JEE MAIN - Mathematics (2019 - 9th April Evening Slot - No. 2)
If the system of equations 2x + 3y – z = 0, x + ky
– 2z = 0 and 2x – y + z = 0 has a non-trival solution
(x, y, z), then $${x \over y} + {y \over z} + {z \over x} + k$$
is equal to :-
-4
$${3 \over 4}$$
$${1 \over 2}$$
$$-{1 \over 4}$$
Explanation
Given 2x + 3y – z = 0,
x + ky – 2z = 0
2x – y + z = 0
For non trivial solution
$$\Delta = 0 \Rightarrow \left| {\matrix{ 2 & 3 & { - 1} \cr 1 & k & { - 2} \cr 2 & { - 1} & 1 \cr } } \right| = 0$$
$$ \Rightarrow k = {9 \over 2}$$
$$ \therefore $$ Equations are 2x + 3y – z = 0 ...(i)
2x – y + z = 0 ...(ii)
2x + 9y – 4z = 0 ...(iii)
By (i) – (ii) we get,
4y - 2z = 0
$$ \Rightarrow $$ 2y = z .......(iv)
$$ \Rightarrow $$ $${y \over z} = {1 \over 2}$$
From equation (i) and (iv)
2x + 3y - 2y = 0
$$ \Rightarrow $$ 2x + y = 0
$$ \Rightarrow $$ $${x \over y} = - {1 \over 2}$$
$$ \Rightarrow $$ $${x \over y} \times {y \over z} = - {1 \over 2} \times {1 \over 2} = - {1 \over 4}$$
$$ \Rightarrow $$ $${z \over x} = - 4$$
$$ \therefore $$ $${x \over y} + {y \over z} + {z \over x} + k = {{ - 1} \over 2} + {1 \over 2} - 4 + {9 \over 2}$$ = $${1 \over 2}$$
x + ky – 2z = 0
2x – y + z = 0
For non trivial solution
$$\Delta = 0 \Rightarrow \left| {\matrix{ 2 & 3 & { - 1} \cr 1 & k & { - 2} \cr 2 & { - 1} & 1 \cr } } \right| = 0$$
$$ \Rightarrow k = {9 \over 2}$$
$$ \therefore $$ Equations are 2x + 3y – z = 0 ...(i)
2x – y + z = 0 ...(ii)
2x + 9y – 4z = 0 ...(iii)
By (i) – (ii) we get,
4y - 2z = 0
$$ \Rightarrow $$ 2y = z .......(iv)
$$ \Rightarrow $$ $${y \over z} = {1 \over 2}$$
From equation (i) and (iv)
2x + 3y - 2y = 0
$$ \Rightarrow $$ 2x + y = 0
$$ \Rightarrow $$ $${x \over y} = - {1 \over 2}$$
$$ \Rightarrow $$ $${x \over y} \times {y \over z} = - {1 \over 2} \times {1 \over 2} = - {1 \over 4}$$
$$ \Rightarrow $$ $${z \over x} = - 4$$
$$ \therefore $$ $${x \over y} + {y \over z} + {z \over x} + k = {{ - 1} \over 2} + {1 \over 2} - 4 + {9 \over 2}$$ = $${1 \over 2}$$
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