Under which one of the following conditions does the equation (\cos\beta-1)x^{2}+(\cos\beta)x+\sin \beta=0 in x have a real root for \beta\in[0,\pi] ?
- A. 1-\cos \beta \lt 0
- B. 1-\cos \beta \leq 0
- C. 1-\cos \beta \gt 0
- D. 1-\cos \beta \geq 0 ✓
Correct Answer: D. 1-\cos \beta \geq 0
Explanation
For the equation to have real roots, the condition depends on the nature of \cos \beta. Since -1 \leq \cos \beta \leq 1 for all real \beta, the expression 1 - \cos \beta is <strong>ALWAYS</strong> non-negative. Thus, 1 - \cos \beta \geq 0 is universally true, covering the cases whether the equation is quadratic or linear (when \cos \beta = 1).
Related questions on Algebra
- How many four-digit natural numbers are there such that <strong>ALL</strong> of the digits are odd?
- What is \sum_{r=0}^{n}2^{r}C(n,r) equal to ?
- If different permutations of the letters of the word 'MATHEMATICS' are listed as in a dictionary, how many words (with or without meaning) a...
- Consider the following statements : 1. If f is the subset of Z\times Z defined by f=\{(xy,x-y);x,y\in Z\}, then f is a function from...
- For how many quadratic equations, the sum of roots is equal to the product of roots?