alleryStart learning →Slope of normal line to y = f(x) at x = a (if f'(a) ≠ 0):
- -1/f'(a) (perpendicular to tangent)
- f'(a)
- 1/f(a)
- -f'(a)
Show answer and solution
Correct answer: A. -1/f'(a) (perpendicular to tangent)
Normal ⊥ tangent.
Slope of normal = -1/f'(a).
Indian Navy 10+2 B.Tech Cadet Application of Derivatives — practice questions
Explore 30 free Application of Derivatives questions in Indian Navy 10+2 B.Tech Cadet Mathematics. Open each question or exercise to read its content, reference answers and available explanation.
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Showing questions 1–30 of 30.
Slope of normal line to y = f(x) at x = a (if f'(a) ≠ 0):Read the answer and worked solution →Approximation by differentials: Δy ≈ f'(x) Δx. Using this, sqrt(25.5) is approximately:Read the answer and worked solution →Maximum area of a rectangle inscribed in a semicircle of radius R (base on diameter):Read the answer and worked solution →Find minimum value of f(x) = (sin x)² + (cos x)² + 2 (sin x)(cos x):Read the answer and worked solution →Find the rate at which the surface area of a balloon (sphere) is changing when radius is 10 cm and dr/dt = 0.1 cm/s: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 →Order the first derivative testRead the answer and worked solution →Match each sign change to its verdictRead the answer and worked solution →The second derivative testRead the answer and worked solution →True or false: maxima and minimaRead the answer and worked solution →Order the working rule for absolute maxima and minimaRead the answer and worked solution →Tap every point you must test for an absolute maximumRead the answer and worked solution →Local or absolute?Read the answer and worked solution →Complete the rate of change statementRead the answer and worked solution →The chain rule for related ratesRead the answer and worked solution →Assertion and reasonRead 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 →Three of these are wrong. Find the one that holds.Read the answer and worked solution →True or false: increasing and decreasingRead the answer and worked solution →Flip and pair: symbol with meaningRead the answer and worked solution →Match each situation to the right toolRead 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 →Which test is being described?Read the answer and worked solution →Assertion and reason: rates of changeRead the answer and worked solution →Spot the odd oneRead the answer and worked solution →Complete the second derivative testRead the answer and worked solution →True or false: scope of this chapterRead the answer and worked solution →
Worked practice questions
Approximation by differentials: Δy ≈ f'(x) Δx. Using this, sqrt(25.5) is approximately:
- 5.5
- 5.05
- 5.10
- 5.15
Show answer and solution
Correct answer: B. 5.05
Let f(x) = sqrt(x), x = 25, Δx = 0.5. f'(x) = 1/(2 sqrt(x)) = 1/10 at x = 25. So sqrt(25.5) ≈ 5 + 0.1 × 0.5 = 5.05. (Actual value ≈ 5.0498.)
Maximum area of a rectangle inscribed in a semicircle of radius R (base on diameter):
- R²
- πR²
- R²/2
- 2 R² / sqrt(2) = R² sqrt(2)
Show answer and solution
Correct answer: D. 2 R² / sqrt(2) = R² sqrt(2)
Width = 2x, height = sqrt(R² - x²).
Area A = 2x sqrt(R² - x²).
dA/dx = 0 gives x = R/sqrt(2).
We have A_max = 2(R/sqrt(2)) sqrt(R² - R²/2).
That is A_max = sqrt(2) R × (R/sqrt(2)) × sqrt(2).
So A_max = R² sqrt(2).
Wait, recompute: A = 2 × (R/√2) × √(R²/2) = 2 × (R/√2) × (R/√2) = R².
So max area = R².
(Width = R√2, height = R/√2.)
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