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

Let x1 and x2 be the real roots of the equation $${x^2} - (k - 2)x + ({k^2} + 3k + 5) = 0$$, then maximum value of $$x_1^2 + x_2^2$$ is

A
19
B
22
C
18
D
17
2
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
$${{{\pi ^3}} \over 8}$$
3
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
$${{3i + 1} \over 2}$$
4
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, $$\pi$$)
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