Quadratic and Trigonometry for JEE #1

Algebra Level 5

The number of integers $b$ that satisfy the following condition is $n:$

The equation $\mid \sin x \mid ^{2} + \mid \sin x \mid + b =0$ has two distinct real roots in the interval [0, $\pi$ ].

Find $n^2 + n$ .

The answer is 6.

1 solution

Tom Engelsman
Jun 19, 2016

Let u = |sin(x)| in the original quadratic equation such that:

u = [-1 (+/-) sqrt(1 - 4b)] / 2

are the corresponding roots (via the Quadratic Formula). If the roots are to be real and distinct, then we require the discriminant to be positive, or b < 1/4. Replacing back |sin(x)| for u gives:

|sin(x)| = [-1 (+/-) sqrt(1 - 4b)] / 2 (i)

however the absolute value is strictly non-negative, which eliminates the negative root and leaves us with:

|sin(x)| = [-1 + sqrt(1 - 4b)] / 2 (ii)

and 0 <= |sin(x)| <= 1 for all x, hence we now obtain the inequalities:

0 <= [-1 + sqrt(1 - 4b)] / 2 <= 1;

or 1 <= sqrt(1 - 4b) <= 3;

or 1 <= 1 - 4b <= 9;

or -8 <= 4b <= 0;

or -2 <= b <= 0.

thus b = {0, -1, -2} for integer values. At b = 0, the original quadratic equation has distinct roots at x = 0 & pi. At b = -1, we get distinct roots at x = arcsin [ (-1+sqrt(5))/2 ] = arcsin (0.618) = 0.212 pi & 0.788 pi. At b = -2, we obtain a double root at x = pi/2 which is disregarded.

So n = 2 integers, and n^2 + n = 6.

Quadratic and Trigonometry for JEE #1

The answer is 6.

1 solution

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