The given expression of the form ∞-∞ So we first we first write in the rational form \(\cfrac { f(x) }{ g(x) } \)
So that it reduces to either \(\cfrac { 0 }{ 0 } \) form \(\cfrac { \infty }{ \infty } \) form. 
\(\therefore \underset { x\rightarrow \infty }{ lim } \sqrt { x } \left\{ \sqrt { x+c } -\sqrt { x } \right\} =\underset { x\rightarrow \infty }{ lim } \cfrac { \sqrt { x } \left\{ \sqrt { x+c } -\sqrt { x } \right\} \left\{ \sqrt { x+c } +\sqrt { x } \right\} }{ \left\{ \sqrt { x+c } +\sqrt { x } \right\} } \) 
= \(\underset { x\rightarrow \infty }{ lim } \cfrac { \sqrt { x } \left( x+c-x \right) }{ \sqrt { x } +c+\sqrt { x } } \) 
= \(\underset { x\rightarrow \infty }{ lim } \cfrac { c\sqrt { x } }{ \sqrt { x+c } +\sqrt { x } } \) 
= \(\underset { x\rightarrow \infty }{ lim } \cfrac { c }{ \sqrt { 1+\frac { c }{ x } } +1 } \) 
= \(\underset { x\rightarrow \infty }{ lim } \cfrac { c }{ \sqrt { 1+\frac { c }{ x } } +1 } =\cfrac { c }{ 2 } \)

\(=\int 25 x e^{-5 x} d x \)
\(=25 x\left(\frac{e^{-5 x}}{-5}\right)-25\left(\frac{e^{-5 x}}{-5 x-5}\right)+c \)
\(=-5 x e^{-5 x}-e^{-5 x}+c\)

\(\int e^{-x^2}xdx\)
Putting x² = u then 2x dx = du
Therefore, \(\int e^{-x^2}xdx=\int e^{-u}{du\over 2}\)
\(={1\over2}\int e^{-u}du={1\over2}(-e^{-u})+c=-{1\over2}e^{-u}+c=-{1\over2}e^{-x^2}+c\)

Let A be a 3 \(\times\) 3 skew symmetric matrix. Thus we have \(A^T=-A\) (by defined)
\(\text {If } \mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{\mathrm{n} \times \mathrm{n}}\) is a skew symmetric matrix, then a_ij = -a_ji for all i anddj.
This means that all the diagonal elements of a skew symmetric matric are 'zero.
The general form of 3 \(\times\) 3 skew symmetric matric are 'zero.
The general form of 3 \(\times\) 3 skew symmetric matrix is
\(A=\left[\begin{array}{ccc} 0 & a_{12} & a_{13} \\ a_{21} & 0 & a_{23} \\ a_{31} & a_{32} & 0 \end{array}\right]\)
We&now that
\(\operatorname{det}(A)=\operatorname{det}\left(A^T\right) \quad \text { [by property (1)] }\)
\(=\operatorname{det}(-A)\)
since A is skew symmetric
\(=(-1)^3 \operatorname{det} A\ [by \ property (2)]\)
\(\operatorname{det} A=-\operatorname{det} A\)
\(\operatorname{det} A+\operatorname{det} A=0\)
\(2 \operatorname{det} A=0\)
\(\operatorname{det} A=0\)
Hence proved.

In general, a 2 \(\times\) 3 matrix is given by A = \(\begin{bmatrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \end{bmatrix}\)
By definition of a_ij, we easily have a₁₁ = \({\sqrt{3}\over 2}|2-3|={\sqrt{3}\over 2} \) and other entries of the matrix
A may be computed similarly. Thus, the required matrix A is\(\begin{bmatrix} {\sqrt{3}\over 2} & 2\sqrt{3} & {7\sqrt{3}\over 2} \\ {\sqrt{3}\over 2}&\sqrt{3} &{5\sqrt{3}\over 2} \end{bmatrix}\)

\(f(x)={1\over 2}x^2\) causes the graph of the function \(f(x)=x^2\) stretches towards x-axis since the multiplying factor is \({1\over 2}\) which is less than one. \(f(x)=2x^2\)causes the graph of the function \(f(x)=x^2\) compresses towards the y-axis that is, moves away from the x-axis since the multiplying factor is 2 which is greater than one.

The given equation is equivalent to the system (x + 4) ≥ 0 and 6 - 4x - x² = (x + 4)²
This implies x ≥ -4 and x² + 6x + 5 = 0. Thus x = -1, -5.
But only x = -1 satisfies both the conditions. Hence, x = -1.

On factorizing the quadratic polynomial we get 3(x + 2)\((x-\frac{1}{3})\le0\) . Draw the number line.

Mark the critical points -2 and \(\frac{1}{3}\) where the factors vanish. On each sub-interval check the sign of (x + 2) \((x-\frac{1}{3})\) . To do this pick an arbitrary point in anywhere in the interval. Whatever sign the resulting value has, the polynomial has the same sign throughout the whole corresponding interval. (Otherwise, there would be another critical point within the interval). This process is easily organized in the following table.

You can see the inequality is satisfied in \([-2,\frac{1}{3}]\)

Let there be n sides of the polygon. We know that the number of diagonals of n sided polygon is \(\frac { n(n-3) }{ 2 } \)
⇒ Given \(\frac { n(n-3) }{ 2 } =90\)
⇒ n²-2n = 180
⇒ n²-3n-180 = 0
⇒ (n-15) (n+12) = 0
⇒  n = 15 or n = -12
⇒ There are 15 sides for the polygon which has 90 diagonals.

Given that P (-3,1) lie on the locus of x² - 5x + ky = 0.
⇒ (-3)²-5 (-3)+k(1) = 0
⇒ 9 +15 + k = 0
⇒ k = -24
Also, it is given that (2, b) lie on the locus of x² - 5x + ky = 0.
⇒ 2² - 5 (2) + kb = 0
⇒ 4 - 10 - 24 (b) = 0
⇒ - 6 - 24b = 0
⇒ -24b = 6
⇒ b = \(\frac{-6}{24}=\frac{-1}{4}\)

The function does not exist for one-toone but not onto.
Since f = A\(\rightarrow\)  B, f is one-one \(\Rightarrow\) f must be onto   [\(\therefore\) n(A) = n(B)]

cos 2\(\theta\)=\(\frac { \sqrt { 5+1 } }{ 4 } \)=cos 36⁰
\(\Rightarrow\) cos2\(\theta\)=cos\(\left( \frac { \pi }{ 5 } \right) \)
\(\Rightarrow 2\theta =2n\pi +\frac { \pi }{ 5 } ,n\epsilon z\)
\(\Rightarrow \theta =n\pi \pm \frac { \pi }{ 10 } ,n\epsilon z\)

\({{3x+1}\over{(x-2)(x+1)}}={{A}\over{(x-2)}}+{{B}\over{(x+1)}}\)
\(\Rightarrow\)  \({{3x+1}\over{(x-2)(x+1)}}={{A(x+1)+B(x-2)}\over{(x-2)(x+1)}}\)
\(\Rightarrow\)  3x + 1 = A(x + 1)B(x - 2)
Putting x = -1 in (1) we get,
-3 + 1 = B(-3)
\(\Rightarrow\)  -2 = -3B \(\Rightarrow\)  \(\boxed{B={{2}\over{}3}}\)
Putting x = 2 in (1) we get,
6 + 1 = A(2 + 1)
\(\Rightarrow\) 7 = 3A \(\Rightarrow\)  \(\boxed{{{7}\over{3}}=A}\)
 \(\therefore\)   \({{3x+1}\over{(x-2)(x+1)}}={{3}\over{x-2}}+{{3}\over{x+1}}={{7}\over{3(x-2)}}+{{2}\over{3(x+1)}}\)

x³ - x²- 17x = 22
⇒ x³ - x² - 17x - 22 = 0
Since x = -2 is one root of the equation, (x + 2) is a divisor of  - x² - 17x - 22 = 0
∴ Using synthetic division,

∴ Consider x² - 3x - 11
Here a = 1, b = -3, c = -11
\(∴ x = {3\pm\sqrt{(.3)^2-4(1)(-11)}\over 2}\)
\(={3\pm\sqrt{9+44}\over2}={3\pm\sqrt{53}\over 2}\)
Hence the other roots are \({3+\sqrt{53}\over 2},{3-\sqrt{53}\over 2}\)

Given in equality is -x² + 3x - 2 > 0.
⇒ x² - 3x + 2 < 0 [∴ a > b ⇒ -a < -b].
⇒ (x -1) (x -2) < 0.
The critical numbers are 1 and 2 and the possible intervals are (-∞ ,1) (1, 2) and (2, ∞)

The in equality x² - 3 x + 2 < 0 is satisfied only in the interval (1, 2).
∴ Solution set is (1, 2).

Given cost of one meal = Rs. 100
Number of customers is given by the function D(x) = 200 - x
∴ Day cost function = (200 - x)(100)
= 20000 - 100x

Since 0 < x<\(\frac{\pi}{2}\) and y = 0\(\frac{\pi}{2}\) , x and y lies in the I quadrant.
ஃ All the trigonometric ratios are positive.
\(\sqrt { { 17 }^{ 2 }-{ 15 }^{ 2 } }\) \(\sqrt { { 13 }^{ 2 }-12^{ 2 } } \)
\(\sqrt { 289-225 } \) = 5
= \(\sqrt { 64 } \)
= 8

\(sin\quad x=\frac { 15 }{ 17 } \quad sin\quad y=\frac { 5 }{ 13 } \)
\(cos\quad x=\frac { 8 }{ 17 } \quad cos\quad y=\frac { 12 }{ 13 } \)
\(tan\quad x=\frac { 15 }{ 8 } \quad tan\quad y=\frac { 5 }{ 12 } \)
= sin x cos y + cos x sin y
=\(\frac { 15 }{ 17 } +\frac { 12 }{ 13 } +\frac { 8 }{ 17 } .\frac { 5 }{ 13 } \)
=\(\frac { 180 }{ 221 } +\frac { 40 }{ 221 } =\frac { 220 }{ 221 } \)

\(y=\left(x^2+5\right) \log (1+e) e^{-3 x}\)
\( \frac{d y}{d x}=\frac{d}{d x}\left[\left(x^2+5\right)\right] \log (1+x) e^{-3 x} +\left(x^2+5\right) \frac{d}{d x}[\log (1+x)] e^{-3 x} +\left(x^2+5\right) \log (1+x) \frac{d}{d x}\left(e^{-3 x}\right) \)
\(\begin{array}{r} =(2 x+0) \log (1+x) e^{-3 x}+\left(x^2+5\right) \cdot\left[\frac{1}{1+x}\right] e^{-3 x} +\left(x^2+5\right) \log (1+x)\left[e^{-3 x}(-3)\right] \end{array}\)
\(\begin{aligned} =e^{-3 x}[2 x \log (1+x)+& \frac{\left(x^2+5\right)}{(1+x)} \left.-3\left(x^2+5\right) \log (1+x)\right] \end{aligned}\)

Given \(C(t)={30t\over 200+t}\)
\(lim_{t\rightarrow \infty}C(t)={30t\over 200+t}\)
\(lim_{t\rightarrow \infty}C(t)=lim_{t\rightarrow \infty}{30t\over 200+t}\)

\(=lim_{t\rightarrow \infty}{30\over {200\over t}+1}=lim_{{1\over t}\rightarrow 0}{30\over {{200\over t}+1}}\) \([\because \ when \ t \rightarrow \infty,{1\over t}\rightarrow 0]\)
\(={30\over 0+1}=30\)
when t \(\rightarrow \infty\), the concentration of salt is 30 gram.

According to the given information, we have the following diagram.

Distance travelled to bring first ball = 24 + 24 = 2 \(\times\) 24 = 48 m
Distance travelled to bring second ball = 2 (24 + 4) = 2(28) = 56 m
Distance travelled to bring third ball = 2 (24 + 4 + 4) = 2(32) = 64 m
\(\therefore\) The series of distances are 48, 56, 64 ...
Here a = 48, d = 56 - 48 = 8 and n = 20.
To find the total distance that he run in bringing back all balls, we have to find the sum of 20 terms of the above series
\(\therefore\) \({ S }_{ 20 }=\frac { 20 }{ 2 } \left[ 2\left( 48 \right) +19\left( 8 \right) \right] \)
= 10[96 + 152]
= 10[248]
S₂₀ = 2480 m.

sin² (A + B) - sin² (A - B) = sin 2A sin 2B
LHS = sin² (A + B) - sin² (A - B)
= sin (A + B + A - B) sin (A +B - A + B)
= sin (2A), sin (2B) = RHS
Hence proved.

\((i) a=5, b=7, P(A)=\frac{a}{a+b}=\frac{5}{5+7}=\frac{5}{1 \cdot 2}\)
\( (ii) P(B)=\frac{2}{5}=\frac{a}{a+b}\)
\(a=2, a+b=5\)
\(b=3\)
The odds that the event B occurs is 2 to 3.

Given P(A) = 0.5, P(B) = 0.8
⇒ P(B/A) = 0.8
We kmow P(B/A) = \(\frac{P(A\cap B)}{P(A)}\)
⇒ 0.8 = \(\frac{P(A\cap B)}{0.5}\)
⇒ \(P(A\cap B)=(0.8)(0.5)=0.4\)
(i) Now P(A./B) = \(\frac{P(A\cap B)}{P(B)}=\frac{0.4}{0.8} =\frac{1}{2}=0.5\)
(ii) \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
= 0.5 + 0.8 - 0.4 = 1.3 - 0.4
= 0.9

\( \int\left(5 x^2-4+\frac{7}{x}+\frac{2}{\sqrt{x}}\right) d x =5 \int x^2 d x-4 \int d x+7 \int \frac{1}{x} d x+2 \int \frac{1}{\sqrt{x}} d x\)
\(=5 \frac{x^{2+1}}{2+1}-4 x+7 \log |x|+2 \frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c \)
\(=\frac{5}{3} x^3-4 x+7 \log |x|+4 \sqrt{x}+c \)