Differential Calculus - PYQs | Engineering Mathematics

In examinations, questions from Differential Calculus are frequently asked in the form of limits, continuity, and differentiability checks, as well as derivative-based problems.

Short Question on Differential Calculus

Question 1: For what value of the constant b is the function continuous on (- ∞ , ∞)

f(x) = \begin{cases}b x^2 + 2x & \text{if } x < 2 \\x^3 - b x & \text{if } x \geq 2\end{cases}

Question 2: Verify Rolle’s theorem for the function f(x) = x³ - x² + 6x + 2 on the given interval [0, 3]

Question 3: Verify Lagrange’s mean value theorem for the function f(x) = x + 1/x, [1/2, 2]

Question 4: Find the n^th derivative of y = sin(ax + b).

Question 5: If u = xy² + x²y where x = at², y = 2at, find du/dt.

Question 6: Find the Max and min of f(x,y) = xy subject to x² + y² = 1.

Question 7: If z = x²y + y²,find dz/dx when y = x²

Question 8: If u = e^xy find \dfrac{\partial^2 u}{\partial x \partial y}

Question 9: Find the envelope of the family of straight lines y = mx + 1/m, m ≠ 0.

Long Questions on Differential Calculus

Question 1: Use Lagrange's MVT to prove x/1 + x < log(1 + x) for all x > 0.

Question 2: If y = acos( logx) + bsin(log x), show that x²y_{n+ 2} + (2n + 1)xy_{n + 1} + n(n + 1)y_n = 0

Question 3: Solve this differential equation (x^2 + 1)\frac{dy}{dx} + 2xy = x, \ y(0) = 1.

Question 4: Find the points on the curve y' = 4x³ - 12x² + 12x; where the function attains local maxima and minima. Also determine the nature of each stationary point.

Question 5 : Verify Rolle’s Theorem for the function f(x) = x³ − 3x² + 2 on the interval [0, 2], and find all points c in (0, 2) that satisfy the theorem

Check if you were right - full answer with solution below.

Short Question on Differential Calculus: Answers

Solution 1:

Given function is continous

lim _{x-> 2}- (bx ² + 2x) = lim _{x-> 2} +(x ³ - bx )

4b + 4 = 8 - 2b

4b + 2b = 8 - 4

6b = 4

b = 2/3

Solution 2:

f(x) = x³ - x² + 6x + 2 , [0,3]

f(0) = 2

f(3) = 27 - 9 + 18 + 2 = 38

f(0) ≠ 𝑓(3)

Hence the Rolle’s theorem is not satisfied

Solution 3

• f(x) is continuous on [1\2 ,2].
• f(x) is differentiable on (1/2, 2).

Hence, MVT is applicable.

Take a=1/2,  b=2

f(b) = f(2) = 2+1/2 = 5/2

f(a) = f(2) = 2 + \tfrac{1}{2} = 5/2

f′(x) = 1 − 1/x²

By Lagrange’s MVT, there exists some c∈(a,b)c \in (a,b)c ∈ (a,b) such that

f′(c) = f(b) − f(a)/b−a

Substitute values:

1 − 1/c² = 5 = \dfrac{\dfrac{5}{2} - \dfrac{5}{2}}{2 - \dfrac{1}{2}} =\frac{0}{\frac{3}{2}} = 0

Solution 4:

Let y = sin(ax + b)

Then y₁ = acos(ax + b)

a₁ = asin(ax + b + \frac{\pi}{2})

a₂ = a²cos(ax + b + \frac{\pi}{2})

= a²sin(ax + b + \frac{\pi}{2}+ \frac{\pi}{2})

....

y_n = aⁿsin(ax + b + \frac{n\pi}{2})

Solution 5:

\frac{du}{dt} \;=\; \frac{\partial u}{\partial x}\frac{dx}{dt} \;+\; \frac{\partial u}{\partial y}\frac{dy}{dt}.

Given u = xy ² + x²y , x = at² , y = 2at

\frac{du}{dt} = y² + 2t , \frac{dx}{dt} = 2at , \frac{dy}{dt} = 2a

d/du​ = (y² + 2xy)2at + (2xy+x²)2a

= (4a²t² + 4a²t³)2at + (4a²t³+a²t⁴)2a

= 8a³t³ + 8a³t⁴+8a³t³ + 2a³t⁴

=16a³t³ + 10a³t⁴

=2a³t³(8 + 5t)

Solution 6:

Constraint: g(x,y) = x²+y² − 1 = 0.
∇f = (y,x),∇g = (2x,2y)
So (y,x) = λ(2x,2y)
⇒4λ² = 1
⟹  λ = ±1/2

• For λ=1/2 y=x. With constraint: x²+ x² = 1  ⟹  x = ±1/√2
  Then xy = 1/2
• For λ=−1/2 y = −x. With constraint: x² + x² = 1 ⟹  x = ±1/√2
  Then xy = −1/2

Maximum = 1/2, Minimum = −1/2

Solution 7:

\frac{dx}{dz} \;=\; \frac{\partial x}{\partial z} \;+\; \frac{\partial y}{\partial z}\,\frac{dx}{dy}

\frac{dz}{dx} = 2xy, \frac{dz}{dy} = x² + 2y

Here, y = x²

\frac{dy}{dx} = 2x

so \frac{dz}{dx} =2xy+(x² + 2y)(2x)
Put y = x²dz/dx = 2x(x²)+(x²+2x²)(2x)=2x³+(3x²)(2x).
= 2x³ + 6x³ = 8x³

dz/dx = 8x³

Solution 8:

u = e^xy

First, w.r.t. x:

\frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(e^{xy}) = ye^xy

Now differentiate w.r.t. y:

\frac{\partial^2 u}{\partial x \partial y} = \frac{\partial}{\partial y}(ye^{xy}) = e^xy + xye^xy

\dfrac{\partial^2 u}{\partial x \partial y} = e^{xy} + xye^{xy}

Solution 9:

F(x, y, m) = y − mx − 1/m = 0

Differentiate w.r.t. m:

\frac{\partial F}{\partial m} = -x + \frac{1}{m^2} = 0
m² = \frac{1}{x}
m = \frac{1}{\sqrt{x}}

Substitute into original:
y = mx+1/m    ⇒    y=±√x⋅x  +  ±√x
y² = 4x

Envelope is a parabola.

Long Question on Differential Calculus: Answers

Solution 1:

Let f(x) = log(1 + x)

f(0) = log 1 = 0

f'(x) = \frac{1}{1 + x} or f'(x) = \frac{1}{1+\theta x}

Then by MVT, for interval [0,x]

we have f(x) = f(0) + xf'(θx), 0 < θ < 1

or log(1 + x) =\frac{1}{1+\theta x} , 0 < θ < 1

0 < θx < 1 since x > 0

= 1 < 1 + θx < 1 + x ( 1 + 0 < 1 + θx < 1 + x)

= 1 > \frac{1}{1+\theta x} > \frac{1}{1+ x}

= \frac{1}{1+ x} < \frac{1}{1+\theta x} < x, x > 0

= \frac{x}{1+ x} < \frac{x}{1+\theta x} < x

= \frac{x}{1+ x} , log( 1 + x)

Solution 2:

y=acos(logx) + bsin(logx)

y₁​=\frac{-a}{x}sin(logx) + \frac{b}{x}​cos(logx).

xy₁ = -a sin(log x) + b cos (log x

Differentiating both side w.r.t.x , we get

xy₂ + y ₁ = - a / x cos(log x) + -b/x sin (log x)

= x²y₂ + xy₁ = -{acos {log x} + b sin(logx)}

= - y

= x²y₂ + xy₁ + y = 0

Using leibintz's theorem, we get

( y_{n + 2} x² + nc₁ y_{n + 1}2x + nc₂y_n.2) + (y_{n + 1} x + nc₁y_n.1) + y_n = 0

y_{n + 2} x² + y_{n + 1}2nx + n(n - 1)y_n + y_{n + 1}x + ny_n + y_n = 0

x² y_{n + 2} x² + y_{n + 1}(2n + 1)xy_{n + 1} + (n² - 1)y_n = 0

Solution 3:

Write in standard form: \frac{dy}{dx} + \frac{2x}{x^2 + 1}y = \frac{x}{x^2 + 1}

Find the integrating factor (IF):

\mu(x) = e^{\int \frac{2x}{x^2 + 1} dx} = e^{\ln(x^2 + 1)} = x^2 + 1

(x^2 + 1)\frac{dy}{dx} + 2xy = x \quad \Rightarrow \quad \frac{d}{dx}[y(x^2 + 1)] = x

Integrate both sides:

y(x^2 + 1) = \int x \, dx = \frac{x^2}{2} + C

Solve for y:

y = \frac{\frac{x^2}{2} + C}{x^2 + 1}

Apply the initial condition y(0) = 1

1 = \frac{0 + C}{0 + 1} \quad \Rightarrow \quad C = 1

y = \frac{\frac{x^2}{2} + 1}{x^2 + 1}

Solution 4:

y^' = 4x³ - 12x² + 12x

y' = 4x(x² - 3x + 3)

4x(x² - 3x + 3) = 0

x² - 3x + 3 = 0

x = \frac{3 \pm \sqrt{9-12}}{2} = \frac{3 \pm i \sqrt{3}}{2}

Only real stationary point r = 0

y'' = 12x² - 24x + 12

Evaluate at y" at x = 0

y''(0) = 12 - 0 + 0 = 12 > 0 {local minimum}

y(0) = 0 - 0 + 0 + 5 = 5

Local minimum at (0,5) No local maximum.

Solution 5:

Check the conditions of Rolle's Theorem:

a) Continuity on [0,2]: (x) is a polynomial, hence continuous on [0,2]
b) Differentiability on} (0,2): f(x) is a polynomial, hence differentiable on (0,2)
c) Equal values at endpoints:

f(0) = 0 - 0 + 2 = 2
f(2) = 8 - 12 + 2 = -2

Since f(0) ≠ f(2) so { Rolle theorem is not applicable}

let's consider the interval [0,1]:

• f(0) = 2
• f(1) = 1 - 3 + 2 = 0 {Not equal}

Find roots of f(x) = k so that f(a) = f(b).

f(0) = 2, f(2) = -2

We can shift function: g(x) = f(x) + 2 = x³ - 3x² + 4

• g(0) = 4, g(2) = 4
• g(0) = g(2)

Now apply rolle theorem

g'(x) = 3x² - 6x

Set g'(x) = 0:
3x² - 6x = 0
3x(x-2) = 0
x = 0, 2

Since Rolle's Theorem requires c in (0, 2), we discard endpoints c in (0, 2) satisfies g'(c) = 0