JEE MAIN - Mathematics (2021 - 17th March Morning Shift - No. 5)
The sum of possible values of x for
tan$$-$$1(x + 1) + cot$$-$$1$$\left( {{1 \over {x - 1}}} \right)$$ = tan$$-$$1$$\left( {{8 \over {31}}} \right)$$ is :
tan$$-$$1(x + 1) + cot$$-$$1$$\left( {{1 \over {x - 1}}} \right)$$ = tan$$-$$1$$\left( {{8 \over {31}}} \right)$$ is :
$$-$$$${{{32} \over 4}}$$
$$-$$$${{{33} \over 4}}$$
$$-$$$${{{31} \over 4}}$$
$$-$$$${{{30} \over 4}}$$
Explanation
tan$$-$$1(x + 1) + cot$$-$$1$$\left( {{1 \over {x - 1}}} \right)$$ = tan$$-$$1$$\left( {{8 \over {31}}} \right)$$
$$ \Rightarrow $$ tan$$-$$1(x + 1) + tan$$-$$1(x - 1) = tan$$-$$1$$\left( {{8 \over {31}}} \right)$$
$$ \Rightarrow $$ $${\tan ^{ - 1}}\left( {{{\left( {x + 1} \right) + \left( {x - 1} \right)} \over {1 - \left( {x + 1} \right)\left( {x - 1} \right)}}} \right)$$ = tan$$-$$1$$\left( {{8 \over {31}}} \right)$$
$$ \Rightarrow $$ $${{(1 + x) + (x - 1)} \over {1 - (1 + x)(x - 1)}} = {8 \over {31}}$$
$$ \Rightarrow {{2x} \over {2 - {x^2}}} = {8 \over {31}}$$
$$ \Rightarrow 4{x^2} + 31x - 8 = 0$$
$$ \Rightarrow x = - 8,{1 \over 4}$$
but at $$x = {1 \over 4}$$
$$LHS > {\pi \over 2}$$ and $$RHS < {\pi \over 2}$$
So, only solution is x = $$-$$ 8 = $$-$$$${{{32} \over 4}}$$
$$ \Rightarrow $$ tan$$-$$1(x + 1) + tan$$-$$1(x - 1) = tan$$-$$1$$\left( {{8 \over {31}}} \right)$$
$$ \Rightarrow $$ $${\tan ^{ - 1}}\left( {{{\left( {x + 1} \right) + \left( {x - 1} \right)} \over {1 - \left( {x + 1} \right)\left( {x - 1} \right)}}} \right)$$ = tan$$-$$1$$\left( {{8 \over {31}}} \right)$$
$$ \Rightarrow $$ $${{(1 + x) + (x - 1)} \over {1 - (1 + x)(x - 1)}} = {8 \over {31}}$$
$$ \Rightarrow {{2x} \over {2 - {x^2}}} = {8 \over {31}}$$
$$ \Rightarrow 4{x^2} + 31x - 8 = 0$$
$$ \Rightarrow x = - 8,{1 \over 4}$$
but at $$x = {1 \over 4}$$
$$LHS > {\pi \over 2}$$ and $$RHS < {\pi \over 2}$$
So, only solution is x = $$-$$ 8 = $$-$$$${{{32} \over 4}}$$
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