Given system of equations
\(x + y + az = 2 \)…(i)
\(3x + y + z = 4\) …(ii)
\(x + 2z = 1\) …(iii)
Solving (i), (ii) and (iii), we get
\(x = 1,\) \(y = 1\) , \(z = 0\) (and for unique solution \(a ≠–3\))
Now, \((α, 1), (1, α)\) and \((1, –1)\) are collinear
\(∴\) \(\begin{vmatrix} \alpha&1 & 1\\1&\alpha&1\\1&-1 &1 \end{vmatrix}=0\)
⇒ \(α(α + 1) – 1(0) + 1(–1 – α) = 0\)
\(⇒ α^2 – 1 = 0\)
\(∴ α = ±1\)
\(∴\) Sum of absolute values of \(α = 1 + 1 = 2\)
Hence, the correct option is (C): \(2\)
If the domain of the function \( f(x) = \frac{1}{\sqrt{3x + 10 - x^2}} + \frac{1}{\sqrt{x + |x|}} \) is \( (a, b) \), then \( (1 + a)^2 + b^2 \) is equal to:
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: