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← JEE Main 2014 · 11 Apr · Shift 1
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JEE Main 2014 · 11 Apr · Shift 1, Mathematics question 70

If
lim⁡x→2tan⁡(x−2){x2+(k−2)x−2k}x2−4x+4=5,\lim_{x\to2}\frac{\tan(x-2)\left\{x^2+(k-2)x-2k\right\}}{x^2-4x+4}=5,
then kk is equal to :

Choose one option
Answer and worked solution

Official answer: D

x2+(k−2)x−2k=(x−2)(x+k)x^2+(k-2)x-2k=(x-2)(x+k), so the expression is
tan⁡(x−2)x−2 (x+k)→2+k.\frac{\tan(x-2)}{x-2}\,(x+k)\to2+k.
Setting 2+k=52+k=5 gives k=3k=3.

Topics:Limits
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