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ISC Class 12 Continuity, Differentiability & Derivatives — practice questions — page 3

Explore 115 free Continuity, Differentiability & Derivatives questions in ISC Class 12 Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Continuity, Differentiability & Derivatives questions and exercises with answers

Showing questions 61–90 of 115.

For a particle moving along x(t) = t³ − 3t² + 5, the velocity at t = 2 is:Read the answer and worked solution →At a critical point c where f'(c) = 0, if f''(c) > 0, then c is a:Read the answer and worked solution →To find the dimensions of a rectangle with maximum area + perimeter 20, set up:Read the answer and worked solution →The critical points of f(x) = x³ − 3x + 1 are found by solving f'(x) = 0:Read the answer and worked solution →What does the sign of f ′(x) tell you?Read the answer and worked solution →Complete the definition of a critical pointRead the answer and worked solution →Match each sign change to its verdictRead the answer and worked solution →The chain rule for related ratesRead the answer and worked solution →True or false: increasing and decreasingRead the answer and worked solution →Tap every statement true of a local maximum at cRead the answer and worked solution →Reading the tests backRead the answer and worked solution →Spot the odd oneRead the answer and worked solution →Complete the second derivative testRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →A function f(x) is INCREASING on an interval if for all x in the interval:Read the answer and worked solution →The tangent to y = x² at the point (1, 1) has slope:Read the answer and worked solution →Two are true, one is notRead the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →True or false: scope of this chapterRead the answer and worked solution →Match each situation to the right toolRead the answer and worked solution →Continuity bingoRead the answer and worked solution →To differentiate an implicit relation you use the:Read the answer and worked solution →A function that fails the limit test at a point is:Read the answer and worked solution →Theorem 2 deals with which kind of function?Read the answer and worked solution →A function whose graph can be drawn without lifting the pen is:Read the answer and worked solution →Does continuity at a point guarantee differentiability?Read the answer and worked solution →If f is not continuous at c, then c is called a point of:Read the answer and worked solution →True or false: continuity and differentiabilityRead the answer and worked solution →Complete the definition of continuityRead the answer and worked solution →The derivative of $\sin(3x)$ is:Read the answer and worked solution →

Worked practice questions

For a particle moving along x(t) = t³ − 3t² + 5, the velocity at t = 2 is:

  1. $0$ (incorrectly assumed at rest)
  2. $3$ m/s (omitted negative term)
  3. $2$ m/s (wrong calculation)
  4. $0$ m/s (3(4) − 6(2) = 0)
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Correct answer: D. $0$ m/s (3(4) − 6(2) = 0)

We have v(t) = dx/dt. So v(t) = 3t² − 6t. At t = 2: v = 3(4) − 6(2) = 12 − 12 = 0. So at t = 2, the particle is instantaneously at rest. Could be turning point in motion.

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At a critical point c where f'(c) = 0, if f''(c) > 0, then c is a:

  1. Local MINIMUM (concave up at c)
  2. Local MAXIMUM (concave down)
  3. Inflection point (no extremum)
  4. Saddle point (multivariable)
Show answer and solution

Correct answer: A. Local MINIMUM (concave up at c)

**Second derivative test:** f''(c) > 0 → concave UP → c is a local MINIMUM. f''(c) < 0 → concave DOWN → local MAXIMUM. f''(c) = 0 → test fails, use first-derivative test.

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To find the dimensions of a rectangle with maximum area + perimeter 20, set up:

  1. Maximise A = xy with x + y = 20 (perimeter as sum)
  2. Maximise A = 2x × 2y with no constraint
  3. Maximise A = xy with x + y = 10 (half perimeter)
  4. Maximise A = x² with x + y = 10 fixed
Show answer and solution

Correct answer: C. Maximise A = xy with x + y = 10 (half perimeter)

Perimeter = 2x + 2y = 20 → x + y = 10. Maximise A = xy = x(10−x) = 10x − x². dA/dx = 10 − 2x = 0 at x = 5, y = 5. Square gives max area = 25 sq units.

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