Advertisements
Advertisements
Question
Slope of the tangent to the curve y = 6 – x2 at (2, 2) is ______.
Options
4
–4
–2
–1
Solution
Slope of the tangent to the curve y = 6 – x2 at (2, 2) is –4.
APPEARS IN
RELATED QUESTIONS
Find the derivative of the following function from first principle.
(x – 1) (x – 2)
Find the derivative of the following function from first principle:
−x
Find the derivative of the following function from first principle:
(–x)–1
Find the derivative of the following function from first principle:
`cos (x - pi/8)`
Find the equation of tangent and normal to the curve at the given points on it.
y = 3x2 - x + 1 at (1, 3)
Find the equation of tangent and normal to the curve at the given points on it.
2x2 + 3y2 = 5 at (1, 1)
Find the equations of tangent and normal to the curve y = 3x2 - 3x - 5 where the tangent is parallel to the line 3x − y + 1 = 0.
Choose the correct alternative.
The equation of tangent to the curve y = x2 + 4x + 1 at (-1, -2) is
Choose the correct alternative.
If elasticity of demand η = 1, then demand is
Choose the correct alternative.
If 0 < η < 1, then demand is
Fill in the blank:
The slope of tangent at any point (a, b) is called as _______.
Fill in the blank:
If f(x) = x - 3x2 + 3x - 100, x ∈ R then f''(x) is ______
Fill in the blank:
If f(x) = `7/"x" - 3`, x ∈ R x ≠ 0 then f ''(x) is ______
State whether the following statement is True or False:
x + 10y + 21 = 0 is the equation of normal to the curve y = 3x2 + 4x - 5 at (1, 2).
Find the equation of tangent and normal to the following curve.
y = x3 - x2 - 1 at the point whose abscissa is -2.
Find the equation of normal to the curve y = `sqrt(x - 3)` which is perpendicular to the line 6x + 3y – 4 = 0.
The slope of the tangent to the curve y = x3 – x2 – 1 at the point whose abscissa is – 2, is ______.
Choose the correct alternative:
Slope of the normal to the curve 2x2 + 3y2 = 5 at the point (1, 1) on it is
State whether the following statement is True or False:
The equation of tangent to the curve y = x2 + 4x + 1 at (– 1, – 2) is 2x – y = 0
Find the equations of tangent and normal to the curve y = 3x2 – x + 1 at the point (1, 3) on it
Find the equation of tangent to the curve y = x2 + 4x at the point whose ordinate is – 3
y = ae2x + be-3x is a solution of D.E. `(d^2y)/dx^2 + dy/dx + by = 0`
Find the equations of tangent and normal to the curve y = 6 - x2 where the normal is parallel to the line x - 4y + 3 = 0
