JEE MAIN - Mathematics (2021 - 20th July Evening Shift - No. 21)
If $$\mathop {\lim }\limits_{x \to 0} {{\alpha x{e^x} - \beta {{\log }_e}(1 + x) + \gamma {x^2}{e^{ - x}}} \over {x{{\sin }^2}x}} = 10,\alpha ,\beta ,\gamma \in R$$, then the value of $$\alpha$$ + $$\beta$$ + $$\gamma$$ is _____________.
Answer
3
Explanation
$$\mathop {\lim }\limits_{x \to 0} {{\alpha x\left( {1 + x + {{{x^2}} \over x}} \right) - \beta \left( {x - {{{x^2}} \over 2} + {{{x^3}} \over 3}} \right) + \gamma {x^2}(1 - x)} \over {{x^3}}}$$
$$\mathop {\lim }\limits_{x \to 0} {{x(\alpha - \beta ) + {x^2}\left( {\alpha + {\beta \over 2} + \gamma } \right) + {x^3}\left( {{\alpha \over 2} - {\beta \over 3} - \gamma } \right)} \over {{x^3}}} = 10$$
For limit to exist
$$\alpha - \beta = 0,\alpha + {\beta \over 2} + \gamma = 0$$$${\alpha \over 2} - {\beta \over 3} - \gamma = 10$$ ..... (i)
$$\beta = \alpha ,\gamma = - 3{\alpha \over 2}$$
Put in (i)
$${\alpha \over 2} - {\alpha \over 3} + {{3\alpha } \over 2} = 10$$
$${\alpha \over 6} + {{3\alpha } \over 2} = 10 \Rightarrow {{\alpha + 9\alpha } \over 6} = 10$$
$$ \Rightarrow \alpha = 6$$
$$\alpha$$ = 6, $$\beta$$ = 6, $$\gamma$$ = $$-$$9
$$\alpha$$ + $$\beta$$ + $$\gamma$$ = 3
$$\mathop {\lim }\limits_{x \to 0} {{x(\alpha - \beta ) + {x^2}\left( {\alpha + {\beta \over 2} + \gamma } \right) + {x^3}\left( {{\alpha \over 2} - {\beta \over 3} - \gamma } \right)} \over {{x^3}}} = 10$$
For limit to exist
$$\alpha - \beta = 0,\alpha + {\beta \over 2} + \gamma = 0$$$${\alpha \over 2} - {\beta \over 3} - \gamma = 10$$ ..... (i)
$$\beta = \alpha ,\gamma = - 3{\alpha \over 2}$$
Put in (i)
$${\alpha \over 2} - {\alpha \over 3} + {{3\alpha } \over 2} = 10$$
$${\alpha \over 6} + {{3\alpha } \over 2} = 10 \Rightarrow {{\alpha + 9\alpha } \over 6} = 10$$
$$ \Rightarrow \alpha = 6$$
$$\alpha$$ = 6, $$\beta$$ = 6, $$\gamma$$ = $$-$$9
$$\alpha$$ + $$\beta$$ + $$\gamma$$ = 3
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