If t is a real number and k=(t2 - t + 1/t2 + t + 1), then the system of equations 3x-y+4z=3 x+2y-3z=-2 6x+5y+kz=-3 for any allowable value of k, has

Question

If t is a real number and k=(t2 - t + 1/t2 + t + 1), then the system of equations 3x-y+4z=3 x+2y-3z=-2 6x+5y+kz=-3 for any allowable value of k, has (A) a unique solution (B) infinite solutions (C) no solution (D) 2 solutions

Tardigrade Admin · Accepted Answer

k=(t2 - t + 1/t2 + t + 1)⇒ t2(k - 1)+t(k + 1)+k-1=0 as t∈ R D≥ 0⇒ (k + 1)2-4(k - 1)2≥ 0 ⇒ k∈ [(1/3) , 3] for the given equation, Δ =| 3 -1 4 1 2 -3 6 5 k |=7(k + 5)>0 ∀ k∈ [(1/3) , 3] hence, unique solution

Q. If $t$ is a real number and $k = \frac{t ^{2} - t + 1}{t ^{2} + t + 1},$ then the system of equations

$3 x - y + 4 z = 3$

$x + 2 y - 3 z = - 2$

$6 x + 5 y + k z = - 3$ for any allowable value of $k,$ has

a unique solution

infinite solutions

no solution

$2$ solutions

Q. If t is a real number and k=t2+t+1t2−t+1​, then the system of equations 3x−y+4z=3 x+2y−3z=−2 6x+5y+kz=−3 for any allowable value of k, has

a unique solution

infinite solutions

no solution

2 solutions

k=t2+t+1t2−t+1​⇒t2(k−1)+t(k+1)+k−1=0 as t∈R D≥0⇒(k+1)2−4(k−1)2≥0 ⇒k∈[31​,3] for the given equation, Δ=∣∣​316​−125​4−3k​∣∣​=7(k+5)>0∀k∈[31​,3] hence, unique solution

Q. If $t$ is a real number and $k = \frac{t ^{2} - t + 1}{t ^{2} + t + 1},$ then the system of equations

$3 x - y + 4 z = 3$

$x + 2 y - 3 z = - 2$

$6 x + 5 y + k z = - 3$ for any allowable value of $k,$ has

$2$ solutions