Let I \( =\int_{0}^{4}|x-1| d x \)
Clearly \( |x-1| =\left\{\begin{array}{rr} (x-1) \text { if } x \geq 1 -(x-1) \text { if } x<1 \end{array}\right. \)
\(I =\int_{0}^{1}|x-1| d x+\int_{1}^{4}|x-1| d x \)
\( =\int_{0}^{1}-(x-1) d x+\int_{1}^{4}(x-1) d x \)
\( =\left[-\frac{x^{2}}{2}+x\right]_{0}^{1}+\left[\frac{x^{2}}{2}-x\right]_{1}^{4}\)
\( =\left[-\frac{1}{2}+1-0\right]+\left[\frac{16}{2}-4-\frac{1}{2}+1\right] \)
\( =\frac{1}{2}+\frac{15}{2}-3 \)
\( =\frac{1+15-6}{2}=\frac{10}{2}=5 \)

\( \therefore q =1-p=1-\frac{1}{5}=\frac{4}{5} \)
\(\mathrm{P}(\mathrm{X}=x) =n \mathrm{C}_{x} p^{x} q^{n-x}, x=0,1,2, \ldots \ldots n \)
\(\mathrm{P}(\mathrm{X}=k) =\mathrm{P}(\mathrm{X}=4) \)
\(=10 \mathrm{C}_{4}\left(\frac{1}{5}\right)^{4}\left(\frac{4}{5}\right)^{10-4}=210\left(\frac{1}{5^{4}}\right)\left(\frac{4^{6}}{5^{6}}\right)=210\left(\frac{1}{5}\right)^{4}\left(\frac{4}{5}\right)^{6} \)

\(\frac { 13 }{ 3 } \)

\(\frac { 1 }{ loga } -\frac { 1 }{ logb } \)

a *b = min (a, b) on A = {1,2,3,4, 5}  Let a,b ∈A
A = {1,2, 3, 4, 5}
a*b = min {(a, b)}
Now, 1,2 ∈ A \(\Rightarrow\)1 * 2 = 1 ∈ A
3, 5 ∈ A \(\Rightarrow\) 3 * 5 = 3 ∈ A
Hence * is a binary operation on A

In this problem a ∗ b is in the quotient form. Since the division by 0 is undefined, the denominator b -1 must be nonzero.
It is clear that b −1 = 0 if b = 1. As 1∈Q, ∗ is not a binary operation on the whole of Q. However it can be found that by omitting 1 from Q, the output a ∗b exists in Q\{1}. Hence ∗ is a binary operation on Q\{1}.

Since × is binary operation on Z, a,b ∈ Z⇒ a × b = ab∈Z and b × b = b²∈Z ...(1)
The fact that + is binary operation on Z and (1) ⇒ 3ab = (ab + ab + ab) ∈Z and 5b²= (b²+b²+b²+b²+b²)∈Z     ...(2)
Also a∈Z and 3ab ∈Z implies a+3ab∈Z   ...(3)
(2),(3), the closure property of -on Z yield a * b = (a+3ab-5b²)∈Z. Since a * b belongs to Z, * is a binary operation on Z.

\({ { \left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) } }^{ 2 }+{ \left( \frac { dy }{ dx } \right) }^{ 2 }=xsin\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) \)
The highest derivative is 2
∴ Order 2
The given differential equation is not a polynomial equation in its derivative and so its degree is not defined.

\({ \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ \frac { 2 }{ 3 } }-3\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +5\frac { dy }{ dx } +4=0\)
Given differential equation is
\({ \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ \frac { 2 }{ 3 } }-3\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) +5\frac { dy }{ dx } +4=0\)
\(\Rightarrow { \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ \frac { 2 }{ 3 } }=3\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) -5\left( \frac { dy }{ dx } \right) -4\)
Taking power 3 both sides,
\(\Rightarrow { \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ 2 }={ \left( 3\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) -5\left( \frac { dy }{ dx } \right) -4 \right) }^{ 3 }\)
The highest derivative is 3 and its power is 2.
∴ Order 3, degree 2.

Given adj (A) =\(\left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \)
We know that A^-1 = ±\(\frac { 1 }{ \sqrt { |adjA| } } \) (adj A) ...............(1)
|adj A| = 0 + 2\(\left| \begin{matrix} 6 & -6 \\ -3 & 6 \end{matrix} \right| \) + 0
[Expanded along R₁]
= 2(36-18) = 2(18) = 36
∴ A^-1 = \(\pm \frac { 1 }{ \sqrt { 36 } } \left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \)
= \(\pm \frac { 1 }{ 6 } \left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \).

Given z₁= 2 - i and z₂= -4+3i
z₁z₂ = (2-i)(-4+3i)
= -8 + 6i + 4i - 3i²
= -8 +10i - 3(-1)
= -8 +10i + 3 = -5 +10i
Inverse of z₁z₁ is \(\frac { 1 }{ { z }_{ 1 }{ z }_{ 2 } } \)
= \(\frac { 1 }{ -5+10i } \times \frac { -5-10i }{ -5-10i } \)
= \(\frac { -5-10i }{ (-5)^{ 2 }-(10i)^{ 2 } } \)
= \(\frac { -5-10i }{ 25-100i^{ 2 } } \)
= \(\frac { -5-10i }{ 25+100 } \)  [∵ i² = -1]
\(=\frac{\not{5}(-1-2 i)}{\not{5}\langle(2 5)}=\frac{-1-2 i}{25}\)
∴ Inverse of z₁z₂ is \(\frac { 1 }{ 25 } \) (-1-2i)
Inverse of \(\frac { { z }_{ 1 } }{ { z }_{ 2 } } \) is \(\frac { 1 }{ \frac { { z }_{ 1 } }{ { z }_{ 2 } } } =\frac { { z }_{ 2 } }{ { z }_{ 2 } } \)
∴ Inverse of \(\frac { { z }_{ 1 } }{ { z }_{ 2 } } =\frac { { z }_{ 2 } }{ { z }_{ 1 } } =\frac { -4+3i }{ 2-i } \times \frac { 2+i }{ 2+i } \)
= \(\frac { -4+3i }{ 2-i } \times \frac { 2+i }{ 2+i } \)
= \(\frac { -8-4i+6i+3i^{ 2 } }{ 2^{ 2 }-({ i }^{ 2 }) } \)
= \(\frac { -8+2i-3 }{ 4+1 } =\frac { -11+2i }{ 5 }\)
\( =\frac { 1 }{ 5 } \)(-11 + 2i)
∴ Inverse of \(\frac { { z }_{ 1 } }{ { z }_{ 2 } } \) is \(\frac { -11+2i }{ 5 }\)or \(\frac { 1 }{ 5 } \)(-11 + 2i)

\(Re\left( \frac { 1 }{ z } \right) \)
= \(Re\left( \frac { 1 }{ x+iy } \right) =Re\left( \frac { 1 }{ x+iy } \times \frac { x-iy }{ x-iy } \right) \)
= \(Re(\left( \frac { x-iy }{ { x }^{ 2 }({ i }^{ 2 }{ y }^{ 2 }) } \right) =Re\left( \frac { x-iy }{ { x }^{ 2 }+{ y }^{ 2 } } \right) \)
∴ Real part is \(\frac { x }{ { x }^{ 2 }+{ y }^{ 2 } } \)

Let cos⁻¹\(\left( \frac { \sqrt { 3 } }{ 2 } \right) \) = y. Then, cos y = \( \frac { \sqrt { 3 } }{ 2 } \)
The range of the principal values of y = cos⁻¹x is [0, \(\pi\)].
So, let us find y in [0, \(\pi\)] such that cos y =\( \frac { \sqrt { 3 } }{ 2 } \)
But, cos\(\frac{\pi}{6}=\frac{\sqrt3}{2} and \frac{\pi}{6}\in[0,\pi]\). Therefore, y = \(\frac{\pi}{6}\)
Thus, the principal value of cos^-1 \(\left( \frac { \sqrt { 3 } }{ 2 } \right) is\frac { \pi }{ 6 } \)

 We know that sin^-1: [-1, 1] \(\rightarrow \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)is given by
sin⁻¹x = y if and only if x = sin y for −1\(\le x\le \) and -\(\frac { \pi }{ 2 } \le y \le \frac { \pi }{ 2 } \). Thus,
 \({ Sin }^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) \)= \(\frac{\pi}{4}\), Since \(\frac{\pi}{4}\)\(\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)and sin \(\frac { \pi }{ 4 } =\frac { 1 }{ \sqrt { 2 } } \)

Equation of the circle in standard form is (xr − h)² + (y − k)² = r²
\( \Rightarrow (x-(-3))^{2}+(y-(-4))^{2} =3^{2} \)
\( \Rightarrow (x+3)^{2}+(y+4)^{2} =3^{2} \)
\( \Rightarrow x^{2}+y^{2}+6 x+8 y+16 =0 .\)

\(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 4 }x{ cos }^{ 6 }xdx=\frac { (6-1) }{ (6+4) } .\frac { (6-3) }{ (6+4-2) } .\frac { (6-5) }{ (6+4-4) } .\frac { (4-1) }{ (4) } .\frac { (4-3) }{ (4-2) } .\frac { \pi }{ 2 } } \)
\(=\frac { (5) }{ (10) } \frac { (3) }{ (8) } \frac { (1) }{ (6) } \frac { (3) }{ (4) } \frac { (1) }{ (2) } \frac { \pi }{ 2 } =\frac { 3\pi }{ 512 } \)
Also, \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 4 }x{ cos }^{ 6 }xdx } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 6 }x{ cos }^{ 4 }xdx } =\frac { (3) }{ (10) } \frac { (1) }{ (8) } \frac { (5) }{ (6) } \frac { (3) }{ (4) } \frac { (1) }{ (2) } \frac { \pi }{ 2 } =\frac { 3\pi }{ 512 } \)

Observe that the function g is defined for all (x, y)∈R² It is easy to check, as in the above examples, that g is continuous at all point (x, y) ≠ (0, 0). Next, we shall check the continuity of g at (0, 0). For that we see if g has a limit L at (0, 0) and if L = g(0, 0) = 0. So we consider
\(\left| g\left( x,y \right) -g\left( 0,0 \right) \right| =\left| \frac { 2{ x }^{ 2 }y }{ { x }^{ 2 }+{ y }^{ 2 } } -0 \right| =\frac { 2\left| { x }^{ 2 }y \right| }{ \left| { x }^{ 2 }+{ y }^{ 2 } \right| } =\frac { 2\left| xy \right| \left| x \right| }{ { x }^{ 2 }+{ y }^{ 2 } } \le \frac { \left( { x }^{ 2 }+{ y }^{ 2 } \right) \left| x \right| }{ { x }^{ 2 }+{ y }^{ 2 } } \le \left| x \right| \)   ...(9)
Note that in the final step above we have used 2 \(\left| xy \right| \) \(\le \) x² + y² (which follows by considering 0\(\le \) (x - y)² for all  x, y∈ R . Note that (x, y)→(0, 0) implies |x| → 0. Then from (9) it follows that \(\underset { (x,y)\longrightarrow (0,0) }{ lim } \) \(\frac { 2{ x }^{ 2 }y }{ { x }^{ 2 }+{ y }^{ 2 } } \) = 0 = g (0, 0) which proves that g is continuous at (0, 0). So g is continuous at every point of R²

\(\int _{ -1 }^{ 1 }{ \frac { dx }{ { x }^{ 2 }+2x+5 } } \)
\(=\int _{ -1 }^{ 1 }{ \frac { dx }{ { x }^{ 2 }+2x+1+4 } } =\int _{ -1 }^{ 1 }{ \frac { dx }{ { (x+1) }^{ 2 }{ 2 }^{ 2 } } } \)
\(\left[ \because \int { \frac { dx }{ { x }^{ 2 }-{ a }^{ 2 } } } =\frac { 1 }{ 2a } log\left| \frac { x-a }{ x+a } \right| \right] \)
\(={ \left[ \frac { 1 }{ 2 } { tan }^{ -1 }\left( \frac { x+1 }{ 2 } \right) \right] }_{ -1 }^{ 1 }\)
\(=\frac { 1 }{ 2 } \left[ { tan }^{ -1 }(1)-{ tan }^{ -1 }(0) \right] \)
\(\\ =\frac { 1 }{ 2 } \left[ \frac { \pi }{ 4 } \right] =\frac { \pi }{ 8 } \)

Sample space s = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
n(S) = 8
Let X be the random variable denoting the number of heads
X = {0, 1, 2, 3}
Values of randomn variabl 0 1 2 3 Total Number of elements in inverse image 1 3 3 1 8 Probability mass function 
x 0 1 2 3 f(x) \(\cfrac { 1 }{ 8 } \) \(\cfrac { 3 }{ 8 } \) \(\cfrac { 3 }{ 8 } \)  \(\cfrac { 1 }{ 8 } \) \(\therefore f(x)=\begin{cases} \begin{matrix} \frac { 1 }{ 8 } & \begin{matrix} for & x=0,3 \end{matrix} \end{matrix} \\ \begin{matrix} \frac { 3 }{ 8 } & \begin{matrix} for & x=1,2 \end{matrix} \end{matrix} \end{cases}\)

Given s = 2t² + 3t
s(3) = 2 \(\times\) 3² + 3 (3)
= 2\(\times\)9+9
= 27 m ....(1)
s(6) = 2\(\times\) 6² + 3 (6)
= 72 + 18 = 90m ... (2)
Average velocity = \(\frac { s(6)-s(3) }{ 6-3 } \)
= \(\frac { 90-27 }{ 3 } \) = 21 m/s
(ii) Instantaneous Velocity V(t) = \(\frac { ds }{ dt } \)
Instantaneous Velocity at t = 3
= V(3) = 15 m/sec  
Instantaneous Velocity at t = 6
= V(6) = 27m/sec

Given adj A =\(\left[ \begin{matrix} 2 & -4 & 2 \\ -3 & 12 & -7 \\ -2 & 0 & 2 \end{matrix} \right] \)
We know that A = \(\pm \frac { 1 }{ \sqrt { |adjA| } } \) adj (adj A)..(1)
|adj A| = \(2\left| \begin{matrix} 12 & -7 \\ 0 & 2 \end{matrix} \right| +4\left| \begin{matrix} -3 & -7 \\ -2 & 2 \end{matrix} \right| +2\left| \begin{matrix} -3 & 12 \\ -2 & 0 \end{matrix} \right| \)
[Expanded along R₁]
= 2(24-0)+4(-6-14)+2(0+24)
= 2(24)+4(-20)+2(24) = 48-80+48
= 96-80 = 16
Now, adj (adj A)
=\(\left[ \begin{matrix} +\left| \begin{matrix} 12 & -7 \\ 0 & 2 \end{matrix} \right| & -\left| \begin{matrix} -3 & -7 \\ -2 & 2 \end{matrix} \right| & +\left| \begin{matrix} -3 & 12 \\ -2 & 0 \end{matrix} \right| \\ -\left| \begin{matrix} -4 & 2 \\ 0 & 2 \end{matrix} \right| & +\left| \begin{matrix} 2 & 2 \\ -2 & 2 \end{matrix} \right| & -\left| \begin{matrix} 2 & -4 \\ -2 & 0 \end{matrix} \right| \\ +\left| \begin{matrix} -4 & 2 \\ 12 & -7 \end{matrix} \right| & -\left| \begin{matrix} 2 & 2 \\ -3 & -7 \end{matrix} \right| & +\left| \begin{matrix} 2 & -4 \\ -3 & 12 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(24-0)-(6-14)+(0+24) \\ -(-8-0)+(4+4)-(0-8) \\ +(28-24)-(-14+6)+(24-12) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 24 & 20 & 24 \\ 8 & 8 & 8 \\ 4 & 8 & 12 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 24 & 8 & 4 \\ 20 & 8 & 8 \\ 24 & 8 & 12 \end{matrix} \right] \)
= \(4\left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] \)
Substituting (2) and (3) in (1) we get,
A = \(\frac { 1 }{ \sqrt { 16 } } .4\left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] \)
A = \(\pm \frac { 4 }{ 4 } \left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] =\pm \left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] \)

The roots of x³+2x2+3x+4 = 0 are ∝, β, ૪
∴ ∝+β+૪ = -co-efficient of x² = -2 ...(1)
∝β + β૪ + ૪∝ = co-effficient of x = 3 ....(2)
-∝β૪ = +4 ⇒ ∝β૪ = -4 ...(3)
Form a cubic equation whose roots are 2∝, 2β, 2૪
2∝+2β+2૪ = 2(∝+β+૪) = 2(-2) = -4 [from (1)]
4∝β+4β૪+4૪∝ = 4(∝β+β૪+୪∝) = 4(3) = 12 [from (2)]
(2∝)(2β)(2૪) = 8(∝β૪) = 8(-4) = -32 [from (3)]
∴ The required cubic equation is
x³-(2∝+2β+2૪)x² + (2∝β+2β૪+2୪∝)x - (2∝)(2β)(2૪) = 0
⇒ x³+(-4)x²+12x+32 = 0
⇒ x³+4x²+12x+32 = 0

With usual notations in triangle ABC, let \(\vec { BC } =\vec { a } ,\vec { CA } =\vec { b } \)  and  \(\vec { AB } =\vec { c } \).
Then \(\left| \vec { BC } \right| =a\) , \(\left| \vec { CA } \right| =b\), \(\left| \vec { AB} \right| =c\), and \(\vec { BC } +\vec { CA } +\vec { AB } =\vec { 0 } \)
So, \(\vec { BC } =-\vec { CA } -\vec { AB } \)
Applying dot product, we get
\(\vec { BC } .\vec { BC } =-\vec { BC } .\vec { CA }-\vec { BC }. \vec { AB } \)
⇒ \({ \left| \vec { BC } \right| }^{ 2 }=-\left| \vec { BC } \right| \left| \vec { CA } \right| \) cos(兀-c)-\(\left| \vec { BC } \right| \left| \vec { AB} \right| \)cos(兀-B)
⇒ a² = ab cos C + ac cos B
Therefore a = b cos C + c cos B
The results (ii) and (iii) are proved in a similar way

With usual notations in triangle ABC, we have \(\vec { BC } =\vec { a } ,\vec { CA } =\vec { b } \) \(\vec { AB } =\vec { c } \).
Then \(\left| \vec { BC } \right| =a\) , \(\left| \vec { CA } \right| =b\), \(\left| \vec { AB} \right| =c\), and \(\vec { BC } +\vec { CA } +\vec { AB } =\vec { 0 } \)
So, \(\vec { BC } =-\vec { CA } -\vec { AB } \)
Then applying dot product, we get
\(\vec { BC } .\vec { BC } =(-\vec { CA } -\vec { AB } ).(-\vec { CA } -\vec { AB } )\)
⇒ \({ \left| \vec { BC } \right| }^{ 2 }={ \left| \vec { CA } \right| }^{ 2 }+{ \left| \vec { AB } \right| }^{ 2 }+\vec { 2CA } .\vec { AB } \)
⇒ a² = b²+c²+2bc cos (\(\pi\) - A)
⇒ a² = b²+c²−2bc cos A.
The results (ii) and (iii) are proved in a similar way.

Equation of the circle passing through the points of intersection of the chord and circle by
Theorem  is x² + y²−16+\(\lambda \)(3x + y + 5) = 0 .
The chord 3x + y + 5 = 0 is a diameter of this circle if the centre\(\left( \frac { -3\lambda }{ 2 } \frac { -\lambda }{ 2 } \right) \) lies on the chord.
So we have 3\(\left( \frac { -3\lambda }{ 2 } \right) \)-\(\frac { -\lambda }{ 2 } \)+5 = 0,
\(\frac { -9\lambda }{ 2 } \)-\(\frac { \lambda }{ 2 } \)+5 = 0,
−5λ + 5 = 0 ,
λ = 1.
Therefore, the equation of the required circle is x² + y²+3x + y −11 = 0.

Note that the function f takes value 1 on the x, y-axes and 0 everywhere else on R². So let us calculate
\(\frac { \partial f }{ \partial x } (0,0)\) = \(\underset { h\longrightarrow 0 }{ lim } \) \(\frac { f\left( 0+h,0 \right) -f(0,0) }{ h } =\underset { h\longrightarrow 0 }{ lim } \frac { 1-1 }{ h } =0;\)
\(\frac { \partial f }{ \partial y } (0,0)\) = \(\underset { k\longrightarrow 0 }{ lim } \) \(\frac { f\left( 0,0+k \right) -f(0,0) }{ k } =\underset { k\longrightarrow 0 }{ lim } \frac { 1-1 }{ k } =0\)
This completes (i).

Here f (x) = x³, a = 0 and b = 1. Hence, we get
\(\int _{ a }^{ b }{ f(x)dx } =\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ f } \left( \frac { r }{ n } \right) \Rightarrow \int _{ 0 }^{ 1 }{ x^3dx } =\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ \frac { r^3 }{ n^3 } } \)
\(=\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ { n }^{ 4 } } [{ 1 }^{ 3 }+{ 2 }^{ 3 }+...+{ n }^{ 3 }]=\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ { n }^{ 4 } } \frac { { n }^{ 2 }{ (n+1) }^{ 2 } }{ 4 } \)
\(=\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ 4 } { \left( 1+\frac { 1 }{ n } \right) }^{ 2 }=\frac { 1 }{ 4 } \)

Here f (x) = x, a = 0 and b = 1. Hence, we get
\(\int _{ a }^{ b }{ f(x)dx } =\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ f } \left( \frac { r }{ n } \right) \Rightarrow \int _{ 0 }^{ 1 }{ xdx } =\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ \frac { r }{ n } } \)
\(=\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ { n }^{ 2 } } [1+2+...+n]\)
\(=\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ { n }^{ 2 } } \frac { n(n+1) }{ 2 } =\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ 2 } \left( 1+\frac { 1 }{ n } \right) =\frac { 1 }{ 2 } \)

The average rate of change in an interval [a, b] is \(\frac { f(b)-f(a) }{ b-a } \) whereas, the instantaneous rate of change at a point x is f′(x) for the given function. They are respectively, b + a and 2x.

Rate of Changes

a	b	x	Average rate is \(\frac { f(b)-f(a) }{ b-a } \) = b+a	Instantaneous rate is f'(x) = 2x
0	0.5	0.5	0.5	1
0.5	1	1	1.5	2
1	1.5	1.5	2.5	3
1.5	2	2	3.5	4

The discriminant Δ =((4p)² - 4(4)(p+2) = 16(p²-p-2) = 16(p+1)(p-2). So we get
Δ < 0 if -1 < p < 2
Δ = 0 if p = -1 or p = 2
Δ >0 if \(\infty\). Thus the given polynomial has
imaginary roots if -1 < p < 2
equal real roots if p = −1 or p = 2;
distinct real roots if -\(\infty\) < p < -1 or 2 < p < \(\infty\)