Question

Number of values of the parameter `alpha in [0, 2pi]` for which the quadratic function, `(sin alpha) x^2 + 2 cos alpha x + 1/2 (cos alpha +sin alpha)` is the square of a linear function is? (A) 2 (B) 3 (C) 4 (D) 1

Answer 1

See below.

Explanation:

If we know that the expression must be the square of a linear form then

`(sin alpha) x^2 + 2 cos alpha x + 1/2 (cos alpha +sin alpha) = (ax+b)^2`

then grouping coefficients we have

`(alpha^2-sin(alpha))x^2+(2ab-2cos alpha)x+b^2-1/2(sinalpha+cosalpha)=0`

so the condition is

`{(a^2-sin(alpha)=0),(ab-cos alpha=0),(b^2-1/2(sinalpha+cosalpha)=0):}`

This can be solved obtaining first the values for `a,b` and substituting.

We know that `a^2+b^2=sin alpha+1/(sin alpha+cos alpha)` and
`a^2b^2=cos^2 alpha` Now solving

`z^2-(a^2+b^2)z+a^2b^2=0`. Solving and substituting for `a^2= sinalpha` we obtain

`a = b = pm 1/root(4)(2), alpha = pi/4`
`a = pm sqrt(2)/root(4)(5), b = pm 1/(sqrt(2)root(4)(5)), alpha = pi-tan^-1(2)`