Engineering Mathematics Test 2

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

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

It is an indeterminate form.

Thus using log-concept,

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

Applying L-H Rule, 

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

Concept:

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

f(x) is Continuous at x = a ⇔ \(\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a}}^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a}}^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right)\)

Calculation:

For the function to be continuous,

\(\mathop {\lim }\limits_{x \to 2} f\left( x \right) = f\left( 2 \right)\)

f(2) = (x)² = (2)² = 4

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

For continuity of the above function,

Let \(y = \mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{{e^x}}}{\pi }} \right)^{\frac{1}{x}}}\) 

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

Taking log on both sides

\(\log y = \mathop {\lim }\limits_{x \to \infty } \frac{1}{x}\log \left( {\frac{{{e^x}}}{\pi }} \right)\) 

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

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, \(\mathop {\lim }\limits_{x \to a} f(x)\) exists

 \( ⇒ \mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = \;\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = l = \;\mathop {\lim }\limits_{x \to a} f(x)\), where l is a finite value.

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

Calculation:

Given: The function \(f\left( x \right) = \;\left\{ {\begin{array}{*{20}{c}} {3ax + b,\;for\;x > 1}\\ {11,\;for\;x = 1}\\ {5ax - 2b,\;for\;x < 1} \end{array}} \right.\) 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 \(\mathop {\lim }\limits_{x \to a} f(x) = l = f\left( a \right)\), where l is a finite value.

⇒ \(\mathop {\lim }\limits_{x \to 1} f(x) = f\left( 1 \right)\)

Here f(1) = 11

We know that \( ⇒ \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \;\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \;\mathop {\lim }\limits_{x \to 1} f(x) = f(1)\)

Right hand limit: RHL

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

∵ \(\;\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = f(1)\)

⇒ 3a + b = 11 --------(1)

Left hand limit: : LHL

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

∵ \(\;\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = f(1)\)

⇒ 5a - 2b = 11 ------(2)

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

⇒  2a = 3b

Concept:

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

Calculation:

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

\( = 1 \times 1 \times \frac{{{x^m}}}{{{x^n}}} = \frac{{{x^m}}}{{{x^n}}}\) 

Now,

\(\mathop {\lim }\limits_{x \to 0} \frac{{{x^m}}}{{{x^n}}}\left\{ {\begin{array}{*{20}{c}} {\infty \;\;;\;\;m < n}\\ {1\;\;:\;\;m = n}\\ {0\;\;;\;\;m > n} \end{array}} \right.\;\) 

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

Concept:

Integration by parts:

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

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

Calculation:

Here u = x² and v = cosx

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

\( = {x^2}\left( {sinx} \right) - \smallint 2x \times sinx,\) Now we will find \(\smallint xsinx\)

\(\smallint xsinx = x\smallint sinx - \smallint (\;(\frac{{dx}}{{dx}}) \times \smallint sinx)\) 

\(= x\left( { - cosx} \right) - \smallint - cosx\;\; = - xcosx + sinx\)  putting this value into x²cosx

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

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

\(= {x^2}sinx + 2xcosx - 2sinx\) 

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

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

Hence option 1 is the correct answer.