Question

Important Question for jee mains.
mathamatics :- Practice set 1 ​

Answer 1

Answer:

Answer in the attachment

Answer 2

$\large\underline{\sf{Solution-}}$

It is given that

$\rm :\longmapsto\: \alpha , \beta \: are \: roots \: of \: a {x}^{2} + bx + c = 0$

$\bf\implies \:a {x}^{2} + bx + c = a(x - \alpha )(x - \beta ) - - (1)$

Now,

Consider,

$\rm :\longmapsto\:\displaystyle\lim_{x \to \alpha } {\bigg(1 + a {x}^{2} + bx + c \bigg) }^{\dfrac{1}{x - \alpha } }$

can be rewritten as

$\rm \: = \:\displaystyle\lim_{x \to \alpha } {\bigg(1 + a (x - \alpha )(x - \beta ) \bigg) }^{\dfrac{1}{x - \alpha } }$

can be further rewritten as

$\rm \: = \:\displaystyle\lim_{x \to \alpha } {\bigg(1 + a (x - \alpha )(x - \beta ) \bigg) }^{\dfrac{a(x - \alpha )(x - \beta )}{a(x - \alpha )(x - \beta )(x - \alpha)} }$

$\rm \: = \:\displaystyle\lim_{x \to \alpha } {\bigg(1 + a (x - \alpha )(x - \beta ) \bigg) }^{\dfrac{a(x - \beta )}{a(x - \alpha )(x - \beta )} }$

We know,

$\boxed{ \sf{ \:\displaystyle\lim_{x \to 0} {(1 + x)}^{ \frac{1}{x} } = e}}$

So, using this result we get

$\rm \:  =  \:  \: {e} \: \: ^{\displaystyle\lim_{x \to \alpha }a(x - \beta )}$

$\rm \:  =  \:  \: {e} \: ^{a( \alpha - \beta )}$

Hence,

$\rm :\longmapsto\:\displaystyle\lim_{x \to \alpha } {\bigg(1 + a {x}^{2} + bx + c \bigg) }^{\dfrac{1}{x - \alpha } }   =  \:  \: {e} \: ^{a( \alpha - \beta )}$

Additional Information :-

$\boxed{ \sf{ \:\displaystyle\lim_{x \to 0} \frac{sinx}{x} = 1}}$

$\boxed{ \sf{ \:\displaystyle\lim_{x \to 0} \frac{tanx}{x} = 1}}$

$\boxed{ \sf{ \:\displaystyle\lim_{x \to 0} \frac{tan^{ - 1} x}{x} = 1}}$

$\boxed{ \sf{ \:\displaystyle\lim_{x \to 0} \frac{sin^{ - 1} x}{x} = 1}}$

$\boxed{ \sf{ \:\displaystyle\lim_{x \to 0} \frac{log(1 + x)}{x} = 1}}$

$\boxed{ \sf{ \:\displaystyle\lim_{x \to 0} \frac{ {e}^{x} - 1}{x} = 1}}$

$\boxed{ \sf{ \:\displaystyle\lim_{x \to 0} \frac{ {a}^{x} - 1}{x} = loga}}$