Test: Difference of Parametric Functions - JEE MCQ

y = sin²(θ²+1)
v = θ²
dy/d(v) = dydθ/dvdθ
dy/dthη = sin²(V+1)
= 2sin(V+1)⋅cos(V+1)dv/dθ
= 2sin(θ²+1)cos(θ²+1)
= sin2(θ²+1).

Difference equation are the equations used in discrete time systems and difference equations are similar to the differential equation in continuous systems.

Together the equations x = at² and y = 2at (where t is the parameter) are called the parametric equations of the parabola y² = 4ax.

x = a(t+sin t) 
⟹dx/dt = a(1+cos t)
And y = a(1−cos t)
⟹dy/dt = a[0−(−sin t)]
=a sin t
Therefore, dy/dx = a sin t/(a(1+cos t))
= 2sin t/2 cos t/2)/(2cos^2 t/2)
=tan(t/2)
At π/2

tan(π/4) = 1

x = cos³θ
⇒ dx/dθ = a(3cos²θ)(−sinθ)
= −3sinθcos²θ
y = sin³θ
⇒dy/dθ = a(3sin²θ)(cosθ)
= 3sin²θcosθ
∴dy/dx = (dy/dθ)(dx/dθ)
= (3sin²θcosθ)/(−3asinθcos²θ)
= −sinθ/cosθ
= −tanθ

x = 20(cost + tsint)
differentiate x with respect to t,
dx/dt = 20{d(cost)/dt + d(tsint)/dt]
= 20[-sint + {t. d(sint)/dt + sint.dt/dt}]
= 20[ -sint + tcost + sint]
= -20t.cost
hence, dx/dt = -20t.cost -----(1)
 
y = 20(sint - tcost)
differentiate y with respect to t,
dy/dt = 20[d(sint)/dt - d(tcost)/dt ]
= 20[cost - {t.d(cost)/dt + cost.dt/dt}]
= 20[cost +tsint -cost]
= 20t.sint
hence, dy/dt = 20t.sint -------(2)
 
dividing equations (2) by (1),
dy/dx = 20t.sint/20t.cost 
 dy/dx = tant
now again differentiate with respect to x
d²y/dx² = sec²t. dt/dx ------(3)
now from equation (1),
dx/dt = 20t.cost
so, dt/dx =1/at.cost put it in equation (3),
e.g., d²y/dx² = sec²t. 1/20t.cost
d²y/dx² = sec³t/20t

If −1≤x≤1, then 0≤xsinπx≤1/2
​∴ f(x)=[xsinπx]=0, for −1≤x≤1
If 1<x<1+h, where h is a small positive real number, then
π<πx<π+πh
⇒−1<sinπx<0
⇒−1<xsinπx<0
∴ f(x)=[xsinπx]=−1 in the right neighbourhood of x=1.
Thus, f(x) is constant and equal to zero in [−1,1] and so f(x) is differentiable and hence continuous on (−1,1).

Differentiation of sinx³ with respect to x³ is
d(sin(x³)/dx = cosx³