Differentiability MCQ Quiz - Objective Question with Answer for Differentiability - Download Free PDF

Explanation:

Equation of curve p = f(r)

Length of chord of curvature

= 2ρ cos ϕ

= \(2{r\over f'(r)}\sqrt{1-\sin^2ϕ}\)

= \(2{r\over f'(r)}\sqrt{1-{p^2\over r^2}}\) (p = r sin ϕ)

= \( \frac{2}{f'(r)}\sqrt{r^2-p^2}\)

= \(\rm \frac{2}{f'(r)}\sqrt{r^2-[f(r)]^2}\)

Option (2) is true.

The correct answer is 3)

For any 5000 increment in steps/day, the largest risk reduction occurs on going from 0 to 5000.

Explanation:

• From the graph, the mortality risk of cardiovascular disease decreases rapidly at lower step counts and slows down at higher step counts.
• The steepest part of the curve is from 0 to 5000 steps/day, indicating the largest reduction in risk.
• Beyond 5000 steps/day, the rate of reduction diminishes, showing smaller incremental benefits as the number of steps increases further.

Explanation:

x = x2 - x y + y3,

x = rcosθ, y = rsinθ

Apply Chain Rule:

(dz/dr) = (dz/dx) (dx/dr) + (dz/dy) (dy/dr)

dz/dx = 2x - y

dz/dy = -x + 3y²

dx/dr = cos(θ)

dy/dr = sin(θ)

Now,  (dz/dr) = (2x - y) cos(θ) + (-x + 3y²) sin(θ)

Since x = r cos(θ) and y = r sin(θ), we can find r and θ at x = 1 and y = 1:

r = √(x² + y²) = √(1² + 1²) = √2

θ = tan⁻¹(y/x) = tan⁻¹(1/1) = π/4

Now, substitute x = 1, y = 1, r = √2, and θ = π/4 into the expression for (dz/dr):

(dz/dr) = (2(1) - 1) cos(π/4) + (-1 + 3(1)²) sin(π/4)

(dz/dr) = 1 (√2/2) + 2 (√2/2)

(dz/dr) = 3√2/2

Therefore, (dz/dr) at x = 1 and y = 1 is 3/√2 

Hence, the correct answer is option 1.

Explanation:

Option 1: f(x) = |x - 1|

The function |x - 1| represents an absolute value function. It has a sharp corner

at x = 1 , so the function is not differentiable at this point.

Option 2: f(x) = [x] (Greatest Integer Function)

The greatest integer function (step function) is not continuous at integer values,

and discontinuity implies that the function is not differentiable at x = 1 (or any

integer value).

Option 3: f(x) = \(1 + (1 - x)^{1/3}\)

The function f(x) = \(1 + (1 - x)^{1/3}\) is differentiable at x = 1 because the cube root

function is continuous and smooth (no sharp corners or discontinuities) at x = 1

Both the absolute value function |x - 1| and the greatest integer function [x] are

not differentiable at x = 1 . 

Hence option 4 is correct.

Explanation:

At  $x=0$, $|x|$ has a kink (non-differentiability)

Let's compute LHD and RHD at $x=0$:

For $x>0$:

\(f(x)= \frac{x}{1+x} ​ ⟹f ′ (x)= -\frac{1}{(1+x) ^ 2} ​ .\)

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\(f _ + ′ ​ (0)= \frac{1}{(1+0)^2} ​ =1. \)

For $x<0:$

\(f(x)= \frac{x}{1-x} ​ ⟹f ′ (x)= \frac{1}{(1-x) ^ 2}\)

At $x=0$, the LHD is:

\(f _ − ′ ​ (0)= \frac{1}{(1-0)^2} ​ =1\)

The function is differentiable for all $x\in (-\infty ,\infty )$.

Therefore, option 2 is correct.

Concept:

A function f(x, y) is said to be homogeneous of degree n in x and y, it can be written in the form

f(λx, λy) = λn f(x, y)

Euler’s theorem:

If f(x, y) is a homogeneous function of degree n in x and y and has continuous first and second-order partial derivatives, then

\(x\frac{{\partial f}}{{\partial x}} + y\frac{{\partial f}}{{\partial y}} = nf\)

\({x^2}\frac{{{\partial ^2}f}}{{\partial {x^2}}} + 2xy\frac{{{\partial ^2}f}}{{\partial x\partial y}} + {y^2}\frac{{\partial ^2f}}{{\partial y^2}} = n\left( {n - 1} \right)f\)

If z is homogeneous function of x & y of degree n and z = f(u), then

\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = n\frac{{f\left( u \right)}}{{f'\left( u \right)}}\)

Calculation:

Given, \(u = \ln \left( {\frac{{{x^3} + {x^2}y - {y^3}}}{{x - y}}} \right)\)

\(z = {\frac{{{x^3} + {x^2}y - {y^3}}}{{x - y}}}\)

z is a homogenous function of x & y with a degree 2.

Now, z = eu

Thus, by Euler’s theorem:

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

\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = 2\)

Concept:

In Partial Differentiation, all variables are considered as a constant except the independent derivative variable i.e If f(x,y,z) is a function, then its partial derivative with respect to x is calculated by keeping y and z as constant.

Calculation:

f(x, y, z) = (x² + y² – 2z²)(y² + z²)

\(\frac{{\partial f}}{{\partial x}} = \left( {2x} \right)\left( {{y^2} + {z^2}} \right)\)

At the point, x = 2, y = 1 and z = 3 is

\(\frac{{\partial f}}{{\partial x}} = 2\left( 2 \right)\left( {{1^2} + {3^2}} \right) = 40\)

Concept:

A function is said to be differentiable at x =a if,

Left derivative = Right derivative = Well defined

Analysis:

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {{e^x},\;x < 1}\\ {\log x + a{x^2} + bx,\;x \ge 1} \end{array}} \right.\)

Taking Differentiation,

\(f'\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {{e^x},\;x < 1}\\ {\frac{1}{x} + 2ax + b,\;x \ge 1} \end{array}} \right.\)

f’(1) = e, x < 1

f’ (1) = 1 + 2a + b, x ≥ 1

since f(x) is differentiable at x = 1,

e = 1 + 2a + b → (1)

At x = 1,

f(1) = e, x < 1

f(1) = a + b, x ≥ 1

since f(x) is continuous at x = 1,

e = a + b → (2)

From (1) and (2)

⇒ 1 + 2a + b = a + b

⇒ a = -1

⇒ b = e + 1

Concept:

Chain Rule of derivatives states that, if y = f(u) and u = g(x) are both differentiable functions, then:

\(\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}\)

\(\frac{{d\left( {\ln x} \right)}}{{dx}} = \frac{1}{x},\;for\;x > 0\)

\(\frac{{d\left( {\sin x} \right)}}{{dx}} = \; cosx\)

Calculation:

Given: y = log sinx

Let sin x = u

⇒ y = log u

\(\frac{d}{{dx}}\left( {\log u} \right) = \frac{1}{{u}}\frac{d}{{dx}}\left( {u} \right) \)

\(= \frac{1}{{\sin x}}\left( { cos x} \right) \)

Hence, the value of \(\frac{dy}{dx}\) will be \(\frac{1}{sin~x} cos~x\).

r = x2 + y - z   ---(1)

z3 -xy + yz + y3 = 1    ---(2)

\(\frac{{\partial r}}{{\partial x}} = 2x + \frac{{\partial y}}{{\partial x}} - \frac{{\partial z}}{{\partial x}}\)

Since y is an independent derivative of ‘y’ w.r.t. ‘x’ is 0.

\(\frac{{\partial r}}{{\partial x}} = 2x - \frac{{\partial z}}{{\partial x}}\)      ----(1)

From 2^nd relation:

Z³ – xy + yz + y³ = 1

Differentiate w.r.t x

\(3{Z^2}\frac{{\partial z}}{{\partial x}} - y + y\frac{{\partial z}}{{\partial x}} = 0\)

\(\left( {3{Z^2} + y} \right)\frac{{\partial z}}{{\partial x}} = y\)

\(\frac{{\partial z}}{{\partial x}} = \frac{y}{{3{z^2} + y}}\)      ----(2)

Substitute in (1)

\(\frac{{\partial r}}{{\partial x}} = 2r - \frac{y}{{3{z^2} + y}}\)

At, (2, -1, 1)

\({\left( {\frac{{\partial r}}{{\partial x}}} \right)_{\left( {2,\; - 1,1} \right)}} = 2\left( 2 \right) - \frac{{ - 1}}{{3{{\left( 1 \right)}^2} + \left( { - 1} \right)}}\)

\(= 4 + \frac{1}{2}\)

Concept

if y = xⁿ, then\(\frac{{\partial y}}{{\partial x}} = n{X^{n - 1}}\frac{{{\partial ^2}y}}{{\partial {x^2}}} = n\left( {n - 1} \right){X^{n - 2}}\)

Calculation:

Given:

\(v = {\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{1}{2}}}\)

\(\frac{{\partial v}}{{\partial x}} = - \frac{1}{2}{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{3}{2}}} \times 2x = - x{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{3}{2}}}\)

\(\frac{{{\partial ^2}v}}{{\partial {x^2}}} = - x\left\{ { - \frac{3}{2}{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{ - \frac{5}{2}}} \times 2x} \right\} + {\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{3}{2}}}\left( { - 1} \right)\)

\( \Rightarrow 3{x^2}{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{5}{2}}} - {\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{3}{2}}}\)

Similarly,

\(\frac{{{\partial ^2}v}}{{\partial {y^2}}} = 3{y^2}{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{5}{2}}} - {\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{3}{2}}}\)

\(\frac{{{\partial ^2}v}}{{\partial {z^2}}} = 3{z^2}{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{5}{2}}} - {\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{3}{2}}}\)

Now,\(\frac{{{\partial ^2}v}}{{\partial {x^2}}} + \frac{{{\partial ^2}v}}{{\partial {y^2}}} + \frac{{{\partial ^2}v}}{{\partial {z^2}}} = \left( {3{x^2} + 3{y^2} + 3{z^2}} \right){\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{5}{2}}} - 3{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{3}{2}}}\)

⇒\(3\left( {{x^2} + {y^2} + {z^2}} \right){\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{5}{2}}} - 3{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{3}{2}}}\)

⇒\(3{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{3}{2}}} - 3{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{3}{2}}} = 0\)

Explanation:

\(\mathop {\lim }\limits_{x \to b} \frac{{{b^x} - {x^b}}}{{{x^x} - {b^b}}} \)

The given limit is in 0/0 form, So

Applying L hospitals rule

\(\begin{array}{l} \mathop {\lim }\limits_{x \to b} \frac{{{b^x}\log b - b{x^{b - 1}}}}{{{x^x}\left( {1\; + \;\log x} \right)}} = - 1\\ \Rightarrow \frac{{{b^b}\log b - b.{b^{b - 1}}}}{{{b^b}\left( {1\; + \;\log b} \right)}} = \frac{{{b^b}\left( {\log b - 1} \right)}}{{{b^b}\left( {1\; + \;\log b} \right)}} = - 1 \end{array}\)

⇒ log b - 1 = -1 - log b

⇒ 2 log b = 0

Concept:

A function f(x) is said to be increasing in the given interval if it’s first order differential \(f'\left( x \right) \ge 0\) holds for every point in the given interval.

Calculation:

Function I:

\(f\left( x \right) = {e^{ - x}}\therefore f'\left( x \right) = \; - {e^{ - x}}\)

\(\because {f'}\left( 0 \right) = - 1 < 0\)

Therefore, the function is non increasing.        

Function II:

\(f\left( x \right) = \;{x^2} - \sin x,\;f'\left( x \right) = 2x - \cos x\)

\(f'\left( 0 \right) = \;0 - \cos 0 = - 1 < 0\)

Therefore, this function is also non increasing.

Furthermore, since cosine function is periodic function, it cannot be strictly increasing in the given range

Function III:

\(f\left( x \right) = \sqrt {{x^3} + 1} \)

\({\rm{f'}}\left( {\rm{x}} \right) = \frac{{3{x^2}}}{{2\sqrt {{x^3} + 1} }}\)

\(f'\left( 0 \right) = 0,\;f'\left( 1 \right) > 0\)

Therefore, the function is increasing in the given interval.

Therefore \(f\left( x \right) = \sqrt {{x^3} + 1} \) function is the only increasing everywhere in [0, 1]

Hence Option(1) is the correct answer.

\(f\left( {x,y} \right) = \frac{{a{x^2} + b{y^2}}}{{xy}} = \frac{{ax}}{y} + \frac{{by}}{x}\)

\(\frac{{\partial f}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {\frac{{ax}}{y} + \frac{{by}}{x}} \right]\)

\(\frac{\partial f}{\partial x}=\frac{a}{y} - \frac{{by}}{{{x^2}}}\)

At, x = 1 and y = 2, we get:

\(\frac{\partial f}{\partial x}|_{x=1,y=2}=\frac{a}{2} - \frac{{b\left( 2 \right)}}{{{1^2}}} \)

\(= \frac{a}{2} - 2b\)   ----(1)

Similarly,

\(\frac{{\partial f}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {\frac{{ax}}{y} + \frac{{by}}{x}} \right]\)

\(= - \frac{{ax}}{{{y^2}}} + \frac{b}{x}\)

At, x = 1 and y = 2

\(\frac{\partial f}{\partial y}|_{x=1,y=2}= \frac{{ - a}}{4} + b\)     ----(2)

For the given condition, equating the value of \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) at x = 1 and y = 2, we get:

\(\frac{a}{2} - 2b = \frac{{ - a}}{4} + b\)

\(\frac{a}{2} + \frac{a}{4} = 3b\)

\(\Rightarrow \frac{{2a + a}}{4} = 3b\)

3a = 12b

Calculations:

Given equation is: f (x, y) = x2 + y³ ........... (1)

and

 x = t2 + t3  ............(2)

 y = 1 + t³   .............(3)

Using CHAIN RULE on equation 1,

\(\frac{{df}}{{dt}} = {f_x}\frac{{dx}}{{dt}} + {f_y}\frac{{dy}}{{dt}}\)   .............(4)

Now, finding out different terms of the above equation(4)

x = t2 + t3 

⇒ \(\frac{{dx}}{{dt}} = 2t + 3{t^2}\)

⇒ \(\frac{{dx}}{{dt}}(t = 1) = 2 × 1 + 3 × {1^2} = 5\)

⇒ x (t = 1) = 12 + 13 = 2

y = 1 + t3 

⇒ \(\frac{{dy}}{{dt}} = 3{t^2}\)

⇒ \(\frac{{dy}}{{dt}}(t=1) = 3×{1^2}=3\)

⇒ y (t = 1) = 1 + 13 = 2

f (x, y) = x2 + y3 

⇒ fx = 2x 

So, f_x (t = 1) = 2 × 2 = 4

⇒ f_y = 3y²

So, f_y (t = 1) = 3 × 2² = 12

Putting all values in equation 4, we get,

\(\frac{{df}}{{dt}} = 4 \times 5 + 3 \times 12 = 56\)

So, \(\frac{{df}}{{dt}}(t = 1) = 56\)