Applications of Derivatives Test - 7

y = 5x² - 10x + 7
dy/dx = 10x - 10
(At x = 1) 10(1) - 10  
m1 = 0
As we know that slope, m1m2 = -1  
=>0(m2) = -1
m2 = -1/0 (which is not defined)

y = 1/(x-3)
dy/dx = d/dx(x-3)^-1
dy/dx = (-1) (x-3)^(-1-1) . d(x-3)/dx
dy/dx = -(x-3)^(-2)
dy/dx = - 1/(x-3)²
Given, slope = 2, dy/dx = 2
- 1/(x-3)² = 2
⇒-1 = 2(x-3)²
2(x-3)² = -1
(x-3)² = -½
We know that square of any number is always positive So, (x-3)² >0
(x-3)² = -½is not possible
No tangent to the curve has slope 2.

Clearly,  dx/dy =tan q = slope of normal = −cotq
Equation of normal at θ'is y −a(sinq −qcosq) = −cotq(x −a(cosq + qsinq)
= ysinq −asin² q + aqcosqsinq  
= −xcosq + acos² q + aqsinqcosq
= xcosq + ysinq = a

h²   = 4k  
slope of normal=−1/(dy/dx) = −2h
equation of normal(y −k)= −2h(x −h)
k = 2 + 2/h(1 −h)
(h² ) / 4 = 2 + 2/h (1 −h)
h = 2, k = 1
equation of line (y - 1)= -1(x - 2)
x + y = 3

y=ax³ +bx² +cx+5
(−z,0) lies on curve y=ax³ + bx² + cx + 5d
⇒0= −8a + 4b −2c + 5.....(1)
Also at (−2,0) curve touches x-axis ie dy/dx at (−2,0) is 0  
∴dy/dx=3ax² +2bx+c
dy/dx (−2,0)=0=12a = −4b + c....(2)
Since curve crosses y-axis at Q, x-coordinate of point Q must be zero  
⇒y = 0 + 0 + 0 + 5
⇒y = 5
∴coordinate of Q = (0,5)
According to Question  dy/dx =3 at Q(0,5)
⇒3= 0 + 0 + c
⇒c = 3
putting the value of c in eqn(1) &eq n(2)and gurthare solving we get  
a=−1/2 &b=-3/4
​∴Eqn of cover is y = −(x³ ) / 2 −3/4x²   + 3x + 5

If  p,q  and  r are the roots,
pq + qr + pr = b  and  pqr = 6  
Since  A.M.≥G.M . applying it to  3  numbers  pq,qr  and  pr  
(pq + qr + pr)/3 ≥(p² q² r² )^1/3  i.e.
(pq + qr + pr)/3 ≥(36)^1/3  
pq + qr + pr ≥3 (36)^1/3  
So minimum value of  b  is  (36)^1/3

y = x³ + 2x + 6
Slope of tangent = m1 = dy/dx = 3x³ + 2  
Slope of perpendicular line = m2 = −1/14
m1 . m2  = -1
m1 = 14
3x² + 2 = 14
x = ±2
Therefore the curve has tangents at x = 2 and x = -2
and these points also lie on the given curve
Equation of tangent - y = 14x + c  
Coordinates of points of tangency
At x = 2 , y = 23 + 2(2) + 6 = 18  
At x = -2 , y = (-2)3 + 2(-2) + 6 = -6
18 = 14(2) + c
c = -10  
-6 = 14(-2) + c
c = 22
Equation of tangent - y = 14x - 10 and y = 14x + 22
Hence, 14x - y +22 = 0

Let P(α, β) be any point on the curve

Now, the equation of the tangent at P is



Hence, the point of contacts are
(2, 2 + 2 √3) and (2, 2 –2 √3)
Slope of the tangents are 2 √3 , –2 √3
Hence, the equations of tangents are
y –(2 + 2 √3) = 2 √3 (x –2)
and
y –(2 –2 √3) = –2 √3 (x –2)