1
BITSAT 2023
MCQ (Single Correct Answer)
+3
-1

Let $$\alpha, \beta$$ be the roots of the equation $$x^2-p x+r=0$$ and $$\frac{\alpha}{2}, 2 \beta$$ be the roots of the equation $$x^2-q x+r=0$$. Then, the value of $$r$$ is equal to

A
$$\frac{2}{9}(p-q)(2 q-p)$$
B
$$\frac{2}{9}(q-2 p)(2 q-p)$$
C
$$\frac{2}{9}(q-p)(2 p-q)$$
D
$$\frac{2}{9}(2 p-q)(2 q-p)$$
2
BITSAT 2023
MCQ (Single Correct Answer)
+3
-1

$$ \text { If } A=\left[\begin{array}{cc} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{array}\right] \text {, then } A(\operatorname{adj} A)^{-1} \text { equals to } $$

A
$$ \left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right] $$
B
$$ \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] $$
C
$$ \left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] $$
D
$$ \left[\begin{array}{cc} 0 & -1 \\ -1 & 0 \end{array}\right] $$
3
BITSAT 2023
MCQ (Single Correct Answer)
+3
-1

$$ \text { The value of } \lim _\limits{x \rightarrow 0} \frac{(27+x)^{1 / 3}-3}{9-(27+x)^{2 / 3}} \text { equals to } $$

A
$$-\frac{1}{3}$$
B
$$\frac{1}{6}$$
C
$$-\frac{1}{6}$$
D
$$\frac{1}{3}$$
4
BITSAT 2023
MCQ (Single Correct Answer)
+3
-1

If a tangent to the circle $$x^2+y^2=1$$ intersect the co-ordinate axes at distinct points $$P$$ and $$Q$$, then the locus of the mid-point of $$P Q$$ is

A
$$x^2+y^2-2 x y=0$$
B
$$x^2+y^2-2 x^2 y^2=0$$
C
$$x^2+y^2-4 x^2 y^2=0$$
D
$$x^2+y^2-16 x^2 y^2=0$$
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