हिंदी

The slope of the tangent to the curve y = x3 – x2 – 1 at the point whose abscissa is – 2, is ______. - Mathematics and Statistics

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प्रश्न

The slope of the tangent to the curve y = x3 – x2 – 1 at the point whose abscissa is – 2, is ______.

विकल्प

  • – 8

  • 8

  • 16

  • – 16

MCQ
रिक्त स्थान भरें
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उत्तर

The slope of the tangent to the curve y = x3 – x2 – 1 at the point whose abscissa is – 2, is 16.

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अध्याय 1.4: Applications of Derivatives - Q.1

संबंधित प्रश्न

Find the derivative of the following function from first principle.

x3 – 27


Find the derivative of the following function from first principle.

(x – 1) (x – 2)


Find the derivative of the following function from first principle.

`1/x^2`


Find the derivative of the following function from first principle.

`(x+1)/(x -1)`


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.

x2 + y2 + xy = 3 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:

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 ______


Find the equation of tangent and normal to the following curve.

xy = c2 at `("ct", "c"/"t")` where t is parameter.


Find the equation of tangent and normal to the following curve.

y = x2 + 4x at the point whose ordinate is -3.


Find the equation of tangent and normal to the following curve.

x = `1/"t",  "y" = "t" - 1/"t"`,  at t = 2


Find the equation of normal to the curve y = `sqrt(x - 3)` which is perpendicular to the line 6x + 3y – 4 = 0.


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 


Slope of the tangent to the curve y = 6 – x2 at (2, 2) is ______.


y = ae2x + be-3x is a solution of D.E. `(d^2y)/dx^2 + dy/dx + by = 0`


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