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JEE Main 2005
Exam
JEE Main
Year
2005
Subject
Mathematics
Chapter
Determinants

JEE Main 2005 · Mathematics · Determinants

If a^2+b^2+c^2=-2 and f(x)=| array ccc 1+a^2 x & (1+b^2) x &...

If a2+b2+c2=−2a^2+b^2+c^2=-2 and f(x)=\left|1+a2x(1+b2)x(1+c2)x(1+a2)x1+b2x(1+c2)x(1+a2)x(1+b2)x1+c2x\begin{array}{ccc}1+a^2 x & \left(1+b^2\right) x & \left(1+c^2\right) x \\ \left(1+a^2\right) x & 1+b^2 x & \left(1+c^2\right) x \\ \left(1+a^2\right) x & \left(1+b^2\right) x & 1+c^2 x\end{array}\right| then f(x)f(x) is aa polynomial of degree

Options

  1. A1
  2. B0
  3. C3
  4. D2

Correct answer

D. 2

Step-by-step solution

f(x)=∣1+(a2+b2+c2+2)x(1+b2)x(1+c2)x1+(a2+b2+c2+2)x1+b2x(1+c2)x1+(a2+b2+c2+2)x(1+b2)x1+c2x∣, Applying C1→C1+C2+C3=∣1(1+b2)x(1+c2)x11+b2x(1+c2)x1(1+b2)x1+c2x∣∵a2+b2+c2+2=0f(x)=∣0x−1001−xx−11(1+b2)x1+cx∣; Applying R1→R1−R2,R2→R2−R3f(x)=(x−1)2\begin{aligned}& f(x)=\left|\begin{array}{ccc}1+\left(a^2+b²+c²+2\right) x & \left(1+b²\right) x & \left(1+c²\right) x \\ 1+\left(a^2+b²+c²+2\right) x & 1+b² x & \left(1+c²\right) x \\ 1+\left(a^2+b²+c²+2\right) x & \left(1+b²\right) x & 1+c² x \end{array}\right| \text{, Applying }C_1 → C_1+C_2+C_3 \\ & =\left|\begin{array}{ccc}1 & \left(1+b²\right) x & \left(1+c²\right) x \\ 1 & 1+b² x & \left(1+c²\right) x \\ 1 & \left(1+b²\right) x & 1+c² x \end{array}\right| \because a²+b²+c²+2=0 \\ & f(x)=\left|\begin{array}{ccc}0 & x-1 & 0 \\ 0 & 1-x & x-1 \\ 1 & \left(1+b²\right) x & 1+c x \end{array}\right| ; \text{ Applying }R_1 → R_1-R_2, R_2 → R_2-R_3 \\ & f(x)=(x-1)^2 \end{aligned}Hence degree =2=2.
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