JEE MAIN - Mathematics (2020 - 3rd September Morning Slot - No. 9)
The function, f(x) = (3x – 7)x2/3, x $$ \in $$ R, is
increasing for all x lying in :
$$\left( { - \infty ,0} \right) \cup \left( {{3 \over 7},\infty } \right)$$
$$\left( { - \infty ,0} \right) \cup \left( {{{14} \over {15}},\infty } \right)$$
$$\left( { - \infty ,{{14} \over {15}}} \right)$$
$$\left( { - \infty ,{{14} \over {15}}} \right) \cup \left( {0,\infty } \right)$$
Explanation
f(x) = (3x – 7)x2/3
f’(x) = $$\left( {3x - 7} \right){2 \over {3{x^{1/3}}}} + {x^{{2 \over 3}}}.3$$
= $${{6x - 14 - 9x} \over {3{x^{1/3}}}}$$
= $${{15x - 14} \over {3{x^{1/3}}}}$$
As f(x) increasing so f'(x) > 0
$$ \therefore $$ $${{15x - 14} \over {3{x^{1/3}}}}$$ > 0_3rd_September_Morning_Slot_en_9_1.png)
$$ \therefore $$ x $$ \in $$ $$\left( { - \infty ,0} \right) \cup \left( {{{14} \over {15}},\infty } \right)$$
f’(x) = $$\left( {3x - 7} \right){2 \over {3{x^{1/3}}}} + {x^{{2 \over 3}}}.3$$
= $${{6x - 14 - 9x} \over {3{x^{1/3}}}}$$
= $${{15x - 14} \over {3{x^{1/3}}}}$$
As f(x) increasing so f'(x) > 0
$$ \therefore $$ $${{15x - 14} \over {3{x^{1/3}}}}$$ > 0
$$ \therefore $$ x $$ \in $$ $$\left( { - \infty ,0} \right) \cup \left( {{{14} \over {15}},\infty } \right)$$
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