Given \(x_{0}=3, f(x)= \begin{cases}\frac{x^{2}-9}{x-3}, & \text { if } x \neq 3 \\ 5, & \text { if } x=3\end{cases}\)
\(lim_{x\rightarrow 3^-}f(x)=lim_{x\rightarrow 3^-}{x^2-9\over x-3}=lim_{x\rightarrow 3^-}{(x+3)(x-3)\over x-3}=lim_{x\rightarrow 3^-}x+3=6\)
\(lim_{x\rightarrow 3^+}f(x)=lim_{x\rightarrow 3^+}{x^2-9\over x-3}=lim_{x\rightarrow 3^+}(x+3)=6\)
But f(3)=5
\(\therefore lim_{x\rightarrow 3^-}f(x)=lim_{x\rightarrow 3^+}f(x)\neq f(3)\)
\(\therefore f(x)\) continuous at x_o=3

i) Given: f(0) is undefined
\(\lim _{x \rightarrow 0} f(x)=4, f(2)=6, \lim _{x \rightarrow 2} f(x)=3\)
To sketch the graph of y = f(x)
(i) Consider the points (0, 4) and (2, 3)
(ii) Draw a curve straight. line passing through (0, 4) and (2, 3)
(iii) Remove (0, 4), (2, 3) and plot the point (2, 6)

\(|A|=\left|\begin{array}{ccc} 4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x \end{array}\right|\)
Put x = 0
\(|A|=\left|\begin{array}{lll} 4 & 4 & 4 \\ 4 & 4 & 4 \\ 4 & 4 & 4 \end{array}\right|=0 \quad\left(C_1 \cong C_2 \cong C_3\right)\)
Since all the three rows are identical.
\(\therefore(x-0)^2=x^2 \text { is a factor of }|A|\)
\(\text { Applying } C_1 \rightarrow C_1+C_2+C_3\)
\(|A|=\left|\begin{array}{ccc} 12+x & 4+x & 4+x \\ 12+x & 4-x & 4+x \\ 12+x & 4+x & 4-x \end{array}\right|\)
Put x = -12
\(=\left|\begin{array}{lll} 0 & 4+x & 4+x \\ 0 & 4-x & 4+x \\ 0 & 4+x & 4-x \end{array}\right|=0\)
\(\therefore\) (x+12) is an other factor of lA|.
The number of roots must be 3
\(\therefore\) x= 0,0,-12

\(\mathrm{LHS}=\left|\begin{array}{ccc} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{array}\right|\)
\(\text { Applying } C_1 \rightarrow C_1-C_2 \& C_2 \rightarrow C_2-C_3\)
\(=\left|\begin{array}{ccc} a & 0 & 1 \\ -b & b & 1 \\ 0 & -c & 1+c \end{array}\right|\)
\(=a(b+b c+c)+0+1(b c-0)\)
\(=a b+a b c+a c+b c\)
\(=\frac{1}{a b c}\left(\frac{1}{c}+1+\frac{1}{b}+\frac{1}{a}\right)\)
\(=\frac{1}{a b c}\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\mathrm{RHS}\)
Hence proved.

By unrolling the hollow cylinder and flattening it into a rectangle, and with a single reflection allows us to determine the ant's path, as shown the figure. Let the baseline x-axis in cm and the vertical line through A (initial position of the ant) be the y-axis. Let H be the position of honey drop and E be the entry point of ant inside the vessel. From the given information we have
Let A(x₁, y₁) and H(x₂, y₂) be (0, 4) and (12, 13) respectively.
(i) The shortest distance between A and H is
\(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{12^2+9^2}=15\)
the ant would need to crawl 15 cm to get the honey.
(ii) The equation of the path AH is \(\frac{y-4}{13-4}=\frac{x-0}{12-0}\)
y = 0.75x + 4
(iii) At the entry point E, y = 10 ⇒ x = 8
E = (8, 10)

Given that the pair of straight lines,
x²- 2cxy - y² = 0  ...(1)
x²- 2dxy - y² = 0 ...(2)
The equation of the angle bisectors of equation (1) is
\(\frac { { x }^{ 2 }-{ y }^{ 2 } }{ 2 } =\frac { xy }{ -c } \)
⇒ cx² + 2xy - cy² = 0
Given that the angle bisector of equation (1) is (2) Therefore equations
x²- 2dxy - y² = 0
cx² + 2xy - cy² = 0
represent the same equation as the angle bisector of equation (1)
Comparing the like terms of the above two equations, we get
\(\frac { 1 }{ c } =\frac { -2d }{ 2 } =\frac { -1 }{ -c } \)
⇒ cd = -1

Let ABC be a triangle with constant perimeter 2s. Thus s is constant.
We know that \(\triangle =\sqrt{s(s-a)(s-b)(s-c)}\)
Observe  that \(\triangle\) is maximum, when (s - a) (s - b) (s - c) is maximum.
Now, (s-a)(s-b)(s-c)\(\le\)\(({(s-a)+(s-b)+(s-c)\over 3})^3\) = \({s^3\over 27}\) [G.M\(\le\) A.M]
Thus, we get (s - a) (s - b) (s - c) \(\le\) \({s^3\over 27}\)
Equality occurs when s - a = s - b = s - c. That is, when a = b = c maximum of (s - a) (s - b) (s - c) is \({s^3\over 27}\)
Thus, for a fixed perimeter 2s, the area of a triangle is maximum when a = b = c.
Hence, for a fixed perimeter, the equilateral triangle has the maximum area and the maximum area is given by \(\triangle =\sqrt {s(s)^3\over 27}={s^2\over3\sqrt{3}}sq.units\)

\(y=\frac{2\ sin\alpha}{1+cos\alpha+sin\alpha}\)
\(=\frac{2sin\alpha[1+sin\alpha-cos\alpha]}{[(1+sin\alpha)+cos\alpha](1+sin\alpha-cos\alpha)}\)
\(=\frac{2sin\alpha(1-cos\alpha+sin\alpha)}{(1+sin\alpha)^2-cos^2\alpha}\)
\(=\frac{2sin\alpha(1-cos\alpha+sin\alpha)}{1+2sin\alpha+sin^2\alpha-(1-sin^2\alpha)}\)
\(=\frac{2sin\alpha(1-cos\alpha+sin\alpha)}{2sin\alpha+2sin^2\alpha}\)
\(=\frac{2sin\alpha(1-cos\alpha+sin\alpha)}{2sin\alpha(1+sin\alpha)}\)
\(=\frac{1-cos\alpha+sin\alpha}{1+sin\alpha}\)

Three balls must be drawn with atleast one black ball.
The possible choices are as follows

White balls(2)	Black balls(3)	Red balls(4)	Combination
2	1	0	2C2\(\times \)3C1\(\times \)4C0
0	1	2	2C0\(\times \)3C1\(\times \)4C2
1	1	1	2C1\(\times \)3C1\(\times \)4C1
1	2	0	2C1\(\times \)3C2\(\times \)4C0
0	2	1	2C0\(\times \)3C2\(\times \)4C1
0	3	0	2C0\(\times \)3C3\(\times \)4C0

∴ Required number of ways of drawing 3 balls?
= \(2{ C }_{ 2 }\times { 3C }_{ 1 }\times { 4C }_{ 0 }+2C_{ 0 }\times { 3C }_{ 1 }\times { 4C }_{ 2 }+2C_{ 1 }\times { 3C }_{ 1 }\times { 4C }_{ 1 }\)\(\times \) \({ 2C }_{ 1 }\times 3C_{ 2 }\times { 4C }_{ 0 }+{ 2C }_{ 0 }\times { 3C }_{ 2 }\times { 4C }_{ 1 }+{ 2C }_{ 0 }\times { 3C }_{ 3 }\times { 4C }_{ 0 }\)
= 1\(\times \)3\(\times \)1+1\(\times \)3\(\times \)6+2\(\times \)3\(\times \)4+2\(\times \)3\(\times \)1+1\(\times \)3\(\times \)4+1\(\times \)1
= 3 + 18 + 24 + 6 + 12 + 1 = 64

(i) Find the linear relationship between C and F.
By the given data
x₁ (100°C)  y₁(212°F)
x₂(0°C) y₂(32°F)
Using two point form, the linear relationship between C and F is
\(\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
\(\frac { y-212 }{ 32-212 } =\frac { x-100 }{ 0-100 } \)
\(\Rightarrow \quad \frac { y-212 }{ -180 } =\frac { x-100 }{ -100 } \)
\(\Rightarrow \quad \frac { y-212 }{ 9 } =\frac { x-100 }{ 5 } =\frac { 5 }{ 9 } (y-212)=x-100\)
\(\Rightarrow \quad x=\frac { 5 }{ 9 } (y-212)+100\quad \Rightarrow \quad x=\frac { 5 }{ 9 } y-\frac { 5 }{ 9 } \times 212+100\)
\(\Rightarrow \quad x=\frac { 5 }{ 9 } y-118+100\quad \Rightarrow x=\frac { 5 }{ 9 } y-18\)
\(\Rightarrow \quad x=\frac { 5 }{ 9 } (y-32) \Rightarrow C=\frac { 5 }{ 9 } (F-32)\quad .....(1)\)
[\(\because \)  x represents Celsius and y represents Fahrenheit]
Which is the required relationship between C and F.
(ii) Find the value of C for 98.6°F and
Find C when F = 98.6° F
Substituting F = 98.6° in (1) we get,
C = \(\frac{5}{9}(98.6-32)=\frac{5}{9}(66.6)=\frac{333}{9}=37°\)
(iii) Find the value of F for 38°C.
Substituting C = 38° in (1) we get,
38=\(\frac{5}{9}(F-32)\)
⇒ \(\frac{342}{5}+32\) = F
⇒ F = 100.4°C

Clearly, number of bacteria at the end of different hours forms a G.P. with
a = 30 and r = 2.
Number of bacteria present at the end of 2^nd hour
t₃ = a . r² = 30 \(\times\) 2² = 30 \(\times\) 4 = 120.
Number of bacteria present at the end of 4^th hour
t₅ = a . r⁴ = 30 (2⁴) = 30 (16) = 480.
Number of bacteria present at the end of n^th hour
t_n+1 = a . rⁿ = 30 (2ⁿ).

Let p(n) be the statement
\(1^3+2^3+3^3+...+n^3=\left[{n(n+1)\over2}\right]\)
Step1: Put n = 1
\(⇒\ 1^3=\left[1(1+1)\over 2\right]^2=\left[1(2)\over2\right]^2\)
⇒ 1 = 1
∴ p(n) is true
Step 2: Let us assume that p(K) is true
∵ \(1^3 + 2^3 + 3^3 + ... +K^3​​​​​​​ =​​​​​​​\left[K(K+1)\over 2\right]^2\)
Step 3: To prove that p(K+1) is true
i.e to P.T \(1^3+2^3+...+K^3+(K+1)^3=\left[(K+1)(K+2)\over 2\right]^2\)
LHS = 1³ + 2³ + .... + K³ + (K + 1)³
\(=\left[K{(K+1)\over 2}\right]^2+(K+1)^3\)
\(=(K+1)^2\left[{K^2\over 4}+K+1\right]=(K+1)^2\left[K^2+4K+4\over 4\right]\)
\(={(K+1)^2(K+2)^2\over4}=\left[(k+1)(k+2)\over 2\right]=RHS\)
∵ p(K+1) is true
∵ By the principle of mathematical induction, p(n) is true for all values of n.

If x - 2y = 0

x	0	2	-2
y	0	1	-1

2x - y = -2

x	0	-1
y	2	0

x > 0, y > 0 represents the portion in the I quadrant only.
Hence, OABC is the required shaded region. 

Dividing by \(\sqrt { 2 } \) we get,
\(\frac { 1 }{ \sqrt { 2 } } \)sin \(\theta\)  + \(\frac { 1 }{ \sqrt { 2 } } \)cos\(\theta\) = 1
sin\(\frac { \pi }{ 4 } \) sin \(\theta\) + cos\(\frac { \pi }{ 4 } \) cos\(\theta\) = 1
\(\Rightarrow cos\left( \theta -\frac { \pi }{ 4 } \right) =1=cos\theta \)
\(\Rightarrow cos\left( \theta -\frac { \pi }{ 4 } \right) =cos\theta \)
\(\Rightarrow \theta -\frac { \pi }{ 4 } =2n\pi \pm 0,n\in z\)
\(\Rightarrow \theta =2n\pi +\frac { \pi }{ 4 } ,,n\in z\)
\(\Rightarrow \theta =\frac { 8n\pi +\pi }{ 4 } ,n\in z\)
\(\Rightarrow \theta =(8n+1)\frac { \pi }{ 4 } ,n\in z\)

\(2cos\left( \frac { \theta +3\theta }{ 2 } \right) .cos\left( \frac { 3\theta -\theta }{ 2 } \right) -2cos2\theta =0\)
2cos2θ.cosθ-2cos2θ = 0
cos2θ(cosθ-1) = 0
cos2θ = 0
or cosθ = 1
cos2θ = 0 or cosθ = 1
cos2θ = 0
2θ = (2n + 1)\(\frac{\pi}{2}\), n∈z
θ = (2n + 1)\(\frac{\pi}{4}\), n∈z
θ = (2n + 1)\(\frac{\pi}{4}\)
cos θ = 1
cos θ = cos0
θ = 2nπ土0, n∈z
θ = 2nπ, n∈z
ஃ The solutions are θ = (2n+1)\(\frac{\pi}{4}\)or 2nπ, n∈z

Given inequality is \({{2x-3}\over{(x-2)(x-4)}}<0.\)
The critical numbers are x = \(\frac { 3 }{ 2 }\), 2, 4
∴ The possible intervals are \(\left(-\infty, {3\over 2}\right)\left({3\over 2},2\right)(2, 4)\) and \( (4, ∞)\)

Intervals	Sign of 2x -3	Sign of x-1	Sign of x-2	Sign of \(2x-3\over (x-2)(x-4)\)
\(\left(-\infty,{3\over 2}\right)\) Say x = 0	-	-	-	-
\(\left(-\infty,{3\over 2}\right)\) Say \(x={7\over 4}\)	+	+	-	-
(2,4) Say x = 3	+	+	+	+
\( (4, ∞)\) Say x = 5	+	+	+	+

The inequality \({2x-3\over (x-2(x-4))}<0\)  is satisfied by the intervals \(\left(-\infty,{3\over 2} \right)\) and \((2,4)\)

LHS = cos 5\(\theta\) = cos (3\(\theta\) + 2\(\theta\))
= cos 3\(\theta\) cos 2\(\theta\) - sin 3\(\theta\) sin 2\(\theta\)
= (4 cos³ \(\theta\) - 3 cos \(\theta\)) (2 cos² \(\theta\) -1) - [3 sin\(\theta\) - 4 sin³ \(\theta\)] 2 sin \(\theta\) cos \(\theta\)
= 8 cos⁵ \(\theta\) - 10 cos³ \(\theta\) + 3 cos \(\theta\) -6 cos \(\theta\) + 6 cos³ \(\theta\) + 8 cos \(\theta\) (1 + cos⁴ \(\theta\) - 2 cos² \(\theta\))
= 16 cos⁵ \(\theta\) - 20 cos³ \(\theta\) + 5 cos \(\theta\)
= RHS
Hence proved.

f(-4) = +4 + 4    [\(\therefore\) f(x) = -x + 4 when x = -4]
=8
f(1) = 1-1² [\(\therefore\) f(x) = x-x² when x = 1]
f(1) = 0
f(-2) = (-2)²-(-2) [\(\therefore\) f(x) = x²-x when x = -2]
= 4+2 = 6
f(7) = 0   [\(\therefore\) f(x) = 0 when x = 7]
f(0 = 0²-0  [\(\therefore\) f(x) = x² - x when x = 0]
=0
\(\therefore\) f(-4) = 8, f(1) = 0, f(-2) = 6, f(7) = 0 and f(0) = 1

Given a³= cosec θ - sin θ =\(\frac{1}{sinθ}-sinθ\)=\(\frac{1-sin^2θ}{sinθ}=\frac{cos^2θ}{sinθ}\)
a = \((\frac{cos^2θ}{sinθ})^{1/3}\)
and b³ = sec θ - cos θ =\(\frac{1}{cosθ}-cosθ=\frac{1-cos^2θ}{cosθ}=\frac{sin^2θ}{cosθ}\)
b = \((\frac{sin^2θ}{cosθ})^{1/3}\)
Now, LHS = a²b²(a²+ b²)
= \((\frac{cos^2θ}{sinθ})^{2/3}.(\frac{sin^2θ}{cosθ})^{2/3}[(\frac{cos^2θ}{sinθ})^{2/3}+(\frac{sin^2θ}{cosθ})^{2/3}]\)
= \(\frac { { cos }_{ \theta }^{ 4/3 } }{ { sin }_{ \theta }^{ 2/3 } } .\frac { { sin }_{ \theta }^{ 4/3 } }{ { cos }_{ \theta }^{ 2/3 } } \left[ \frac { { cos }_{ \theta }^{ 4/3 } }{ { sin }_{ \theta }^{ 2/3 } } +\frac { { sin }_{ \theta }^{ 4/3 } }{ { cos }_{ \theta }^{ 2/3 } } \right] \)
= \({ cos }_{ \theta }^{ 4/3+2/3 }+{ sin }_{ \theta }^{ 4/3+2/3 }\)
cos²\(\theta\) + sin²\(\theta\) = 1 = RHS

y = sin |x|

We know \(|x| =\begin{cases}x\ if\ x\ge0 \\-x\ if\ x<0 \end{cases}\)
∴ sin |x| = sin x if x > 0 and sin |x| = sin (-x) = -sin x if x < 0.
The graph of y = sin (-x) = - sin x is the reflection of the graph of sin x about Y-axis.

(i) y = sin (-x)

Let y = sin x.
Then sin(-x) is the reflection of the graph of sin x, about y-axis.
(ii) y = -sin(-x)

-sin(-x) is the reflection of the graph of sin(-x) about the x-axis.
(iii) \(y=sin\left( {\pi\over 2}+x\right)\)
Let y = sinx.
Then \(sin\left( {\pi\over 2}+n\right)\)causes the shift to the left for \(\pi\over 2\) unit to the sin x curve.
(iv) \(y=sin\left({\pi\over 2}-x \right)\)

Let y = sin x. Then \(sin\left( {\pi\over 2}-n\right)\) causes the shift to the left for \(\pi\over 2\) unit to the sin (-x) curve.

Given functions are f(x) = x³ and g(x) = \(x^{ ^{ \frac { 1 }{ 3 } } }\)
Now, f o g(x) = f(g(x)
= \(f\left( { x }^{ \frac { 1 }{ 3 } } \right) \)
= \(\left( { x }^{ 3 } \right) ^{ \frac { 1 }{ 3 } }=x\)

Since f o g(x) = x is symmetric about the line y = x, g(x) is the inverse of f(x)
∴ g(x) = f^-1(x).
where f o g(x) = g o f(x) = x so that
(i) f o g is bijective.
(ii) both f(x) and g(x) also bijective.
(iii) f and g are symmetrical about y = x

Given f(x) = 3x - 4
Let y = 3x - 4 ⇒ y + 4 = 3x
\(⇒ x={y+4\over 3}\)
Let g(y) = \(y+4\over 3\)
Now gof(n) = g(f(n)) = g(3\(\times\) -4) = \({3x-4+4\over 3}={3x\over 3}=x\)
and fog(y) = f(g(y)) = \(f\left(y+4\over 4\right)=3\left(y+4\over 3\right)-4=y+4-4=y\)
Thus, gof(x) = I_x and fog (y) = I_y
This implies that f and g are bijections and inverses to each other
Hence f is bijection and \(f^{-1} (x)={y+4\over 3}\)
Replacing y by x, we get f^-1 (x) = \(\frac { x+4 }{ 3 } \)

Hence, the graph of y = f^-1(x) is the reflection of the graph of f in y = x

3x² + 2xy - y² = 0
Consider 3x² + 2xy - y² = 0
\(\Rightarrow\) (x +y) (3x - y) = 0   [By factorizing]

Hence the separate equations are
x + y = 0 and 3x - y = 0

Clearly
\(-1\le\cos x\le 1 \\ \Rightarrow -3\ge -3\cos x \ge-3\)
\(\Rightarrow -3\le-3\cos x\le3 \\ \Rightarrow 1-3\le 1-3\ \cos x\le 1+3\)
Thus we get \(-2\le 1-3\ \cos\ x\) and \(1-3\cos\ x\le 4.\)
By taking reciprocals, we get \({1 \over 1-3\cos x}\le-{1\over2}\) and \({1\over 1-3\cos\ x}\ge{1\over 4}.\)
Hence the range of f is \(\left( -\infty, {1\over 2} \right]\cup[{1\over4},\infty).\)