Question

Number of integral values of ‘b’ for which the origin and the point $$(1,1)$$lie on the same
side of the straight line $$a^{2}x+aby+1=0$$, for all $$a\in R-\{0\}$$ is
(a) 4
(b) 3
(c) 5
(d) None of these

Answer 1

Hint: Substitute the given points into the straight line equation, check whether it's greater than 0 and find the discriminant which will be less than 0 to satisfy the given equation.

The given equation of straight line is,
$$a^{2}x+aby+1=0$$
We know two given points $$(x_1,y_1)$$and $$(x_2,y_2)$$will lie on the same side of the line $$ax+by+c=0$$ if $$a{x}_1+b{y}_1+c$$ and
$$a{x}_2+b{y}_2+c$$ will have the same signs.
Now we will substitute $$(0,0)$$ in the given equation of line, we get
$$a^2(0)+ab(0)+1=1>0$$

Now as we need to find whether point $$(1,1)$$ lie on same side as the origin, so, we substitute the value
of point $$(1,1)$$ in line equation, then it should be greater than zero, i.e.,
$$a^2(1)+ab(1)+1>0..................(i)$$
This forms a quadratic equation.
We know in case of equation inequality, for any quadratic equation to be greater than zero, its
discriminant should be less than zero. Because if discriminant is less than zero, the roots of the equation
will be imaginary and the inequality of the quadratic equation will be fulfilled.
The standard form of the quadratic equation is $$a{x}^2+bx+c=0$$. Comparing this with equation (i), we get,
$$a=1,b=b,c=1$$
Discriminant formula is, $$D=b^2-4ac$$.
So the discriminant of equation (i) will be,
$$D=b^2-4(1)(1)$$
For the equation (i) to be greater than zero, its discriminant should be less than zero, i.e.,
$$b^2-4<0$$
$$b^2<4$$
Taking square on both sides, we get
$$b<\pm 2$$
So, we get the range as, $$b\in (-2,2)$$.
So, the number of integral values of ‘b’ are $$(-1,0,1)$$ .
Hence, the correct answer is option (b).

Note: The alternative method is if two given points A and B will lie on the same side of the line L, then
$$L_A.L_B>0$$
In this method also we will get the same answer as above.