JEE MAIN - Mathematics (2020 - 8th January Evening Slot - No. 9)
If $$I = \int\limits_1^2 {{{dx} \over {\sqrt {2{x^3} - 9{x^2} + 12x + 4} }}} $$, then :
$${1 \over 16} < {I^2} < {1 \over 9}$$
$${1 \over 8} < {I^2} < {1 \over 4}$$
$${1 \over 9} < {I^2} < {1 \over 8}$$
$${1 \over 6} < {I^2} < {1 \over 2}$$
Explanation
Let f(x) = $${1 \over {\sqrt {2{x^3} - 9{x^2} + 12x + 4} }}$$
f'(x) = $$ - {1 \over 2}{{ - 6{x^2} - 18x + 12} \over {2{{\left( {2{x^3} - 9{x^2} + 12x + 4} \right)}^{3/2}}}}$$
= $${{ - 6\left( {x - 1} \right)\left( {x - 2} \right)} \over {2{{\left( {2{x^3} - 9{x^2} + 12x + 4} \right)}^{3/2}}}}$$
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fmax = f(1)
fmin = f(2)
f(1) = $${1 \over {\sqrt {2 - 9 + 12 + 4} }}$$ = $${1 \over {\sqrt 9 }}$$ = $${1 \over 3}$$
f(2) = $${1 \over {\sqrt {16 - 36 + 24 + 4} }}$$ = $${1 \over {\sqrt 8 }}$$
$$ \therefore $$ $${1 \over 3} < {I} <$$ $${1 \over {\sqrt 8 }}$$
So $${1 \over 9} < {I^2} < {1 \over 8}$$
f'(x) = $$ - {1 \over 2}{{ - 6{x^2} - 18x + 12} \over {2{{\left( {2{x^3} - 9{x^2} + 12x + 4} \right)}^{3/2}}}}$$
= $${{ - 6\left( {x - 1} \right)\left( {x - 2} \right)} \over {2{{\left( {2{x^3} - 9{x^2} + 12x + 4} \right)}^{3/2}}}}$$
fmax = f(1)
fmin = f(2)
f(1) = $${1 \over {\sqrt {2 - 9 + 12 + 4} }}$$ = $${1 \over {\sqrt 9 }}$$ = $${1 \over 3}$$
f(2) = $${1 \over {\sqrt {16 - 36 + 24 + 4} }}$$ = $${1 \over {\sqrt 8 }}$$
$$ \therefore $$ $${1 \over 3} < {I} <$$ $${1 \over {\sqrt 8 }}$$
So $${1 \over 9} < {I^2} < {1 \over 8}$$
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