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Maharashtra HSC Mathematics & Statistics - Arts & Science (MSBSHSE) study materials
Maharashtra HSC Mathematics & Statistics - Arts & Science (MSBSHSE) study resources

Maharashtra HSC Mathematics & Statistics - Arts & Science (MSBSHSE) practice questions

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Question 1Differentiation

What is the derivative of sin x with respect to x?

  1. −sin x
  2. sec^2 x
  3. −cos x
  4. cos x Correct answer.

Explanation

The standard derivative formula for the sine function is d/dx(sin x) = cos x. This is a fundamental result in differentiation of trigonometric functions.

Question 2Differentiation

If y = e^(3x), what is dy/dx?

  1. e^(3x)/3
  2. e^(3x)
  3. 3e^(3x) Correct answer.
  4. 3x e^(3x)

Explanation

For y = e^(kx), the derivative is k e^(kx). Here k = 3, so dy/dx = 3e^(3x).

Question 3Differentiation

What is the derivative of ln x for x > 0?

  1. 1/x Correct answer.
  2. ln x
  3. 1/ln x
  4. x

Explanation

The natural logarithm function has derivative d/dx(ln x) = 1/x for x > 0. This follows from the inverse relationship between ln x and e^x.

Question 4Differentiation

If f(x) = 5x^2 − 3x + 7, what is f'(x)?

  1. 10x^2 − 3
  2. 10x − 3 Correct answer.
  3. 5x − 3
  4. 10x + 7

Explanation

Differentiate term by term: d/dx(5x^2) = 10x, d/dx(−3x) = −3, and d/dx(7) = 0. Hence f'(x) = 10x − 3.

Question 5Differentiation

The derivative of a constant function f(x) = c is:

  1. c
  2. 0 Correct answer.
  3. 1
  4. undefined

Explanation

A constant function has the same value for all x, so its graph is a horizontal line with slope zero. Therefore its derivative is 0 everywhere.

Question 6Differentiation

If y = x^2 + 3x, what is the value of dy/dx at x = 1?

  1. 2
  2. 4
  3. 3
  4. 5 Correct answer.

Explanation

Differentiating gives dy/dx = 2x + 3. Substituting x = 1 yields 2(1) + 3 = 5.

Question 7Differentiation

What is the derivative of cos x with respect to x?

  1. sin x
  2. −cos x
  3. −sin x Correct answer.
  4. tan x

Explanation

The standard derivative formula is d/dx(cos x) = −sin x. The negative sign is essential and distinguishes it from the derivative of sin x.

Question 8Differentiation

If y = (2x + 1)^5, what is dy/dx?

  1. (2x + 1)^4
  2. 10(2x + 1)^4 Correct answer.
  3. 2(2x + 1)^4
  4. 5(2x + 1)^4

Explanation

Apply the chain rule: differentiate the outer power to get 5(2x + 1)^4, then multiply by the derivative of the inner function 2x + 1, which is 2. This gives 10(2x + 1)^4.

Question 9Differentiation

If y = sin(3x), what is dy/dx?

  1. cos(3x)
  2. 3 sin(3x)
  3. 3 cos(3x) Correct answer.
  4. −3 cos(3x)

Explanation

By the chain rule, d/dx[sin(u)] = cos(u) · du/dx. With u = 3x, du/dx = 3, so dy/dx = 3 cos(3x).

Question 10Differentiation

If y = x^2 e^x, what is dy/dx?

  1. e^x(x^2 + 2x) Correct answer.
  2. x^2 e^x
  3. e^x(x^2 − 2x)
  4. 2x e^x

Explanation

Use the product rule: (uv)' = u'v + uv'. Here u = x^2, v = e^x, so u' = 2x and v' = e^x. Thus dy/dx = 2x e^x + x^2 e^x = e^x(x^2 + 2x).

Question 11Differentiation

If y = (x^2 + 1)/(x − 1), what is dy/dx?

  1. (2x)/(x − 1)^2
  2. (x^2 − 2x − 1)/(x − 1)^2 Correct answer.
  3. (x^2 + 2x + 1)/(x − 1)^2
  4. (2x)/(1)

Explanation

By the quotient rule, dy/dx = [(2x)(x − 1) − (x^2 + 1)(1)]/(x − 1)^2. Expanding the numerator gives 2x^2 − 2x − x^2 − 1 = x^2 − 2x − 1, so dy/dx = (x^2 − 2x − 1)/(x − 1)^2.

Question 12Differentiation

If x^2 + y^2 = 25, what is dy/dx?

  1. x/y
  2. −y/x
  3. 2x + 2y
  4. −x/y Correct answer.

Explanation

Differentiate both sides implicitly with respect to x: 2x + 2y(dy/dx) = 0. Solving for dy/dx gives dy/dx = −x/y.

Question 13Differentiation

If y = log_10 x, what is dy/dx?

  1. 1/x
  2. 1/(x ln 10) Correct answer.
  3. ln 10 / x
  4. 10/x

Explanation

The derivative of log_a x is 1/(x ln a). For a = 10, this gives dy/dx = 1/(x ln 10).

Question 14Differentiation

If y = tan x, what is dy/dx?

  1. cot x
  2. −csc^2 x
  3. sec x tan x
  4. sec^2 x Correct answer.

Explanation

The standard derivative formula for tan x is sec^2 x. This can be derived by writing tan x = sin x / cos x and applying the quotient rule.

Question 15Differentiation

If y = e^(sin x), what is dy/dx?

  1. e^(cos x)
  2. e^(sin x) cos x Correct answer.
  3. e^(sin x) sin x
  4. cos x

Explanation

By the chain rule, d/dx[e^(u)] = e^(u) · du/dx. With u = sin x, du/dx = cos x, so dy/dx = e^(sin x) cos x.

Question 16Differentiation

If y = x ln x, what is dy/dx?

  1. 1/x
  2. 1 + x
  3. ln x + 1 Correct answer.
  4. ln x

Explanation

Use the product rule with u = x and v = ln x: u' = 1, v' = 1/x. Then dy/dx = 1·ln x + x·(1/x) = ln x + 1.

Question 17Differentiation

If y = sin^2 x, what is dy/dx?

  1. sin 2x Correct answer.
  2. 2 cos x
  3. cos^2 x
  4. 2 sin x

Explanation

Using the chain rule with u = sin x, dy/dx = 2 sin x · cos x. By the double-angle identity, this equals sin 2x.

Question 18Differentiation

If y = (x + 1)/(x − 1), what is dy/dx at x = 2?

  1. 2
  2. −2 Correct answer.
  3. −1
  4. 1

Explanation

By the quotient rule, dy/dx = [(1)(x − 1) − (x + 1)(1)]/(x − 1)^2 = −2/(x − 1)^2. At x = 2 this equals −2/1 = −2.

Question 19Differentiation

If y = sqrt(x^2 + 1), what is dy/dx?

  1. 1/(2√(x^2 + 1))
  2. x√(x^2 + 1)
  3. 2x/√(x^2 + 1)
  4. x/√(x^2 + 1) Correct answer.

Explanation

Write y = (x^2 + 1)^(1/2). The chain rule gives dy/dx = (1/2)(x^2 + 1)^(−1/2) · 2x = x/√(x^2 + 1).

Question 20Differentiation

If y = x^x for x > 0, what is dy/dx?

  1. x^x
  2. x^x (1 + ln x) Correct answer.
  3. x·x^(x−1)
  4. x^x ln x

Explanation

Take ln y = x ln x and differentiate implicitly: (1/y)(dy/dx) = ln x + 1. Hence dy/dx = y(1 + ln x) = x^x(1 + ln x).

Question 21Differentiation

For the curve y = x^3 − 3x, at which values of x is the tangent horizontal?

  1. x = ±1 Correct answer.
  2. x = 0 only
  3. x = 3 only
  4. x = ±√3

Explanation

A horizontal tangent occurs where dy/dx = 0. Here dy/dx = 3x^2 − 3 = 3(x^2 − 1), which is zero at x = ±1.

Question 22Differentiation

If y = sin^(−1) x for −1 < x < 1, what is dy/dx?

  1. 1/√(1 − x^2) Correct answer.
  2. −1/√(1 − x^2)
  3. √(1 − x^2)
  4. 1/(1 + x^2)

Explanation

The derivative of the inverse sine function is 1/√(1 − x^2) for −1 < x < 1. This follows from differentiating x = sin y implicitly.

Question 23Differentiation

If x = a(θ − sin θ) and y = a(1 − cos θ), what is dy/dx?

  1. cot(θ/2) Correct answer.
  2. tan(θ/2)
  3. sin θ
  4. 1 − cos θ

Explanation

Parametric differentiation gives dy/dx = (dy/dθ)/(dx/dθ) = [a sin θ]/[a(1 − cos θ)] = sin θ/(1 − cos θ). Using half-angle identities this simplifies to cot(θ/2).

Question 24Differentiation

If y = e^(2x) sin 3x, what is dy/dx?

  1. e^(2x)(2 sin 3x + 3 cos 3x) Correct answer.
  2. e^(2x)(2 sin 3x − 3 cos 3x)
  3. 2e^(2x) sin 3x
  4. 3e^(2x) cos 3x

Explanation

Apply the product rule: (uv)' = u'v + uv' with u = e^(2x) and v = sin 3x. Then u' = 2e^(2x) and v' = 3 cos 3x, so dy/dx = e^(2x)(2 sin 3x + 3 cos 3x).

Question 25Differentiation

If y = a^x, where a > 0 and a ≠ 1, what is dy/dx?

  1. a^x
  2. x a^(x−1)
  3. a^x ln a Correct answer.
  4. a^x / ln a

Explanation

For an exponential function with constant base a, the derivative is the function itself multiplied by the natural logarithm of the base. This follows from writing a^x = e^(x ln a) and applying the chain rule. The power rule applies only when the exponent is a constant, not a variable.

Question 26Differentiation

If y = cot x, what is dy/dx?

  1. −cosec x cot x
  2. cosec^2 x
  3. sec^2 x
  4. −cosec^2 x Correct answer.

Explanation

The derivative of cot x is −cosec^2 x, which can be obtained by writing cot x = cos x / sin x and applying the quotient rule. The negative sign is essential because cot x is decreasing on each interval where it is defined.

Question 27Differentiation

If y = log(sin x) for sin x > 0, what is dy/dx?

  1. cosec x
  2. cot x Correct answer.
  3. tan x
  4. sec x

Explanation

Using the chain rule, differentiate the outer logarithm to get 1/sin x, then multiply by the derivative of sin x, which is cos x. This gives cos x / sin x = cot x. The condition sin x > 0 ensures the logarithm is defined.

Question 28Differentiation

If x = a(cos θ + θ sin θ) and y = a(sin θ − θ cos θ), where a is a positive constant, what is dy/dx?

  1. cot θ
  2. tan θ Correct answer.
  3. sec θ
  4. θ tan θ

Explanation

Differentiate parametrically: dx/dθ = a(−sin θ + sin θ + θ cos θ) = aθ cos θ, and dy/dθ = a(cos θ − cos θ + θ sin θ) = aθ sin θ. Dividing gives dy/dx = (aθ sin θ)/(aθ cos θ) = tan θ, for θ ≠ 0 and cos θ ≠ 0.

Question 29Applications of Derivatives

Find the slope of the tangent to the curve y = x³ − 3x at the point where x = 2.

  1. 12
  2. 3
  3. 6
  4. 9 Correct answer.

Explanation

The slope of the tangent is the value of dy/dx at the given point. Here dy/dx = 3x² − 3, so at x = 2 the slope equals 3(4) − 3 = 9.

Question 30Applications of Derivatives

A function f is defined on an interval containing c. Which condition guarantees that f has a local maximum at c?

  1. f′(c) = 0 and f″(c) > 0
  2. f′(c) < 0 and f″(c) < 0
  3. f′(c) > 0 and f″(c) = 0
  4. f′(c) = 0 and f″(c) < 0 Correct answer.

Explanation

The second derivative test states that if f′(c) = 0 and f″(c) < 0, the curve is concave downward at c, so f has a local maximum there. A stationary point with negative concavity is a peak, not a trough.

Question 31Applications of Derivatives

The radius r of a circle increases at the constant rate of 2 cm/s. At what rate is the area increasing when r = 5 cm?

  1. 20π cm²/s Correct answer.
  2. 25π cm²/s
  3. 10π cm²/s
  4. 4π cm²/s

Explanation

The area is A = πr², so dA/dt = 2πr · dr/dt by the chain rule. Substituting r = 5 and dr/dt = 2 gives dA/dt = 2π(5)(2) = 20π cm²/s.

Question 32Applications of Derivatives

For the function f(x) = x² − 4x + 7, at which value of x does f have a stationary point?

  1. x = −2
  2. x = 7
  3. x = 4
  4. x = 2 Correct answer.

Explanation

Stationary points satisfy f′(x) = 0. Here f′(x) = 2x − 4, which vanishes at x = 2. At that point f has a local minimum.

Question 33Applications of Derivatives

The derivative of a differentiable function is positive on (a, b). What can be concluded about f on that interval?

  1. f is strictly decreasing on (a, b)
  2. f has a local maximum at every point of (a, b)
  3. f is constant on (a, b)
  4. f is strictly increasing on (a, b) Correct answer.

Explanation

A positive derivative on an interval means the function rises as x increases, so f is strictly increasing there. This is the first derivative test for monotonicity.

Question 34Applications of Derivatives

Evaluate the limit lim(x→0) (sin 3x)/x.

  1. 1
  2. 0
  3. 1/3
  4. 3 Correct answer.

Explanation

Using the standard limit lim(t→0)(sin t)/t = 1, write (sin 3x)/x = 3 · (sin 3x)/(3x). As x→0 the fraction tends to 1, so the limit is 3.

Question 35Applications of Derivatives

A particle moves along a line with position s(t) = t³ − 6t² + 9t. At which time is its acceleration zero?

  1. t = 2 Correct answer.
  2. t = 3
  3. t = 1
  4. t = 0

Explanation

Velocity is s′(t) = 3t² − 12t + 9 and acceleration is s″(t) = 6t − 12. Setting 6t − 12 = 0 gives t = 2.

Question 36Applications of Derivatives

Which of the following is an indeterminate form to which l'Hôpital's rule may be applied directly?

  1. 0 · 5
  2. 2/3
  3. 0/0 Correct answer.
  4. 1/0

Explanation

L'Hôpital's rule applies to the indeterminate forms 0/0 and ∞/∞. A form such as 1/0 is not indeterminate but tends to infinity in magnitude.

Question 37Applications of Derivatives

Find the equation of the tangent to the curve y = x² + 2x at the point (1, 3).

  1. y = 2x + 1
  2. y = 3x
  3. y = 4x − 1 Correct answer.
  4. y = 4x + 1

Explanation

Here dy/dx = 2x + 2, so the slope at x = 1 is 4. Using point-slope form with (1, 3): y − 3 = 4(x − 1), which simplifies to y = 4x − 1.

Question 38Applications of Derivatives

The function f(x) = x³ − 3x² + 4 is examined for concavity. On which interval is the curve concave upward?

  1. (−∞, ∞)
  2. (1, ∞) Correct answer.
  3. (−∞, 1)
  4. (0, 2)

Explanation

Compute f′(x) = 3x² − 6x and f″(x) = 6x − 6. The curve is concave upward where f″(x) > 0, that is 6x − 6 > 0, giving x > 1.

Question 39Applications of Derivatives

A ladder 13 m long rests against a vertical wall. Its foot slides away from the wall at 2 m/s. How fast is the top sliding down when the foot is 5 m from the wall?

  1. 12/5 m/s
  2. 5/6 m/s Correct answer.
  3. 2 m/s
  4. 6/5 m/s

Explanation

With x the foot distance and y the height, x² + y² = 169. Differentiating gives 2x dx/dt + 2y dy/dt = 0, so dy/dt = −(x/y)(dx/dt). When x = 5, y = 12, giving dy/dt = −(5/12)(2) = −5/6, so the top descends at 5/6 m/s.

Question 40Applications of Derivatives

Evaluate lim(x→0) (e^x − 1 − x)/x².

  1. 1/2 Correct answer.
  2. 0
  3. 1
  4. 2

Explanation

The form is 0/0, so apply l'Hôpital's rule twice. First derivative ratio: (e^x − 1)/(2x), still 0/0; second: e^x/2, which tends to 1/2 as x→0.

Question 41Applications of Derivatives

The function f(x) = 2x³ − 9x² + 12x has critical points at x = 1 and x = 2. Which statement correctly classifies them?

  1. Local maximum at both x = 1 and x = 2
  2. Local minimum at x = 1, local maximum at x = 2
  3. Local maximum at x = 1, local minimum at x = 2 Correct answer.
  4. Local minimum at both x = 1 and x = 2

Explanation

Here f′(x) = 6x² − 18x + 12 and f″(x) = 12x − 18. At x = 1, f″ = −6 < 0, a local maximum; at x = 2, f″ = 6 > 0, a local minimum.

Question 42Applications of Derivatives

Find the point of inflection of the curve y = x³ − 6x² + 5x.

  1. (0, 0)
  2. (3, −6)
  3. (1, 0)
  4. (2, −6) Correct answer.

Explanation

y′ = 3x² − 12x + 5 and y″ = 6x − 12. Setting y″ = 0 gives x = 2, and substituting gives y = 8 − 24 + 10 = −6, so the inflection point is (2, −6).

Question 43Applications of Derivatives

A rectangular garden is to be fenced on all four sides using 40 m of fencing. What is the maximum possible area?

  1. 100 m² Correct answer.
  2. 400 m²
  3. 80 m²
  4. 160 m²

Explanation

Let the sides be x and y with 2x + 2y = 40, so y = 20 − x. Area A = x(20 − x) = 20x − x², and A′ = 20 − 2x = 0 gives x = 10. Then y = 10 and the maximum area is 100 m².

Question 44Applications of Derivatives

For what values of x is f(x) = x³ − 3x strictly decreasing?

  1. x < −1 or x > 1
  2. −1 < x < 1 Correct answer.
  3. x < 0
  4. x > 0

Explanation

f′(x) = 3x² − 3 = 3(x − 1)(x + 1), which is negative between the roots −1 and 1. Hence f is strictly decreasing on (−1, 1).

Question 45Applications of Derivatives

The volume of a sphere increases at 36π cm³/s. At what rate is the radius increasing when r = 3 cm?

  1. 3 cm/s
  2. 2 cm/s
  3. 1 cm/s Correct answer.
  4. 4 cm/s

Explanation

V = (4/3)πr³, so dV/dt = 4πr² · dr/dt. Substituting dV/dt = 36π and r = 3 gives 36π = 36π · dr/dt, hence dr/dt = 1 cm/s.

Question 46Applications of Derivatives

Using differentials, find the approximate value of √(36.6).

  1. 6.01
  2. 6.60
  3. 6.10
  4. 6.05 Correct answer.

Explanation

Take f(x) = √x with a = 36 and dx = 0.6. Then f′(x) = 1/(2√x) gives f′(36) = 1/12, so dy ≈ (1/12)(0.6) = 0.05 and √36.6 ≈ 6 + 0.05 = 6.05.

Question 47Applications of Derivatives

Evaluate lim(x→∞) (3x² + 2x + 1)/(x² − 5x + 4).

  1. 3 Correct answer.
  2. 0
  3. ∞
  4. 1

Explanation

This is of the form ∞/∞, so l'Hôpital's rule applies, or divide numerator and denominator by x². Either way the limit equals the ratio of leading coefficients, 3/1 = 3.

Question 48Applications of Derivatives

A function f satisfies f′(x) = (x − 1)²(x − 3). Which of the following correctly describes the behaviour at x = 1 and x = 3?

  1. No extremum at x = 1; local maximum at x = 3 Correct answer.
  2. Local minimum at x = 1; local maximum at x = 3
  3. Local maximum at x = 1; local minimum at x = 3
  4. No extremum at either point

Explanation

At x = 1 the factor (x − 1)² does not change sign, so f′ keeps the same sign and there is no extremum (a stationary inflection). At x = 3 the factor (x − 3) changes sign from negative to positive, giving a local maximum as f′ goes from negative to positive.

Question 49Applications of Derivatives

A cone has height 12 cm and base radius 6 cm, with water flowing in at 8 cm³/s. How fast is the water level rising when the depth is 4 cm?

  1. 2/(3π) cm/s Correct answer.
  2. 3/(2π) cm/s
  3. 8/(9π) cm/s
  4. 1/(3π) cm/s

Explanation

By similar triangles the water radius is r = h/2, so V = (1/3)π(h/2)²h = πh³/12. Then dV/dt = (πh²/4) dh/dt, and at h = 4 this gives 8 = 4π · dh/dt, so dh/dt = 2/(3π) cm/s.

Question 50Applications of Derivatives

The sum of two positive numbers is 20. What is the minimum value of the sum of their squares?

  1. 400
  2. 180
  3. 200 Correct answer.
  4. 100

Explanation

Let the numbers be x and 20 − x. Then S = x² + (20 − x)² = 2x² − 40x + 400, and S′ = 4x − 40 = 0 gives x = 10. The minimum sum of squares is 10² + 10² = 200.