2] → R be a continuous function such that $\frac{1}{2} .\int^2_0 f(x)\,dx < f(2)$.

## Look at the knowledge graph.

[/et_pb_text][et_pb_image src="https://www.cheenta.com/wp-content/uploads/2020/01/Untitled-Diagram-3.png" _builder_version="4.1" hover_enabled="0" align="center"][/et_pb_image][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.1" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.1"]I.S.I. B.Stat. Entrance 2017, UGA Problem 20[/et_pb_accordion_item][et_pb_accordion_item title="Key Competency" _builder_version="4.1" open="off"]Maximum and minimum property of  function [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.1" open="off"]Medium[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" open="off"]Mathematical Circles[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="4.1" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" hover_enabled="0" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="4.1" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="4.1"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.1"]See that from the given result we have  $\frac{1}{2}.\int^2_0 f(x)\,dx < f(2) \Rightarrow \int^2_0 f(x)\,dx < 2.f(2) \Rightarrow \int^2_0 f(x)\,dx < \int^2_0 f(2)\,dx$.[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.1"]Now if f(2) is minimum then f(x)>f(2) for all x belong to [0,2] . Therefore, $\int^2_0 f(x)\,dx > \int^2_0 f(2)\,dx$, which gives contradiction to the given result $\int^2_0 f(x)\,dx < \int^2_0 f(2)\,dx$.    [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.1"]From this we can't say that f must be strictly increasing or f must attain a maximum value at x = 2.The only thing we can say that  f cannot have a minimum at x = 2.[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

# Similar Problems

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