JEE MAIN - Mathematics (2020 - 4th September Morning Slot - No. 12)
Let f be a twice differentiable function on (1, 6). If f(2) = 8, f’(2) = 5, f’(x) $$ \ge $$ 1 and f''(x) $$ \ge $$ 4, for all x $$ \in $$ (1, 6), then :
f(5) $$ \le $$ 10
f(5) + f'(5) $$ \ge $$ 28
f(5) + f'(5) $$ \le $$ 26
f'(5) + f''(5) $$ \le $$ 20
Explanation
Given, $$f'(x) \ge 1$$
$$ \therefore $$ $$\int_2^5 {f'(x)} dx\, \ge \,\int_2^5 {dx} $$
$$ \Rightarrow f(5) - f(2) \ge 3$$
$$ \Rightarrow f(5) - 8 \ge 3$$
$$ \Rightarrow f(5) \ge 11$$ ...(1)
Also, $$f''(x) \ge 4$$
$$ \therefore $$ $$\int_2^5 {f''(x)} dx\, \ge \,\int_2^5 {4dx} $$
$$ \Rightarrow f'(5) - f'(2) \ge 4(3)$$
$$ \Rightarrow f'(5) - 5 \ge 12$$
$$ \Rightarrow f'(5) \ge 17$$ ...(2)
From (1) and (2),
$$f'(5) + f'(5) \ge 11 + 17$$
$$ \Rightarrow f'(5) + f'(5) \ge 28$$
$$ \therefore $$ $$\int_2^5 {f'(x)} dx\, \ge \,\int_2^5 {dx} $$
$$ \Rightarrow f(5) - f(2) \ge 3$$
$$ \Rightarrow f(5) - 8 \ge 3$$
$$ \Rightarrow f(5) \ge 11$$ ...(1)
Also, $$f''(x) \ge 4$$
$$ \therefore $$ $$\int_2^5 {f''(x)} dx\, \ge \,\int_2^5 {4dx} $$
$$ \Rightarrow f'(5) - f'(2) \ge 4(3)$$
$$ \Rightarrow f'(5) - 5 \ge 12$$
$$ \Rightarrow f'(5) \ge 17$$ ...(2)
From (1) and (2),
$$f'(5) + f'(5) \ge 11 + 17$$
$$ \Rightarrow f'(5) + f'(5) \ge 28$$
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