Question

A region in the x-y plane is bounded by the curve $$y=\sqrt{\left( 25-x^{2}\right) }$$ and the line y=0. If the point (a,a+1) lies in the interior of the region, then
A) $a\in \left( -4,3\right) $
B) $a\in \left( -\infty ,-1\right) \cup \left( 3,\infty \right) $
C) $a\in \left( -1,3\right) $
D) None of these

Answer 1

Hint: In this question it is given that a region in the x-y plane is bounded by the curve $$y=\sqrt{\left( 25-x^{2}\right) }$$ and the line y=0. If the point (a,a+1) lies in the interior of the region, then we have to find the range for a.
The given equation $$y=\sqrt{\left( 25-x^{2}\right) }$$, which also can be written as, $x^{2}+y^{2}-5^{2}=0$, which is a equation of a circle.
So to find the solution we have to know that if any point (p,q) lies inside the circle $x^{2}+y^{2}-r^{2}=0$ then we can write, $p^{2}+q^{2}-r^{2}<0$ …………….(1).
So from here we get one condition of a and also the region is bounded by y = 0,i.e, from the region y > 0 we will get another condition of a, and by intersecting we will get our required solution.

Complete step-by-step solution:
Since it is given that point (a,a+1) lies inside the circle.
So by (1) we can write,
$$p^{2}+q^{2}-r^{2}<0$$
$$\Rightarrow a^{2}+\left( a+1\right)^{2} -5^{2}<0$$
$$\Rightarrow a^{2}+\left( a^{2}+2a+1\right) -25<0$$
$$\Rightarrow 2a^{2}+2a-24<0$$
$$\Rightarrow 2\left( a^{2}+a-12\right) <0$$
$$\Rightarrow \left( a^{2}+a-12\right) <0$$ [dividing both side by 2]
Now by middle term factorisation,
$$\Rightarrow a^{2}+4a-3a-12<0$$
$$\Rightarrow a\left( a+4\right) -3\left( a+4\right) <0$$
Now by taking (a+4) common, we can write,
$$\Rightarrow \left( a-3\right) \left( a+4\right) <0$$........(2)
Whenever multiplication of two terms less than zero, i.e ab<0, then
Either a < 0 and b > 0,
Or, a > 0 and b < 0,

So by this we can write (2) as,
Either, (a-3) < 0 and (a+4) > 0
Which implies, a < 3 and a > -4.
So the range of ‘a’ is -4 < a <3.
Also this can be written as $$a\in \left( -4,3\right) $$

Or, (a-3) > 0 and (a+4) < 0
Which implies, a > 3 and a < -4
So there is no common region by this above condition.
So have got one condition of a i.e, $$a\in \left( -4,3\right) $$

Now another condition is that the region is also bounded by y=0.
So the region is y > 0
Which gives (a+1) > 0 , [since we are finding conditions for ‘a’, where the coordinate is (a,a+1) , i.e y = a+1]
Which implies a > -1.
So another condition is $a\in \left( -1,\infty \right) $
So we get $$a\in \left( -4,3\right) \cap \left( -1,\infty \right) $$
$$\therefore a\in \left( -1,3\right) $$.
Which is our required solution.
Hence the correct option is option C.

Note: To solve this type of question you should keep in mind that, choose proper conditions according to the position of a point.
And also while solving any inequation always remember that whenever multiplication of two terms less than zero, i.e ab < 0, then either a < 0 and b > 0, Or, a > 0 and b < 0 and for sets or intervals the meaning of “and” is intersection.