JEE MAIN - Mathematics (2019 - 10th April Morning Slot - No. 20)
If $$\mathop {\lim }\limits_{x \to 1} {{{x^4} - 1} \over {x - 1}} = \mathop {\lim }\limits_{x \to k} {{{x^3} - {k^3}} \over {{x^2} - {k^2}}}$$, then k is :
$${3 \over 2}$$
$${8 \over 3}$$
$${4 \over 3}$$
$${3 \over 8}$$
Explanation
If $$\mathop {\lim }\limits_{x \to 1} {{{x^4} - 1} \over {x - 1}} = \mathop {\lim }\limits_{x \to K} \left( {{{{x^3} - {k^3}} \over {{x^2} - {k^2}}}} \right)$$
L路H路S路
$$\mathop {Lt}\limits_{x \to 1} {{{x^4} - 1} \over {x - 1}} = \left( {{0 \over 0}form} \right)$$
$$ \Rightarrow \mathop {Lt}\limits_{x \to 1} {{4{x^3}} \over 1} = 4$$
Now, $$\mathop {\lim }\limits_{x \to K} \left( {{{{x^3} - {k^3}} \over {{x^2} - {k^2}}}} \right)$$ = 4
$$ \Rightarrow \mathop {\lim }\limits_{x \to K} {{3{x^2}} \over {2x}} = 4$$
$$ \Rightarrow {3 \over 2}k = 4 \Rightarrow k = {8 \over 3}$$
L路H路S路
$$\mathop {Lt}\limits_{x \to 1} {{{x^4} - 1} \over {x - 1}} = \left( {{0 \over 0}form} \right)$$
$$ \Rightarrow \mathop {Lt}\limits_{x \to 1} {{4{x^3}} \over 1} = 4$$
Now, $$\mathop {\lim }\limits_{x \to K} \left( {{{{x^3} - {k^3}} \over {{x^2} - {k^2}}}} \right)$$ = 4
$$ \Rightarrow \mathop {\lim }\limits_{x \to K} {{3{x^2}} \over {2x}} = 4$$
$$ \Rightarrow {3 \over 2}k = 4 \Rightarrow k = {8 \over 3}$$
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