JEE Advanced 2016 · Paper 2 · Mathematics Q49

← JEE Advanced 2016 · Paper 2

MathematicsMultiple correct+4 / -2Question 49 of 54

Let $a,\lambda,\mu\in\mathbb R$ . Consider the system of linear equations
$ax+2y=\lambda$
$3x-2y=\mu$
Which of the following statement(s) is(are) correct?

a=-3

, then the system has infinitely many solutions for all values of

\lambda

and

\mu

a\ne-3

, then the system has a unique solution for all values of

\lambda

and

\mu

\lambda+\mu=0

, then the system has infinitely many solutions for

a=-3

\lambda+\mu\ne0

, then the system has no solution for

a=-3

Answer and worked solution

Official answer: B, C, D

Determinant. $\begin{vmatrix}a&2\\3&-2\end{vmatrix}=-2a-6$ .

(B) For $a\ne-3$ it is non-zero, so the solution is unique. True.
For $a=-3$ : adding the equations gives $0=\lambda+\mu$ $0 = λ + μ$ .
- (C) If $\lambda+\mu=0$ , there are infinitely many solutions. True.
- (D) If $\lambda+\mu\ne0$ , there are none. True.
- (A) is false.

(B), (C), (D)

Topics:Systems of linear equations

← Previous question Next question →

JEE Advanced 2016 · Paper 2, Mathematics question 49