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NDA 2026
Exam
NDA
Year
2026
Subject
Mathematics
Chapter
Determinants

NDA 2026 · Mathematics · Determinants

Let p=(x+y+z) and q=xyz . If vmatrix x and 1 and 1 1 and y and 1 1...

Let p=(x+y+z)p=(x+y+z) and q=xyzq=xyz. If ∣x and 1 and 11 and y and 11 and 1 and z∣\begin{vmatrix} x \text{ and } 1 \text{ and } 1 \\ 1 \text{ and } y \text{ and } 1 \\ 1 \text{ and } 1 \text{ and } z \end{vmatrix} is positive, then which one of the following is correct ?

Options

  1. Aq>pq > p
  2. Bq+1>pq+1 > p
  3. Cq+2>pq+2 > p
  4. Dq+2≥pq+2\geq p

Correct answer

C. q+2>pq+2 > p

Step-by-step solution

Expanding the given determinant along the first row and setting it greater than zero:

x(yz−1)−1(z−1)+1(1−y)>0x(yz - 1) - 1(z - 1) + 1(1 - y) > 0

xyz−x−z+1+1−y>0xyz - x - z + 1 + 1 - y > 0

xyz−(x+y+z)+2>0xyz - (x + y + z) + 2 > 0

Substituting p=x+y+zp = x + y + z and q=xyzq = xyz:

q−p+2>0q - p + 2 > 0

q+2>pq + 2 > p

Answer: q+2>pq+2 > p
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