1
BITSAT 2020
MCQ (Single Correct Answer)
+3
-1

Circle centered at origin and having radius $$\pi$$ units is divided by the curve y = sin x in two parts. Then area of upper parts equals to

A
$${{{\pi ^2}} \over 2}$$
B
$${{{\pi ^3}} \over 4}$$
C
$${{{\pi ^3}} \over 2}$$
D
$${{{\pi ^3}} \over 8}$$
2
BITSAT 2020
MCQ (Single Correct Answer)
+3
-1

The root of the equation $$2(1 + i){x^2} - 4(2 - i)x - 5 - 3i = 0$$, where $$i = \sqrt { - 1} $$, which has greater modulus, is

A
$${{3 - 5i} \over 2}$$
B
$${{5 - 3i} \over 2}$$
C
$${{3 + i} \over 2}$$
D
$${{3i + 1} \over 2}$$
3
BITSAT 2020
MCQ (Single Correct Answer)
+3
-1

The equation $$(\cos \beta - 1){x^2} + (\cos \beta )x + \sin \beta = 0$$ in the variable x has real roots, then $$\beta$$ lies in the interval

A
(0, 2$$\pi$$)
B
($$-$$$$\pi$$, 0)
C
$$\left( { - {\pi \over 2},{\pi \over 2}} \right)$$
D
(0, $$\pi$$)
4
BITSAT 2020
MCQ (Single Correct Answer)
+3
-1

An ordered pair ($$\alpha$$, $$\beta$$) for which the system of linear $$(1 + \alpha )x + \beta y + z = 2$$, $$\alpha x + (1 + \beta )y + z = 3$$, $$\alpha x + \beta y + 2z = 2$$ has a unique solution.

A
(1, $$-$$3)
B
($$-$$3, 1)
C
(2, 4)
D
($$-$$4, 2)
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