If A=[a_{i j}]_{2 \\times 2}, B=[b_{i j}]_{2 \\times2} such that a_{i j}=f(x^{2 i-j-2}) and b_{i j}=f(x^{2 i-j-1}) and A+B^{T}=(\table -x^{2} f(x+f(1)) , (1-{1}/{x^{4}}) f({1}/{x^{2}}) ; f(\frac{1}{x})(1-{1}/{x^{2}}) , -{1}/{x^{2}} f(\frac{1+f(1)}{x}) ), then \\int_{x} ^{1\/ x} f(t) d t is always equal to

JEE Advanced 2021 Full Test 3 Paper 1

PART – III

Mathematics

SECTION – A (One Options Correct Type)
This section contains 06 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE option is correct.

Question : 42 of 54

Marks: +1, -0

If A=[aij]2×2,B=[bij]2×2 such that aij=f(x2i−j−2) and bij=f(x2i−j−1) and

A+BT=(

−x2f(x+f(1))

(1−

)f(

)

)(1−

)

−

1+f(1)

)

, then

1∕x

∫

f(t)dt is always equal to

f(1)
f(1)−1
f(1)+1
None of these

Validate

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