The following properties are true in calculus:

• If a function is differentiable at any point, then it is necessarily continuous at the point.
• But the converse of this statement is not true i.e. continuity is a necessary but sufficient condition for the Existence of a finite derivative.
• Differentiability implies Continuity
• Continuity does not necessarily imply differentiability.

Hence,

Statement P is wrong.

Statement Q is right

Statement R is Right

Concept:

Function f(x) is laid to be differentiable at x = a, if

$\underset{x\to a}{lim}\left(\frac{f\left(x\right)-f\left(a\right)}{a}\right)$ exist.

And value of this limit is called derivative of f(x) at x = a & it is denoted by f’(a)

i.e. $f^{\prime }\left(a\right)=\underset{x\to a}{lim}\left\{\frac{f\left(x\right)-f\left(a\right)}{a}\right\}$ 

Now,

LHD $=\underset{h\to 0}{lim}\left[\frac{f\left(a-h\right)-f\left(a\right)}{-h}\right]$

RHD $=\underset{h\to 0}{lim}\left[\frac{f\left(a+h\right)-f\left(a\right)}{h}\right]$ 

Shortcut method to find LHD and RHD,

If LHL $=\underset{x\to a^{-}}{lim}f\left(x\right)$  &  RHL $=\underset{x\to a^{+}}{lim}f\left(x\right)$ 

Then,

LHD = LHL of $f^{\prime }\left(x\right)=\underset{x\to a^{-}}{lim}\left[f^{\prime }\left(x\right)\right]$ 

RHD = RHL of $f^{\prime }\left(x\right)=\underset{x\to a^{+}}{lim}\left[f^{\prime }\left(x\right)\right]$ 

Calculation:

$f\left(x\right)=\left|2-3x\right|=\{\begin{array}{c}-\left(2-3x\right),x>\frac{2}{3}\\ 0,x=\frac{2}{3}\\ \left(2-3x\right)x<\frac{2}{3}\end{array}$ 

Exact value of f(x) = 0

L.H.L. $\underset{x\to \frac{2^{-}}{3}}{lim}\left[f\left(x\right)\right]=\underset{x\to \frac{2}{3}}{lim}\left(2-3x\right)$ = 0

R.H.L $=\underset{x\to \frac{2^{+}}{3}}{lim}\left[f\left(x\right)\right]=\underset{x\to \frac{2}{3}}{lim}\left(3x-2\right)$ = 0

∵ L.H.L = R.H.L = Exact value

∴ f(x) is continuous at $x=\frac{2}{3}$

Check for differentiability;

$f\left(x\right)=\{\begin{array}{c}3x-2;x>\frac{2}{3}\\ 0;x=0\\ 2-3x;x<\frac{2}{3}\end{array}$ 

$\therefore f^{\prime }\left(x\right)=\{\begin{array}{c}3;x>\frac{2}{3}\\ 0;x=\frac{2}{3}\\ -3;x<\frac{2}{3}\end{array}$ 

∵ LHD ≠ RHD

∴ at $x=\frac{2}{3}$, f’(x) is Non-Differentiable.

Method II:

Graph of f(x) = |2 – 3x| is

From the graph, at $x=\frac{2}{3}$, function is continuous but since there is a kink at $x=\frac{2}{3}$ , the function is not Differentiable at $x=\frac{2}{3}$

Let $y=\underset{x\to 0}{lim}{\left[\tan \left(\frac{\pi }{4}+x\right)\right]}^{1/x}$ 

It is an indeterminate form.

Thus using log-concept,

log y $=\underset{x\to 0}{lim}\left[\frac{\tan \left(\frac{\pi }{4}+x\right)}{x}\right]$ 

Applying L-H Rule, 

$\log y=\underset{x\to 0}{lim}\left[\frac{\frac{1}{\tan \left(\frac{\pi }{4}+x\right)}\times {\sec }^2\left(\frac{\pi }{4}+x\right)}{1}\right]$ = 2

Concept:

A function f(x) is said to be continuous at a point x = a, in its domain if $\underset{\mathrm{x}\to \mathrm{a}}{lim}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{a}\right)$ exists or its graph is a single unbroken curve.

f(x) is Continuous at x = a ⇔ $\underset{\mathrm{x}\to {\mathrm{a}}^{+}}{lim}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {\mathrm{a}}^{-}}{lim}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to \mathrm{a}}{lim}\mathrm{f}\left(\mathrm{x}\right)$

Calculation:

For the function to be continuous,

$\underset{x\to 2}{lim}f\left(x\right)=f\left(2\right)$

f(2) = (x)² = (2)² = 4

$\underset{x\to 2}{lim}\left(e^{ax}\right)=e^{2a}$ 

For continuity of the above function,

Concept:

In multiple Integral Involving variables as limits, first draw the curves

y = x²

y = 2 - x

Calculation:

$\underset{0}{\stackrel{1}{\int }}x\left(\underset{y=x^2}{\stackrel{y=2-x}{\int }}ydy\right)dx$

$\underset{0}{\stackrel{1}{\int }}x{\left[\frac{y^2}{2}\right]}_{x^2}^{2-x}dx$

$\Rightarrow \frac{1}{2}\underset{0}{\stackrel{1}{\int }}x[{(2-x)}^2-{\left(x^2\right)}^2]dx$

$\Rightarrow \frac{1}{2}\underset{0}{\stackrel{1}{\int }}x(4+x^2-4x-x^4)dx$

$\Rightarrow \frac{1}{2}\underset{0}{\stackrel{1}{\int }}(-x^5+x^3-4x^2+4x)dx$

$\Rightarrow \frac{1}{2}{[-\left(\frac{x^6}{6}\right)+\left(\frac{x^4}{4}\right)-4\left(\frac{x^3}{3}\right)+4\left(\frac{x^2}{2}\right)]}_0^1$

$\Rightarrow \frac{1}{2}[\frac{-1}{6}(1-0)+\frac{1}{4}(1-0)\frac{4}{3}(1-0)+2(1-0)]$

$\Rightarrow \frac{1}{1}[\frac{-1}{6}+\frac{1}{4}-\frac{4}{3}+2]$

$\Rightarrow \frac{1}{2}\left[\frac{-2+3-16+24}{12}\right]$

$\Rightarrow \frac{1}{2}\left[\frac{9}{12}\right]=\frac{9}{24}=\frac{3}{8}$

Let $y=\underset{x\to \mathrm{\infty }}{lim}{\left(\frac{e^x}{\pi }\right)}^{\frac{1}{x}}$ 

It is a ∞° form. To evaluate such limits use log concept.

Taking log on both sides

$\log y=\underset{x\to \mathrm{\infty }}{lim}\frac{1}{x}\log \left(\frac{e^x}{\pi }\right)$ 

$=\frac{x\log e-\log \pi }{x}\left(\frac{\mathrm{\infty }}{\mathrm{\infty }}form\right)$ 

Now applying L-H rule,

$\Rightarrow \log y=\left(\frac{\log e-0}{1}\right)\Rightarrow \log y=1$ 

Concept:

Continuity:

For a function say f, $\underset{x\to a}{lim}f(x)$ exists

 $\Rightarrow \underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a^{+}}{lim}f\left(x\right)=l=\underset{x\to a}{lim}f(x)$, where l is a finite value.

Any function say f is said to be continuous at point say a if and only if $\underset{x\to a}{lim}f(x)=l=f\left(a\right)$, where l is a finite value.

Calculation:

Given: The function $f\left(x\right)=\{\begin{array}{c}3ax+b,forx>1\\ 11,forx=1\\ 5ax-2b,forx<1\end{array}$ is continuous at x = 1.

∵ f(x) is  continuous at x = 1.

As we know that, if a  function say f is said to be continuous at point say a if and only if $\underset{x\to a}{lim}f(x)=l=f\left(a\right)$, where l is a finite value.

⇒ $\underset{x\to 1}{lim}f(x)=f\left(1\right)$

Here f(1) = 11

We know that $\Rightarrow \underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1}{lim}f(x)=f(1)$

Right hand limit: RHL

$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f\left(1+h\right)=\underset{h\to 0}{lim}\left[3a\cdot \left(1+h\right)+b\right]=3a+b$

∵ $\underset{x\to 1^{+}}{lim}f\left(x\right)=f(1)$

⇒ 3a + b = 11 --------(1)

Left hand limit: : LHL

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f\left(1-h\right)=\underset{h\to 0}{lim}\left[5a\cdot \left(1-h\right)-2b\right]=5a-2b$

∵ $\underset{x\to 1^{-}}{lim}f\left(x\right)=f(1)$

⇒ 5a - 2b = 11 ------(2)

From equation (1) and (2) we get,

⇒  2a = 3b

Concept:

For the limit to exists, $\underset{x\to 0}{lim}\left[\frac{\sin \left(x^m\right)}{{\sin }^n\left(x\right)}\right]$ must be a unique finite number.

Calculation:

$\underset{x\to 0}{lim}\left(\frac{\sin \left(x^m\right)}{{\sin }^n\left(x\right)}\right)=\underset{x\to 0}{lim}\left(\frac{\sin \left(x^m\right)}{x^m}\right)\times \underset{x\to 0}{lim}{\left(\frac{x}{\sin x}\right)}^n\times \frac{x^m}{x^n}$ 

$=1\times 1\times \frac{x^m}{x^n}=\frac{x^m}{x^n}$ 

Now,

$\underset{x\to 0}{lim}\frac{x^m}{x^n}\{\begin{array}{c}\mathrm{\infty };m<n\\ 1:m=n\\ 0;m>n\end{array}$ 

For the limit to exist, it must be a finite, value which means

Concept:

Integration by parts:

$\int uv=u\int v-\int ((\frac{du}{dx})\times \int v),$ here u will choose in LIATE order.

L: logs, I: Inverse, A: Algebraic, T: Trigonometry, E: Exponential

Calculation:

Here u = x² and v = cosx

$\int x^{2}cosx=x^2\int cosx-\int ((\frac{d\left(x^2\right)}{dx})\times \int cosx)$ 

$=x^2\left(sinx\right)-\int 2x\times sinx,$ Now we will find $\int xsinx$

$\int xsinx=x\int sinx-\int ((\frac{dx}{dx})\times \int sinx)$ 

$=x\left(-cosx\right)-\int -cosx=-xcosx+sinx$  putting this value into x²cosx

$\int x^{2}cosx=x^2\left(sinx\right)-2\int xsinx$ 

$=x^{2}sinx-2\left(-xcosx+sinx\right)$ 

$=x^{2}sinx+2xcosx-2sinx$ 

${\left[x^{2}sinx+2xcosx-2sinx\right]}_0^{\pi }$ 

$\left[{\pi }^2\times sin\pi +2\times \pi cos\pi -2sin\pi \right]-\left[0+0+2\times sin0\right]$ = - 2π

Hence option 1 is the correct answer.

Concept:

Volume of the solid of rotation obtained by rotating the shaded area by 360° around the x-axis is asked,

there is a direct relation for this;

 ${\mathrm{\forall }}_1=\int \pi y^{2}dx$…1)

Calculation:

Given area;

Using (1); Volume of solid of rotation can be calculated by:

${\mathrm{\forall }}_1={\int }_0^1\pi xdx$

$=\pi {\int }_0^{1}xdx=\pi {\left\{\frac{x^2}{2}\right\}}_0^1$

$=\frac{\pi }{2}\left\{1-0\right\}$

${\mathrm{\forall }}_1=\frac{\pi }{2}units;$

Key points:

In the given problem, the volume is generated by revolving the area by 360° about the x-axis.

But if rotation/revolution is about the y-axis, then the volume of solid of rotation is calculated by:

${\mathrm{\forall }}_2=\int \pi x^{2}dy$ …2)

So, always be careful about which axis rotation is asked.

Depending upon that, you should use either 1) or 2).