Question

If the sum of the zeros of the polynomial $f(x)=2x^{3}-3kx^{2}+4x-5$ is 6, then the value of k is
(a) 2
(b) 4
(c) -2
(d) -4

Answer 1

SOLUTION :

The correct option is (b) : 4 .

Let α  and β are the zeroes of the  polynomial f(x) = 2x³ - 3kx² + 4x - 5  

Given : Sum of the zeroes (α + β + γ) = 6

On comparing with ax³ + bx² + cx + d ,

a = 2, b= -3k , c = 4 , d = - 5

Sum of the zeroes = −coefficient of x² / coefficient of x³

α + β + γ = −b/a

6  = -(- 3k)/2  

6 × 2 = 3k

12 = 3k  

k = 12/3  

k = 4

Hence, the value of k is 4 .

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Answer 2

Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic and (c) non periodic motion? Give period for each case of periodic motion (ω is any positive constant) a. sign ωt - cos ωt b. sin² ωt c. 3 Cos (π/4 - ωt) d. cos ωt + cos 3ωt + cos 5ωt e. exp(-ω²t²) f. 1 + ωt + ω²t²For any arbitrary motion in space, which of the following relations are true: a. $V_{average}= (\frac{1}{2})[v(t_1)+v(t_2)]$ b. $V_{average}=\frac {[r(t_2)-r(t_1)]} {(t_2-t_1)}$ c. v (t) = v (0) + at d. r (t) = r (0) + v (0) t + (1/2) at² e. $a_{average}=\frac {[v(t_2)-v(t_1)]} {(t_2-t_1)}$ (The 'average' stands for average of the quantity over the time interval $t_1 to t_2$)The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For large angles of oscillation, a more involved analysis shows that T is greater than $2\pi \sqrt{\frac {I}{g}}$. Think of a quantitative argument to appreciate this result