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Please solve RD Sharma Class 12 Chapter 29 Linear Programming Exercise Multiple Choice Question Question 3 maths textbook solution.

Answers (1)

Answer: (d) \{(x, y): y \geq 2, y \leq 4\}

Hint:

For convex set, all points should be inside the set.

Given:

\begin{aligned} &A=x^{2}+y^{2} \geq 1, B=y^{2} \geq x, C=3 x^{2}+4 y^{2} \geq 5 \\ &D=y \geq 2, y \leq 4 \end{aligned}

Solution :

\begin{aligned} &A=x^{2}+y^{2} \geq 1 \\ &x^{2}+y^{2}=1 \end{aligned}

 At (0,0) we have0\geq 1 , which is wrong. So the region of set A will not contain the origin.

Here, its all points are outside the set. So option (a) is incorrect

\begin{aligned} &B=y^{2} \geq x \\ &y^{2}=x \end{aligned}

At (0,0) we have 0\geq 0, which is wrong. So the region of set B will not contain the origin.

We observe that option (b) is also incorrect because, it’s all points are outside the set

\begin{aligned} &C=3 x^{2}+4 y^{2} \geq 5 \\ &\Rightarrow \frac{x^{2}}{\frac{5}{3}}+\frac{y^{2}}{\frac{5}{4}} \geq 1 \end{aligned}

At (0,0) we have  0\geq 1, which is wrong. So, the region of set C will not contain the origin. We observe that option (c) also incorrect. Because it fails the condition of convex set

Now, D=y \geq 2, y \leq 4

It is only possible when y=3

So, the correct option is (d)

 

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