1
BITSAT 2022
MCQ (Single Correct Answer)
+3
-1

If $$\left[ {\matrix{ 1 & { - \tan \theta } \cr {\tan \theta } & 1 \cr } } \right]{\left[ {\matrix{ 1 & {\tan \theta } \cr { - \tan \theta } & 1 \cr } } \right]^{ - 1}} = \left[ {\matrix{ a & { - b} \cr b & a \cr } } \right]$$, then

A
a = 1, b = 1
B
$$a = \sin 2\theta ,b = \cos 2\theta $$
C
$$a = \cos 2\theta ,b = \sin 2\theta $$
D
None of these
2
BITSAT 2022
MCQ (Single Correct Answer)
+3
-1

The value of $$\mathop {\lim }\limits_{x \to 0} {{{{(1 + x)}^{{1 \over x}}} - e + {1 \over 2}ex} \over {{x^2}}}$$ is

A
$${{11} \over {24}}e$$
B
$$ - {{11} \over {24}}e$$
C
$${e \over {24}}$$
D
None of these
3
BITSAT 2022
MCQ (Single Correct Answer)
+3
-1

The locus of the mid-point of the chord if contact of tangents drawn from points lying on the straight line $$4x - 5y = 20$$ to the circle $${x^2} + {y^2} = 9$$ is

A
$$20({x^2} + {y^2}) - 36x + 45y = 0$$
B
$$20({x^2} + {y^2}) + 36x - 45y = 0$$
C
$$36({x^2} + {y^2}) - 20x + 45y = 0$$
D
$$36({x^2} + {y^2}) + 20x - 45y = 0$$
4
BITSAT 2022
MCQ (Single Correct Answer)
+3
-1

Let $$f(x) = \int {{{{x^2}dx} \over {(1 + {x^2})(1 + \sqrt {1 + {x^2}} )}}} $$ and $$f(0) = 0$$, then the value of $$f(1)$$ be

A
$$\log (1 + \sqrt 2 )$$
B
$$\log (1 + \sqrt 2 ) - {\pi \over 4}$$
C
$$\log (1 + \sqrt 2 ) + {\pi \over 2}$$
D
None of these
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