JEE Mains · Maths · STD 11 - 9. straight line
Let the points \(\left(\frac{11}{2}, \alpha\right)\) lie on or inside the triangle with sides \(x+y=11, x+2 y=16\) and \(2 x+3 y=29\). Then the product of the smallest and the largest values of \(\alpha\) is equal to :
- A 44
- B 22
- C 33
- D 55
Answer & Solution
Correct Answer
(C) 33
Step-by-step Solution
Detailed explanation
Clearly, \(x=\frac{11}{2}\) intersect \(x+y-11=0\) at \(\left(\frac{11}{2}, \frac{11}{2}\right)\) and \(2 x+3 y-29=0\) at \(\left(\frac{11}{2}, 6\right) \Rightarrow \alpha=\left[\frac{11}{2}, 6\right]\) \(\alpha_{\min } \cdot \alpha_{\max }=\frac{11}{2} \cdot 6=33\)
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