AP ICET 2017 Shift 2 Paper

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Question : 133 of 200

Marks: +1, -0

The rank of the matrix

(\table1,2,0,-1;3,4,1,2;-2,3,2,5)

is

1
2
3
4
None of the above

Solution:

Given :

\lim↙{x→2a}{√{x}-√{2a}+√{x-2a}}/{√{x^{2}-4a^{2}}}

For

x=2a

, it is of

0/0

form. Applying L’ Hospital’s rule,

=\lim↙{x→2a}{{1}/{2√{x}}+{1}/{2√{x-2a}}}{2x}/{2√{x^{2}-4a^{2}}}

=\lim↙{x→2a}{√{x-2a}+√{x}}/{2√{x}√{x-2a}}×{√{x^{2}-4a}}/{x}

=\lim↙{x→2a};{(√{x-2a}+√{x})(√{x+2a})}/{2x√{x}}

={(√{2a})(√{4a})}/{2×2a√{2a}}={2√{a}}/{2×2a}={1}/{2√{a}}

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