1. If the instantaneous rate of change at t = 1, then what will the rate of change be for the function f (t) =te-t + 9?
- f′(t) = te–t(–1)+ e–t
⇒ f′(1)=–e–1 + e–1
=0
2. The function f (x) = x2 – 2x is supposed to strictly decrease in the interval
- f ‘ (x ) = 2x – 2 = 2 ( x – 1) <0 if x < 1, i.e. x x ∈(–∞,1)
Thus, f is strictly decreasing in (–∞,1)
3. What will be the approximate value of (3.02) where f(x) = 3x2 + 5x + 3.
- x=3,Δx=0.02
f(x + Δx) = f(x) + f ′ (x) Δx
f(x+Δx)=(3x2 +5x+3)+(6x+5)×0.02
After putting, x=3,Δx=0.02
f (3.02) = {3 (9) + 5 (3) +3} + {6 (3) + 5} × 0.02 = 45 + 0.46
f (3.02) = 45.46
4. Considering a particle moving along the curve 6y = x3 + 2. Now, given the information, what will be the expected points on the curve at which the y-coordinate is varying 8 times as fast as the x – coordinate.
Given curve is 6y = x3 + 2 …(i)
So, 6dy/dt =3x2 dx/dt
⇒6×8dx/dt =3x2 dx/dt
⇒16=x2
⇒x=±4
Now, putting the value of x in equation (1)
When x = 4
6y = ( 4 )3 + 2
⇒ 6y = 64 + 2
⇒ 6y = 66
∴y=66/6 =11
So, point is (4, 11)
Now, When x = – 4
6y = ( – 4}3 + 2
= – 64 + 2
∴ y = –62/6 = −31/3
Thus, the point is (–4, –31/3)
5. What is the equation of normal to the curve y = tan x at (0, 0)?
- x + y = 0
You can find other similar questions on Application of Derivatives Class 12 Ncert Solutions that need extra attention, problem-solving skills, and a foundation of clear concepts.