Given equation can be rearranged as
x(x⁶ + 3x⁴ - 13x² - 15) = 0
Clearly x = 0 is one of the roots and the other part can be observed by replacing x² = t, from which we get:
t³ + 3t² - 13t - 15 = 0
⇒ (t - 3)(t² + 6t + 5) = 0
So, t = 3, t = -1, t = -5
Now we are getting:
x² = 3, x² = -1, x² = -5
⇒ x = ±√3, x = ±i, x = ±√5i
From the given condition:
|α₁| ≥ |α₂| ≥ .... ≥ |α₆|
We can clearly say that:
|α₁| = 0 and |α₂| = √5 = |α₅| and |α₄| = √3
= |α₃| and |α₂| = 1 = |α₁|
So we can have:
α₁ = √5i, α₂ = -√5i, α₃ = √3i, α₄ = √3, α₅ = i, α₆ = -i
Hence
α₁ - α₂ - α₃ + α₄ + α₅ + α₆ = 1 - (-3) + 5 = 9
Let \( M \) be a \( 7 \times 7 \) matrix with entries in \( \mathbb{R} \) and having the characteristic polynomial \[ c_M(x) = (x - 1)^\alpha (x - 2)^\beta (x - 3)^2, \] where \( \alpha>\beta \). Let \( {rank}(M - I_7) = {rank}(M - 2I_7) = {rank}(M - 3I_7) = 5 \), where \( I_7 \) is the \( 7 \times 7 \) identity matrix.
If \( m_M(x) \) is the minimal polynomial of \( M \), then \( m_M(5) \) is equal to __________ (in integer).
In the given figure, graph of polynomial \(p(x)\) is shown. Number of zeroes of \(p(x)\) is

In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
