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

= $2\frac{r}{f^{\prime }(r)}\sqrt{1-{\sin }^2\varphi }$

= $2\frac{r}{f^{\prime }(r)}\sqrt{1-\frac{p^2}{r^2}}$ (p = r sin ϕ)

= $\frac{2}{f^{\prime }(r)}\sqrt{r^2-p^2}$

= $\frac{2}{{\mathrm{f}}^{\prime }(\mathrm{r})}\sqrt{{\mathrm{r}}^2-[\mathrm{f}(\mathrm{r}){]}^2}$

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 $x>0$:

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

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$f_{+}\prime (0)=\frac{1}{(1+0{)}^2}=1.$

For $x<0:$

$f(x)=\frac{x}{1-x}⟹f\prime (x)=\frac{1}{(1-x{)}^2}$

At $x=0$, the LHD is:

$f_{-}\prime (0)=\frac{1}{(1-0{)}^2}=1$

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

$x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf$

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

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

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

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:

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

Taking Differentiation,

$f^{\prime }\left(x\right)=\{\begin{array}{c}{e}^x,x<1\\ \frac{1}{x}+2ax+b,x\ge 1\end{array}$

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:

$\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$

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

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

Calculation:

Given: y = log sinx

Let sin x = u

⇒ y = log u

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

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

Hence, the value of $\frac{dy}{dx}$ will be $\frac{1}{sinx}cosx$.

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

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

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

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

Substitute in (1)

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

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:

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

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

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

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

Similarly,

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

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

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

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

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

Explanation:

$\underset{x\to b}{lim}\frac{b^x-x^b}{x^x-b^b}$

The given limit is in 0/0 form, So

Applying L hospitals rule

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

⇒ 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 $f^{\prime }\left(x\right)\ge 0$ holds for every point in the given interval.

Calculation:

Function I:

$f\left(x\right)=e^{-x}\therefore f^{\prime }\left(x\right)=-e^{-x}$

$\because f^{\prime }\left(0\right)=-1<0$

Therefore, the function is non increasing.        

Function II:

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

$f^{\prime }\left(0\right)=0-\cos 0=-1<0$

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:

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

${\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=\frac{3x^2}{2\sqrt{x^3+1}}$

$f^{\prime }\left(0\right)=0,f^{\prime }\left(1\right)>0$

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 

⇒ $\frac{dx}{dt}=2t+3t^2$

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

⇒ x (t = 1) = 12 + 13 = 2

y = 1 + t3 

⇒ $\frac{dy}{dt}=3t^2$

⇒ $\frac{dy}{dt}(t=1)=3\times 1^2=3$

⇒ 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, $\frac{df}{dt}(t=1)=56$