Board Boosters

The questions your board exam loves to ask

800 most-asked Class 11 & 12 (+1 / +2) questions across Physics, Chemistry, Maths and Biology — each with a model answer and the exact marking-scheme points examiners reward. Revise smart, walk in calm.

800 board questions•Model answers•Marking-scheme points•Every chapter•CBSE · ISC · State boards

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MathsClass 122 markseasy

Application of Derivatives

State the condition for a function to be increasing or decreasing on an interval.

Reveal model answer + marking points▾

A differentiable function f is increasing on an interval if its derivative f'(x) is greater than or equal to 0 (strictly increasing if f'(x) > 0) for all x in that interval. It is decreasing if its derivative f'(x) is less than or equal to 0 (strictly decreasing if f'(x) < 0) for all x in that interval. So the sign of the first derivative tells us whether the function rises or falls.

f'(x) > 0 increasing; f'(x) < 0 decreasing

Marking-scheme points

  • ✓Increasing: f'(x) >= 0 (strictly increasing if f'(x) > 0)
  • ✓Decreasing: f'(x) <= 0 (strictly decreasing if f'(x) < 0)
  • ✓Sign of the first derivative decides the behaviour
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MathsClass 123 marksmedium

Application of Derivatives

Find the intervals in which the function f(x) = x^2 - 4x + 3 is increasing or decreasing.

Reveal model answer + marking points▾

Find the derivative: f'(x) = 2x - 4 = 2(x - 2). Set f'(x) = 0 to get the critical point: x = 2. For x < 2, f'(x) < 0, so the function is decreasing on (-infinity, 2). For x > 2, f'(x) > 0, so the function is increasing on (2, infinity). Thus f is decreasing on (-infinity, 2) and increasing on (2, infinity).

Marking-scheme points

  • ✓f'(x) = 2x - 4 = 0 gives x = 2
  • ✓x < 2: f'(x) < 0 -> decreasing
  • ✓x > 2: f'(x) > 0 -> increasing
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MathsClass 122 markseasy

Application of Derivatives

How do you find the slope of the tangent and the normal to a curve at a given point?

Reveal model answer + marking points▾

The slope of the tangent to a curve y = f(x) at a point is the value of the derivative dy/dx at that point. The normal is the line perpendicular to the tangent at that point, so its slope is the negative reciprocal of the tangent's slope, that is, slope of normal = -1/(dy/dx). Using the point-slope form with these slopes gives the equations of the tangent and the normal.

slope of normal = -1/(slope of tangent)

Marking-scheme points

  • ✓Slope of tangent = dy/dx at the point
  • ✓Slope of normal = -1/(dy/dx)
  • ✓Use point-slope form for the equations
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