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Concept:

The Squeeze Theorem (The Sandwich Theorem): is used on a function where it will be almost impossible to differentiate.

• The squeeze theorem states that if we define functions such that h(x) ≤ f(x) ≤ g(x) and if \(\rm \displaystyle \lim_{x \to a}h(x) = \lim_{x \to a}g(x) = L\), then \(\rm \displaystyle \lim_{x \to a}f(x) = L\).

 

Calculation:

We know that -1 ≤ sin θ ≤ 1.

⇒ -1 ≤ \(\rm \sin \left(\dfrac1x\right)\) ≤ 1

Since, e^x is a strictly increasing function for all real values of x, we can say that:

⇒ e^-1 ≤ \(\rm e^{\sin \left(\tfrac1x\right)}\) ≤ e¹

Also, since x² ≥ 0, we can say that:

⇒ x²e-1 ≤ \(\rm x^2e^{\sin \left(\tfrac1x\right)}\) ≤ x²e1

⇒ \(\rm \dfrac{x^2}{e}\leq x^2e^{\sin \left(\tfrac1x\right)}\leq x^2e\)

So, we can consider h(x) = \(\rm \dfrac{x^2}{e}\), f(x) = \(\rm \displaystyle \lim_{x \to 0} {x^2}{e^{\sin \left( {\tfrac{1}{x}} \right)}}\) and g(x) = x²e.

Now, \(\rm \displaystyle \lim_{x \to 0}h(x) = \lim_{x \to 0}\dfrac{x^2}{e}=0\).

And \(\rm \displaystyle \lim_{x \to 0}g(x) = \lim_{x \to 0}x^2e=0\).

Since, \(\rm \displaystyle \lim_{x \to 0}h(x) =\lim_{x \to 0}g(x) =0\), we must have \(\rm \displaystyle \lim_{x \to 0}f(x) =0\).

Hence, \(\rm \displaystyle \lim_{x \to 0} {x^2}{e^{\sin \left( {\tfrac{1}{x}} \right)}}=0\).

Concept:

Continuity of a Function:

• A function f(x) is said to be continuous at a point x = a in its domain, if \(\rm \displaystyle \lim_{x\to a}f(x)\) exists or or if its graph is a single unbroken curve at that point.
• f(x) is continuous at x = a ⇔ \(\rm \displaystyle \lim_{x\to a^+}f(x)=\lim_{x\to a^-}f(x)=\lim_{x\to a}f(x)=f(a)\).

 

Differentiability of a Function:

• A function f(x) is differentiable at a point x = a in its domain if its derivative is continuous at a.

This means that f'(a) must exist, or equivalently: \(\rm \displaystyle \lim_{x\to a^+}f'(x)=\lim_{x\to a^-}f'(x)=\lim_{x\to a}f'(x)=f'(a)\).

 

Maxima/Minima:

• If f(x) has a local maximum or a local minimum at a point x = a, then it must be either a critical point [f'(a) = 0] or a point of non-differentiability.

​

Calculation:

Let us check for the continuity and differentiability (maxima/minima) of the function at x = 0.

Continuity:

\(\rm f(x) = \left { \begin{matrix} \rm x^2; & \rm x \leq 0 \ \rm 2\sin x; & \rm x > 0 \end{matrix}\right.\)

\(\rm \displaystyle \lim_{x\to 0^-}f(x)=\lim_{x\to 0^-}x^2\) = 0² = 0.

\(\rm \displaystyle \lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}2\sin x\) = 2 sin 0 = 0.

f(0) = 0² = 0.

∵ \(\rm \displaystyle \lim_{x\to 0^-}f(x)=\lim_{x\to 0^+}f(x)\) = f(0), the function f(x) is continuous at x = 0.

Differentiability:

\(\rm f'(x) = \left { \begin{matrix} \rm 2x; & \rm x \leq 0 \ \rm 2\cos x; & \rm x > 0 \end{matrix}\right.\)

\(\rm \displaystyle \lim_{x\to 0^-}f'(x)=\lim_{x\to 0^-}2x\) = 2×0 = 0.

\(\rm \displaystyle \lim_{x\to 0^+}f'(x)=\lim_{x\to 0^-}2\cos x\) = 2 cos 0 = 2.

∵ \(\rm \displaystyle \lim_{x\to 0^+}f'(x)\neq\lim_{x\to 0^-}f'(x)\), the function is not differentiable at x = 0.

Since, the function is not differentiable at x = 0, let us examine the possibility of maximum/minimum at the point.

The function f(x) = x² is strictly decreasing in (-∞, 0] and its minimum is 0² = 0 at x = 0.

The function f(x) = 2sin x is strictly increasing in \(\rm \left(0,\dfrac{\pi}{2}\right]\) and its minimum is 2 sin 0 = 0 at x = 0.

∴ The function f(x) has a local minimum at x = 0.

Concept:

Definition:

• A function f(x) is said to be continuous at a point x = a in its domain, if \(\rm \displaystyle \lim_{x\to a}f(x)\) exists or if its graph is a single unbroken curve at that point.
• f(x) is continuous at x = a ⇔ \(\rm \displaystyle \lim_{x\to a^+}f(x)=\lim_{x\to a^-}f(x)=\lim_{x\to a}f(x)=f(a)\).

 

Calculation:

For x ≠ 0, the given function can be re-written as:

\(\rm f(x) = \left{ \begin{matrix} \rm \dfrac{1-x}{2}; & \rm x \ne 0 \ \rm K; & \rm x = 0 \end{matrix}\right.\)

Since the equation of the function is same for x < 0 and x > 0, we have:

\(\rm \displaystyle \lim_{x\to 0^+}f(x)=\lim_{x\to 0^-}f(x)=\lim_{x\to 0}\dfrac{1-x}{2}= \dfrac{1-0}{2}=\dfrac{1}{2}\)

For the function to be continuous at x = 0, we must have:

\(\rm \displaystyle \lim_{x\to 0}f(x)=f(0)\)

⇒ K = \(\dfrac{1}{2}\).

Concept:

For a function y = f(x):

• At the relative (local) extrema (maxima or minima), f'(x) = 0.
• In the regions where f(x) is increasing, f'(x) > 0.
• In the regions where f(x) is decreasing, f'(x) < 0.

 

Formula:

• \(\rm \dfrac{d}{dx}x^n=nx^{n-1}\).
• \(\rm \dfrac{d}{dx}e^x=e^x\).
• \(\rm \dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\).

 

Calculation:

Let us first find \(\rm \dfrac{dy}{dx}\) for the given function y = 2x³e^x.

\(\rm \dfrac{d}{dx}(2x^3e^x)=2x^3\left (\dfrac{d}{dx}e^x \right )+2e^x\left (\dfrac{d}{dx}x^3 \right )\)

= 2x3ex + 6x2ex

= 2x2ex(x + 3)

At the extrema, \(\rm \dfrac{dy}{dx}\) = 0.

⇒ 2x2ex(x + 3) = 0

Since, e^x > 0 for all values of x, we get:

⇒ x² = 0 OR x + 3 = 0

⇒ x = 0 OR x = -3.

So, we need to analyze the function in the intervals (-∞, -3), (-3, 0) and (0, ∞).

In the interval (0, ∞):

f'(1) = 2x2ex(x + 3) = 8e > 0.

∴ f(x) is increasing in (0, ∞).

In the interval (-3, 0):

f'(-1) = 2(-1)2e-1(-1 + 3) = \(\rm \dfrac4e\) > 0.

∴ f(x) is increasing in (-3, 0).

​In the interval (-∞, -3):

f'(-4) = 2(-4)2e-4(-4 + 3) = \(\rm -\dfrac{32}{e^4}\) < 0.

∴ f(x) is decreasing in (-∞, -3).

We can see that the function is increasing in the intervals (-3, 0) and (0, ∞).

Concept:​

Derivatives of Trigonometric Functions:

• \(\rm \dfrac{d}{dx}\sin x=\cos x\ \ \ \ \ \ \ \ \ \ \ \ \ \dfrac{d}{dx}\cos x=-\sin x\ \dfrac{d}{dx}\tan x=\sec^2x\ \ \ \ \ \ \ \ \ \ \ \dfrac{d}{dx}\cot x=-\csc^2 x\ \dfrac{d}{dx}\sec x=\tan x\sec x\ \ \ \ \dfrac{d}{dx}\csc x=-\cot x\csc x\)

 

Integration by Parts:

• ∫ f(x) g(x) dx = f(x) ∫ g(x) dx - ∫ [f'(x) ∫ g(x) dx] dx.

 

Calculation:

Integrating by parts by taking csc⁴ x as the first function and sec² x as the second function:

\(\rm \displaystyle \int \sec^2 x \csc^4x\ dx\)

= \(\rm \displaystyle \csc^4 x\int \sec^2 x\ dx-\int \left [\left (\dfrac{d}{dx}\csc^4x \right )\left ( \int \sec^2 x\ dx \right ) \right ]dx+C\)

= \(\rm \displaystyle \csc^4 x\tan x-\int 4\csc^3x(-\cot x\csc x) \tan x\ dx+C\)

= \(\rm \displaystyle \csc^4 x\tan x+4\int \csc^4 x\ dx+C\)

Integrating csc⁴ x by parts and taking csc² x as the first and second functions:

\(\rm \displaystyle\int (\csc^2 x)(\csc^2 x)\ dx\)

= \(\rm \displaystyle \csc^2 x\int \csc^2 x\ dx-\int \left [\left (\dfrac{d}{dx}\csc^2x \right )\left ( \int \csc^2 x\ dx \right ) \right ]dx+C\)

= \(\rm \displaystyle \csc^2 x(-\cot x)-\int 2\csc x(-\cot x\csc x)(-\cot x)\ dx+C\)

= \(\rm \displaystyle -\csc^2 x\cot x-2\int \csc^2 x \cot^2 x\ dx+C\)

Integrating csc2 x cot² x by parts and taking cot2 x as the first and csc² x as the second function:

\(\rm \displaystyle \int \csc^2 x \cot^2 x\ dx\)

= \(\rm \displaystyle \cot^2 x\int \csc^2 x\ dx-\int \left [\left (\dfrac{d}{dx}\cot^2x \right )\left ( \int \csc^2 x\ dx \right ) \right ]dx+C\)

= \(\rm \displaystyle \cot^2 x(-\cot x)-\int 2\cot x (-\csc^2 x)(-\cot x)dx+C\)

= \(\rm \displaystyle -\cot^3 x-2\int \csc^2 x\cot^2 x\ dx+C\)

⇒ \(\rm \displaystyle 3\int \csc^2 x \cot^2 x\ dx=-\cot^3 x+C\)

⇒ \(\rm \displaystyle \int \csc^2 x \cot^2 x\ dx=-\dfrac{\cot^3 x}{3}+C\)

Finally, \(\rm \displaystyle \int \sec^2 x \csc^4x\ dx\)

= \(\rm \csc^4 x\tan x+4\left[-\csc^2 x\cot x-2\left(-\dfrac{\cot^3 x}{3}\right)\right]+C\)

= \(\rm \csc^4 x\tan x-4\csc^2 x\cot x+\dfrac{8}{3}\cot^3 x+C\)

= \(\rm (1+\cot^2 x)^2\tan x-4(1+\cot^2 x)\cot x+\dfrac{8}{3}\cot^3 x+C\)

= \(\rm (1+2\cot^2 x+\cot^4 x)\tan x-4\cot x-4\cot^3 x+\dfrac{8}{3}\cot^3 x+C\)

= \(\rm \tan x+2\cot x+\cot^3 x-4\cot x-\dfrac{4}{3}\cot^3 x+C\)

= \(\rm -\dfrac{1}{3} \cot^3 x + \tan x -2\cot x+ C\)

∴ Comparing \(\rm \left(-\dfrac{1}{3} \cot^3 x + \tan x -2\cot x+ C\right)\) with \(\rm \left(-\dfrac{1}{3} \cot^3 x + k\tan x -2\cot x+ C\right)\) we can say that k = 1.

Concept:

• Derivatives of Trigonometric Functions:

\(\rm \dfrac{d}{dx}\sin x=\cos x\ \ \ \ \ \ \ \ \ \ \ \ \ \dfrac{d}{dx}\cos x=-\sin x\ \dfrac{d}{dx}\tan x=\sec^2x\ \ \ \ \ \ \ \ \ \ \ \dfrac{d}{dx}\cot x=-\csc^2 x\ \dfrac{d}{dx}\sec x=\tan x\sec x\ \ \ \ \dfrac{d}{dx}\csc x=-\cot x\csc x\)

 

• ∫ e^x [f'(x) ​+ f(x)] dx = e^x f(x) + C.

 

Calculation:

\(\rm \displaystyle \int e^x \left(\dfrac{1+\sin x \cos x}{\cos^2 x}\right)dx\)

= \(\rm \displaystyle \int e^x \left(\dfrac{1}{\cos^2 x}+\dfrac{\sin x \cos x}{\cos^2x}\right)dx\)

= ∫ ex [sec² x ​+ tan x] dx 

= ∫ ex [f'(x) ​+ f(x)] dx, where f(x) = tan x.

= e^x f(x) + C

= ex tan x + C.

Concept:

Integration by Parts:

∫ f(x) g(x) dx = f(x) ∫ g(x) dx - ∫ [f'(x) ∫ g(x) dx] dx.

 

Calculation:

Let us first evaluate ∫ (a² - x²)ⁿ dx by integrating it in parts. Let's say f(x) = (a2 - x2)n and g(x) = 1.

∴ I = ∫ [(a2 - x2)n × 1] dx

⇒ I = (a2 - x2)n ∫ 1 dx - ∫ [n(a2 - x2)n-1 × 2x ∫ 1 dx] dx

⇒ I = x(a2 - x2)n - 2n ∫ x²(a2 - x2)n-1 dx

⇒ I = x(a2 - x2)n + 2n ∫ (a2 - x² - a²)(a2 - x2)n-1 dx

⇒ I = x(a2 - x2)n + 2n[ ∫ (a2 - x2)ⁿ dx - a2 ∫ (a2 - x2)n-1 dx]

Since I = ∫ (a2 - x2)n dx, we get:

⇒ I = x(a2 - x2)n + 2nI - 2na2 ∫ (a2 - x2)n-1 dx

⇒ (1 - 2n)I = x(a2 - x2)n - 2na2 ∫ (a2 - x2)n-1 dx

⇒ \(\rm \displaystyle I = \dfrac{x(a^2 - x^2)^n}{1-2n} + \dfrac{2na^2}{2n-1}\int (a^2 - x^2)^{n-1}\ dx\)

⇒ \(\rm \displaystyle I_n = \left [\dfrac{x(a^2 - x^2)^n}{1-2n} \right ]_0^a + \dfrac{2na^2}{2n-1}\int_0^a (a^2 - x^2)^{n-1}\ dx\)

⇒ \(\rm I_n = \left(\dfrac{2na^2}{2n-1}\right)I_{n-1}\).

 

Additional Information

Reduction formula:

• \(\rm \displaystyle \int (a^2 + x^2)^n \ dx=\dfrac{x(a^2 +x^2)^n}{1+2n} + \dfrac{2na^2}{1+2n}\int (a^2 + x^2)^{n-1}\ dx\).
• \(\rm \displaystyle \int (a^2 - x^2)^n \ dx=\dfrac{x(a^2 -x^2)^n}{1-2n} - \dfrac{2na^2}{1-2n}\int (a^2 - x^2)^{n-1}\ dx\).

Concept:

For an odd function f(x): \(\rm \int_{-a}^{\ \ a}f(x)\ dx=0\).

Calculation:

We observe that ax⁵ and bx³ are odd functions of x, 

∵ a(-x)5 = -ax⁵ and b(-x)3 = -bx3.

∴ \(\rm \int_{-2}^{\ \ 2}ax^5\ dx=0\) and \(\rm \int_{-2}^{\ \ 2}bx^3\ dx=0\).

The value of \(\rm \int_{-2}^{\ \ 2}(ax^5 + bx^3 + c)\ dx\) depends on the value of c.

In fact, 

\(\rm \int_{-2}^{\ \ 2}(ax^5 + bx^3 + c)\ dx=\int_{-2}^{\ \ 2}c\ dx\).

= c [2 - (-2)]

= 4c

Additional Information

A function f(x) is:

• Even, if f(-x) = f(x). And \(\rm \int_{-a}^{\ \ a}f(x)\ dx=2\int_{0}^af(x)\ dx\).
• Odd, if f(-x) = -f(x). And \(\rm \int_{-a}^{\ \ a}f(x)\ dx=0\).

Concept:

 

• Two curves f(x, y) = 0 and g(x, y) = 0 cut/touch at a point (a, b) if f(a, b) = g(a, b) = 0.
• The area under the function y = f(x) from x = a to x = b and the x-axis is given by the definite integral \(\rm \displaystyle \left|\int_a^b f(x)\ dx\right|\), for curves which are entirely on the same side of the x-axis in the given range.
• If the curves are on both the sides of the x-axis, then we calculate the areas of both the sides separately and add them.
• Definite integral: If ∫ f(x) dx = g(x) + C, then \(\rm \displaystyle \int_a^b f(x)\ dx = [ g(x)]_a^b=g(b)-g(a).\)
• \(\rm \displaystyle \int x^n\ dx = \dfrac{x^{n+1}}{n+1}+C\).

 

Calculation:

Let's say that the two curves are f(x, y) = x + y - 3 = 0 and g(x, y) = x² - y - 9 = 0.

The points of their intersection are the points where f(x, y) = g(x, y).

⇒ x + y - 3 = x2 - y - 9 = 0

⇒ -x - y + 3 = x2 - y - 9 = 0

⇒ x2 + x - 12 = 0

⇒ x2 + 4x - 3x - 12 = 0

⇒ x(x + 4) - 3(x + 4) = 0

⇒ (x + 4)(x - 3) = 0

⇒ x + 4 = 0 OR x - 3 = 0

⇒ x = -4 OR x = 3.

And, y = 3 - (-4) = 7 OR y = 3 - 3 = 0.

Hence, the curves intersect at the points B(3, 0) and C(-4, 7) as shown in the diagram below:

The points where y = x² - 9 cuts the x-axis (y = 0) are A(-3, 0) and B(3, 0).

The required area is the shaded part ABC = Area of BDC - Area of ADC.

= \(\rm \displaystyle \int_{-4}^3 (3-x)\ dx - \int_{-4}^{-3} (x^2-9)\ dx\)

= \(\rm 3[x]{-4}^{\ \ \ 3}-\left [ \dfrac{x^2}{2} \right ]{-4}^{\ \ \ 3}-\left [ \dfrac{x^3}{3} \right ]{-4}^{-3}+9[ x]{-4}^{-3}\)

= \(\rm 3\left[3-(-4)\right]-\dfrac{1}{2}\left[3^2-(-4)^2\right]-\dfrac{1}{3}\left[(-3)^3-(-4)^3\right]+9[-3-(-4)]\)

= \(\rm 3(7)-\dfrac{1}{2}(-7)-\dfrac{1}{3}(37)+9(1)\)

= \(\rm 21+\dfrac{7}{2}-\dfrac{37}{3}+9\)

= \(\dfrac{127}6\).

The answer is None of these.

Concept:

If \(\rm \vec P, \vec Q, \vec R\) are vectors, then

• \(\rm \vec P \times \vec P = 0\)
• \(\rm \vec P\cdot(\rm \vec P \times \text{any vector}) =0\)
• \(\rm \vec P\cdot (\vec Q\times\vec R) = \vec Q\cdot (\vec R\times\vec P)=\vec R\cdot( \vec P\times\vec Q) =[ \vec P, \vec Q, \vec R]\)
• For dot product \(\rm (\vec P + \vec Q) \cdot \vec R = (\vec P\cdot \vec R) +(\vec Q\cdot \vec R)\)
• For cross product \(\rm (\vec P + \vec Q) \times\vec R = (\vec P\times\vec R) + (\vec Q\times\vec R) \)

 

Calculation:

 \(\rm S=(\vec{a}+\vec{b}+\vec{c}) \cdot[(\vec{a}+\vec{b}) \times( \vec{a}+\vec{c})]\)

⇒ \(\rm S=(\vec{a}+\vec{b}+\vec{c}) \cdot[(\vec{a}\times\vec{a}+\vec{a}\times\vec{c}+\vec{b}\times\vec{a}+\vec{b}\times\vec{c})]\)

⇒ \(\rm S=(\vec{a}+\vec{b}+\vec{c}) \cdot[(0+\vec{a}\times\vec{c}+\vec{b}\times\vec{a}+\vec{b}\times\vec{c})]\)

\(\rm S=\vec{a}\cdot[\vec{a}\times\vec{c}+\vec{b}\times\vec{a}+\vec{b}\times\vec{c}]+\vec{b}\cdot[\vec{a}\times\vec{c}+\vec{b}\times\vec{a}+\vec{b}\times\vec{c}]+\vec{c}\cdot[\vec{a}\times\vec{c}+\vec{b}\times\vec{a}+\vec{b}\times\vec{c}] \)

\(\rm S=\vec{a}\cdot(\vec{a}\times\vec{c})+\vec{a}\cdot(\vec{b}\times\vec{a})+\vec{a}\cdot(\vec{b}\times\vec{c})+\vec{b}\cdot(\vec{a}\times\vec{c})+\vec{b}\cdot(\vec{b}\times\vec{a})+\vec{b}\cdot(\vec{b}\times\vec{c})+\vec{c}\cdot(\vec{a}\times\vec{c})+\vec{c}\cdot(\vec{b}\times\vec{a})+\vec{c}\cdot(\vec{b}\times\vec{c})\)

⇒ \(\rm S=0+0+\vec{a}\cdot(\vec{b}\times\vec{c})+\vec{b}\cdot(\vec{a}\times\vec{c})+0+0+0+\vec{c}\cdot(\vec{b}\times\vec{a})+0\)

⇒ \(\rm S=\vec{a}\cdot(\vec{b}\times\vec{c})+\vec{b}\cdot(\vec{a}\times\vec{c})+\vec{c}\cdot(\vec{b}\times\vec{a}) \)

⇒ \(\rm S=\vec{a}\cdot(\vec{b}\times\vec{c})+\boldsymbol{\rm [-\vec{b}\cdot(\vec{c}\times\vec{a})]+[-\vec{c}\cdot(\vec{a}\times\vec{b})]}\)

⇒ \(\rm S=[\vec{a}, \vec{b}, \vec{c}]-[\vec{a}, \vec{b}, \vec{c}]-[\vec{a}, \vec{b}, \vec{c}]\)

∴ \(\boldsymbol{\rm S=-[\vec{a}, \vec{b}, \vec{c}]} \)