MCQ
If the point $(\lambda,\ \lambda+1)$ lies inside the region bounded by the curve $\text{x}=\sqrt{25-\text{y}^2}$ and y-axis, then $\lambda$ belongs to the interval:
- A$(-1,\ 3)$
- B$(-4,\ 3)$
- C$(-\infty,\ -4)\cup(3,\ \infty)$
- DNone of these
Solution:
The given equation of the curve is x2 + y2 = 25
Since $(\lambda,\ \lambda+1)$ lies inside the region bounded by the curve x2 + y2 = 25 and the y-axis, we have:
$\lambda^2+(\lambda+1)^2<25,$ provided $\lambda+1>0$
$\Rightarrow\lambda^2+\lambda^2+12\lambda<25,\ \lambda>-1$
$\Rightarrow2\lambda^2+2\lambda-24<0,\ \lambda>-1$
$\Rightarrow\lambda^2+\lambda-12<0,\ \lambda>-1$
$\Rightarrow(\lambda-3)(\lambda+4)<0,\ \lambda>-1$
$\Rightarrow-4<\lambda<3,\ \lambda>-1$
$\Rightarrow\lambda\in(-1,\ 3)$
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Events A and B are said to be mutually exclusive if:
Let F1 be the set of parallelograms, F2 the set of rectangles, F3 the set of rhombuses, F4 the set of squares and F5 the set of trapeziums in a plane. Then F1 may be equal to,
$\text{F}_2\cap\text{F}_3$
$\text{F}_3\cap\text{F}_4$
$\text{F}_2\cup\text{F}_5$
$\text{F}_2\cup\text{F}_3\cup\text{F}_4\cup\text{F}_1$
$\frac{5}{4}$
$\frac{4}{5}$
$1$
$0$