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 velocity-time graph of an object moving along a straight line is shown in the figure. What is the distance covered by the object between \( t = 0 \) to \( t = 4s \)?
A bob of mass \(m\) is suspended at a point \(O\) by a light string of length \(l\) and left to perform vertical motion (circular) as shown in the figure. Initially, by applying horizontal velocity \(v_0\) at the point ‘A’, the string becomes slack when the bob reaches at the point ‘D’. The ratio of the kinetic energy of the bob at the points B and C is: