Q50 / 90 · NTA Abhyas (JEE Main) 2020 · Mathematics · Limits

The value of limx→π2⁡11cos2⁡x+21cos2⁡x+.………+101cos2⁡xcos2⁡x is equal to

Correct answer: 10

Solution

Let, L=limx→π2⁡11cos2⁡x+21cos2⁡x+.………+101cos2⁡xcos2⁡x

Putting 1cos2⁡x=y, we get,
L=limy→∞⁡1y+2y+.…….+10y1y
=limy→∞10110y+210y+.…….+1010y1y
=100+0+.……+10=10

Hence, the correct answer is 10.

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