If limx- 2tan x - 2x2 - k - 2x - 2kx2 - 4x - 4 - 5- then k is equal to

The correct option is D 3 lim_x→ 2tan(x 2)(x^2+(k 2)x 2k)(x 2)(x 2)=5 lim_x→ 2tan(x 2)(x 2)(x^2+(k 2)x 2k)(x 2)=5 lim_x→ 21.(x^2+(k 2)x 2k) ...

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Standard XII

Mathematics

Implicit Differentiation

If limx→ 2t...

Question

If $lim x \to 2 \frac{tan (x - 2) {x^{2} + (k - 2) x - 2 k}}{x^{2} - 4 x + 4} = 5$ , then $k$ is equal to

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Solution

The correct option is D $3$
${lim}_{x \to 2} \frac{t a n (x - 2) (x^{2} + (k - 2) x - 2 k)}{(x - 2) (x - 2)} = 5$
${lim}_{x \to 2} \frac{\frac{t a n (x - 2)}{(x - 2)} (x^{2} + (k - 2) x - 2 k)}{(x - 2)} = 5$
${lim}_{x \to 2} \frac{1. (x^{2} + (k - 2) x - 2 k)}{(x - 2)} = 5$
0/0 form,applying D'l Hospital rule
${lim}_{x \to 2} \frac{(2 x + (k - 2))}{(1)} = 5$
$2.2 + (k - 2) = 5$
$4 + k - 2 = 5$
$k = 5 - 2 = 3$
option D is correct.

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If limx→2tan(x−2){x2+(k−2)x−2k}x2−4x+4=5, then k is equal to

The correct option is D 3limx→2tan(x−2)(x2+(k−2)x−2k)(x−2)(x−2)=5limx→2tan(x−2)(x−2)(x2+(k−2)x−2k)(x−2)=5limx→21.(x2+(k−2)x−2k)(x−2)=50/0 form,applying D'l Hospital rulelimx→2(2x+(k−2))(1)=52.2+(k−2)=54+k−2=5k=5−2=3option D is correct.

If $lim x \to 2 \frac{tan (x - 2) {x^{2} + (k - 2) x - 2 k}}{x^{2} - 4 x + 4} = 5$ , then $k$ is equal to