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): \(2\)
The obtuse angle between lines \(2y = x + 1\) and \(y = 3x + 2\) is:
What is the general solution of the equation \( \cot\theta + \tan\theta = 2 \)?
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)), \] for all \( x, y \in \mathbb{R}. \)
If \( f'(0) = \frac{1}{2} \), then the value of \( 24f''\left( \frac{5\pi}{3} \right) \) is: