Take u = g(x) = x² + 1 and f(u) =\(\sqrt{u}\)
\(\therefore F(x)=(fog)(x)=f(g(x))\)
Since \(f'(u)={1\over2}u^{1\over2}={1\over 2\sqrt{u}}\) and
\(g'(x)=2x,\) we get
\(F'(x)=f'(g(x))g'(x)\)
\(={1\over 2\sqrt{x^2+1}}.2x={x\over \sqrt{x^2+1}}\).

Since -x \(\le sin x\le x\) for all x\(\ge 0\)
\(lim_{x\rightarrow 0}(-x)=0 \) and
\(lim_{x\rightarrow 0}(x)=0 \)
By Sandwich theorem
\(lim_{x\rightarrow 0}sinx=0 \)

\(lim_{x\rightarrow{1}}sin \pi x\)
At x = 1, the curve meets the x-axis.
\(\therefore lim_{x\rightarrow{1}}sin \pi x=0\)

tan 120° = tan (180° - 60°)
= - tan(60°) =\(-\sqrt 3\)
(or) write tan 120° as tan (90° + 30°) and find the value.

We start with the assumption f(x) = f(y). Then
\({x \over x^2-1}={y \over y^2-1}\)
\(\Rightarrow x(y^2-1)=(x^2-1)\)
\(\Rightarrow xy^2-x-yx^2+y-0\Rightarrow (y-x)(xy+1)=0\)
This implies that x =y or xy = - 1. So if we select two numbers x and y so that xy = -1, then f(x) = f(y).  \(f(x)=f(y) \cdot\left(2,-\frac{1}{2}\right),\left(7,-\frac{1}{7}\right),\left(-2, \frac{1}{2}\right)\)are some among the infinitely many possible pairs. Thus \(f(2)=f\left(\frac{-1}{2}\right)=\frac{2}{3}\). That is, f(x) = f(y) does not imply x = y. Hence it is not one-to-one.

(A\(\times\) B) = {(1,4) (1,5) (1,6) (1,7) (2,4) (2,5) (2,6) (2,7) (3,4) (3,5) (3,6) (3,7)}
(B\(\times\)A) = {(4,1) (4,2) (4,3) (5,1) (5,2) (5,3) (6,1) (6,2) (6,3) (7,1) (7,2) (7,3)}
LHS = (A\(\times\)B)\(\cap \)(B\(\times\)A) = { }....(1)
(A\(\cap \)B) = { }, (B\(\cap \)A) = { }
\(\therefore\) RHS = (A\(\cap \)B) \(\times\) (B\(\cap \)A) = { }.....(2)
From (1) and (2), LHS = RHS

[a < b ⇒ ay > by for all y< 0]
Here y = -1
Given |3-x| < 7
This means -7 < 3 -x < 7
⇒-7-3 < -x < 7-3
⇒-10 < -x < 4
⇒10 > x > -4
\(\therefore\) The Solution set is \(x\in \left( -\infty ,-4 \right) \cup \left( -4,10 \right) \)

\(\cos { \left( \pi +\theta \right) } =-\cos { \theta } \)
LHS = \(\cos { \left( \pi +\theta \right) } =-\cos { \theta } \)
= RHS
Hence proved.

{x\(\in \)N : x²<121 and x is a prime}
Let A = {x\(\in \)N : x²<121, and x is a prime}
A = {2,3,5,7}.

= \(({ 2 }^{ 4 })^{ -\frac { 3 }{ 4 } ={ 2 }^{ 4\times \frac { 2 }{ 4 } }={ 2 }^{ -3 } }=\frac { 1 }{ { 2 }^{ 3 } } =\frac { 1 }{ 8 } \)

Let\(\sqrt{7-4\sqrt{3}}=a+b\sqrt{3}\) with a,b rationals.
Squaring on both sides, we get 7-4\(\sqrt{3}\) = a²+3b²+ 2ab\(\sqrt{3}\)
So, a²+ 3b² = 7 and 2ab = -4
Therefore a = \(\frac{-2}{b}\)
From a² + 3b² = 7, we get \((\frac{-2}{b})^2+3b^2=7\), which gives \(\frac{4}{b^2}+3b^2=7\) or 3b⁴- 7b² + 4 = 0
Solving for b² we get b² = \(\frac{(7\pm\sqrt{49-48})}{8}\), which gives b² = 1 or b² = \(\frac{4}{3}.\)
Thus, b = 士1 or b = 士 \(\frac{2}{\sqrt{3}}\)
Since b is rational, we have b = 土1 and hence the corresponding values for a are ∓ 2.
Since \(\sqrt{7-4\sqrt{3}}>0\) we have \(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}.\)

cos²A + cos²B + cos²C = \({1\over2}\)[2 cos²A + 2 cos²B + 2 cos²C]
= \({1\over2}\) [(1 + cos 2A) + (1 + cos 2B) + (1 + cos 2C)]
= \({3\over2}\)+\({1\over2}\) [(cos 2A + cos 2B) +cos 2C]
= \({3\over2}\)+\({1\over2}\)[2 cos(A + B) cos (A - B) + (2 cos²C-1)]
= \({3\over2}\)+\({1\over2}\)[-2 cos C cos (A -B) + 2 cos²C -1]  (A + B =\(\pi\)-C)
= \({3\over2}\)-\({1\over2}\)+\({1\over2}\) [-2 cos C (cos (A - B)-cos C)]
= 1- cos C [cos (A - B) - cos C]
= 1 - cos C [cos (A - B) + cos (A + B)]
= 1- cos C [2 cos A cos B]
= 1- 2 cos A cos B cos C

(i) Assume that no three points out of 11 points are collinear. Then we can draw unique straight through any arbitrary pair of points out of the 11 given points. This is a combination of 2 objects taken at a time from a total of 11 and can be done in 11C₂ = 55 ways.
However, it is given that 4 points are collinear. Had they not been collinear the number of unique lines that could have been drawn through them is a combination of 2 objects taken at a time from a total of 5 and can be done in 4C₂ = 6 ways. Since they are collinear we get only one line out of these 4 points instead of 6.
So, the total number of straight lines that can be drawn through 13 points on a plane with 5 of the points being collinear is 55 - 6 + 1 = 50.
(ii) To form a triangle we need 3 points. The following are the choices.
a) If we take one point from 4 collinear points and 2 from remaining 7. So this case will give \({ 4C }_{ 1 }\times { 7C }_{ 2 }\) points = \(4\times 21=84\)
b) If we two points from 4 collinear points and 1 from remaining 7. So this will give \({ 4C }_{ 2 }\times { 7C }_{ 1 }=6\times 7=42\)
c)If we take all the three points from 7 non collinear points. Which will give 7C₃ = 35
∴ Total number of triangles are 84 + 42 + 35 = 161.

Let R(h, k) be the locus of the mid-point of PQ where, P is (2, -7) and Q is a point on (2x² + 9y² = 18)
Given equation is 2x² + 9y² = 18
Dividing by 18 we get,
\(\Rightarrow \quad \frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 2 } =1\)
\(\Rightarrow \quad \frac { { x }^{ 2 } }{ { 3 }^{ 2 } } +\frac { { y }^{ 2 } }{ { \left( \sqrt { 2 } \right) }^{ 2 } } =1\)
\(\Rightarrow \quad a=3\quad and\quad b=\sqrt { 2 } \)
Any point on the ellipse is (a cos \(\theta\), b sin \(\theta\))
\(\therefore\) Q is (3 cos \(\theta\), \(\sqrt { 2 } \) sin \(\theta\)
Since R is the mid-point of PQ, we get, (h,k) = \(\left( \frac { 2+3cos\theta }{ 2 } ,\frac { -7+\sqrt { 2 } sin\theta }{ 2 } \right) \)
\(\Rightarrow \quad h=\frac { 2+3\quad cos\quad \theta }{ 2 } \)
\(\Rightarrow \quad 2h=2+3cos\theta \)
\(\Rightarrow \quad 2h-2=3cos\theta \)
\(\Rightarrow \quad \frac { 2h-2 }{ 3 } =cos\quad \theta \)
\(k=\frac { -7+\sqrt { 2 } sin\theta }{ 2 } \)
\(\Rightarrow \quad 2k=-7+\sqrt { 2 } sin\theta \)
\(\Rightarrow \quad 2k+7\quad =\quad \sqrt { 2 } sin\theta \)
\(\Rightarrow \quad \frac { 2k+7 }{ \sqrt { 2 } } =sin\quad \theta \)
Squaring and adding we get,
\({ \left( \frac { 2h-2 }{ 3 } \right) }^{ 2 }+{ \left( \frac { 2k+7 }{ \sqrt { 2 } } \right) }^{ 2 }={ cos }^{ 2 }\theta +{ sin }^{ 2 }\theta \)
\(\Rightarrow \quad \frac { { 4h }^{ 2 }+4-8h }{ 9 } +\frac { { 4k }^{ 2 }+49+28k }{ 2 } =1\quad [\because { cos }^{ 2 }\theta +{ sin }^{ 2 }\theta =1]\)
\(\Rightarrow \quad 2({ 4h }^{ 2 }+4-8h)=9({ 4k }^{ 2 }+49+28k)=18\)
\(\Rightarrow \quad { 8h }^{ 2 }+8-16h+36{ k }^{ 2 }+441+252k-18=0\)
\(\Rightarrow \quad { 8h }^{ 2 }+36{ k }^{ 2 }-16h+252k+431=0\)
\(\therefore\)   Locus of (h,k) is
8x²+36y²-16x+252y+431=0

Let ABC be the length of the tunnel.
Let a 5, b = 3 and ㄥc = 120°

Using cosine formula,
c² = a² + b² - 2ab cos C
= 25 + 9 - 2(5) (3) cos (120°)
= 34 - 30 cos (180 - 60)
= 34 -30 (-cos 60°) = 34 -\(\left(-1\over 2\right)\) = 34 + 15 = 49
c²= 49
c = 7 km.
Hence, length of the tunnel = 7 km.

Let A be the point on the eastern side and B be the point on the western side.
Given PA = 8, PB = 6
Let a = 6, b = 8 and ㄥC = 60°
Using cosine formula

c² = a² + b² - 2ab cos C
= 36 + 64 - 2(6) (8) cos 60°
= 100 - 12 \(\times\) 8 \(\times\) \(1\over 2\)
= 100 - 48 = 52
\(⇒\ C=\sqrt {52}=\sqrt{4\times13}\)
\(∴\ C=2\sqrt{13}km\)
Hence the width of the river is \(2\sqrt{13}km\)

Given x =\(\sqrt { 2 } +\sqrt { 3 } \)
⇒  x³ =  \((\sqrt { 2 } +\sqrt { 3 } )​​​​^{ 2 }=2+3+2\sqrt { 6 } =5+2\sqrt { 6 } \)
∴ \(\frac { { x }^{ 2 }+1 }{ { x }^{ 2 }-1 } =\frac { 5+2\sqrt { 6 } +1 }{ 5+2\sqrt { 6 } -2 } =\frac { 6+2\sqrt { 6 } }{ 3+2\sqrt { 6 } } \)
⇒ \(\frac { 6+2\sqrt { 6 } }{ 3+2\sqrt { 6 } } \times \frac { 3-2\sqrt { 6 } }{ 3-2\sqrt { 6 } } =\frac { (6+2\sqrt { 6 } )(3-2\sqrt { 6 } ) }{ 9-(2\sqrt { 6 } )^{ 2 } } \)
⇒ \(\frac { 18-12\sqrt { 6 } +6\sqrt { 6 } -4(\sqrt { 6 } )^{ 2 } }{ 9-24 } =\frac { 18-12\sqrt { 6 } -24 }{ -15 } \)
⇒ \(\frac { -6-6\sqrt { 3 } }{ -15 } \)
\(=\frac{2(1+\sqrt{6})}{5}=\frac{2+2 \sqrt{6}}{5}\)

Given f(x) = 1.23x where x represents the number of American dollars.
and g(y) = 50.50y where y represents the number of Singapore dollars.

To convert American dollars to Indian rupees, we have to find out go f(x)
∴ go f(x) = g(f(x))
= g(1.23x)
= 50.50[1.23x]
= 62.115x
∴ The function for exchange rate of American dollars in terms of Indian rupee is g o f (x) = 62.115x.

\(f(x)=|x-1|+|x-3|+\sin x\)
\(\text { Since } \sin x \text { is differentiable everywhere }\)

At x = 1 and x = 3. The graph admit cups.
\(\therefore\) The derivative is not exist.
\(\therefore\) The number of points in R is 2.

\(y =(a x-5) e^{3 x} \)
\(\frac{d y}{d x} =(a x-5) e^{3 x}(3)+e^{3 x}(a) \)
\(-13 =(-5)(3)+a \quad(\text { At } x=0) \)
\(-13 =-15+a \)
\(a =-13+15=2 \)
\(\therefore a =2 \)

\(\text { Let } f(x)=\frac{\sqrt{1-\cos 2 x}}{x}\)
\(=\frac{\sqrt{2 \sin ^{2} x}}{x} \)
\(=\frac{\sqrt{2} \sqrt{\sin ^{2} x}}{x} \)
\(=\frac{\sqrt{2}|\sin x|}{x}\left(\therefore \sqrt{x^{2}}=|x|\right) \)
\(= \begin{cases}\frac{\sqrt{2}(-\sin x)}{x}, & \sin x<0 \\ \sqrt{2}\left(\frac{\sin x}{x}\right), & \sin x>0\end{cases} \)
\(= \begin{cases}-\sqrt{2} \frac{\sin x}{x}, x \in\left(-\frac{\pi}{2}, 0\right) \\ \sqrt{2} \frac{\sin x}{x}, x \in\left(0, \frac{\pi}{2}\right)\end{cases} \)
\(\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{-\sqrt{2} \sin x}{x}=-\sqrt{2}\)
\(\text { and } \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{2} \sin x}{x}=\sqrt{2}\)
\(\text { From (1) and (2) }\)
\(\lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x)\)
\(\therefore \lim _{x \rightarrow 0} f(x) \text { does not exist }\)

\(\overrightarrow{A B}=-\overrightarrow{C D}\)
\(\overrightarrow{A D}=\overrightarrow{B C}=-\overrightarrow{C B}\)
\(\therefore \overrightarrow{A B}+\overrightarrow{A D}+\overrightarrow{C B}+\overrightarrow{C D}=-\overrightarrow{C D}-\overrightarrow{C B}+\overrightarrow{C B}+\overrightarrow{C D}=\overrightarrow{0}\)

\(A \& B \text { are symmetric }\)
\(A^{T}=A \text {, and } B^{T}=B\)
\(\text { 1) }(A+B)^{T}=B^{T}+A^{T}=B+A=A+B \Rightarrow A+B\text { is symmetric }\)
\(\text { 2) }(A B)^{T}=B^{T} A^{T}=B A \neq A B\)
\(\therefore A B \text { is not symmetric }\)
\(\text { 3) } A B=A^{T} B^{T}=(B A)^{T}\)
\(\text { 4) } A^{T} B=A B=A B^{T}\)

\(A^{2}=A \times A=\left[\begin{array}{cc} \lambda & 1 \\ -1 & -\lambda \end{array}\right]\left[\begin{array}{cc} \lambda & 1 \\ -1 & -\lambda \end{array}\right]\)
\(=\left[\begin{array}{cc} \lambda^{2}-1 & \lambda-\lambda \\ -\lambda+\lambda & -1+\lambda^{2} \end{array}\right]=0\)
\(\therefore \lambda^{2} =1 \)
\(\lambda =\pm 1\)

x - y + 5 = 0;
Put x = 0, then y = 5
The point is (0, 5)
Any line perpendicular to x - y + 5 = 0 is
x + y + k = 0,
This passes through (0, 5)
k = -5
Required equation is x + y - 5 = 0

\(1+3+5+7+\ldots \ldots+17 \text { is equal to } 9^{2}=81\)

Let the point be (x, y)
Its distance from origin is \(\sqrt{x^{2}+y^{2}}\)
Given \(x =\frac{1}{2} \sqrt{x^{2}+y^{2}} \)
\(\Rightarrow 2 x =\sqrt{x^{2}+y^{2}} \)
\(4 x^{2} =x^{2}+y^{2} \)
\(3 x^{2}-y^{2}=0 \)  is the required equation of the locus

Number of shake hands = 66
Let there be n persons.
No. of . Shake hands
= (n - 1) + (n - 2) + .......+.2 + 1
\(66 =\frac{(n-1)}{2} n \)
\(n^{2}-n =132 \)
\(n^{2}-n-132 =0 \)
\((n-12)(n+11) =0 \)
\(\mathrm{n}=2 \text { (or) }-11 \text { (not valid) }\)

\(\mathrm{AM}=16, \quad \mathrm{GM}=8, \quad \mathrm{HM}=?\)
\(\frac{a+b}{2}=16 \Rightarrow a+b=32\)
\(\sqrt{a b}=8 \Rightarrow a b=64\)
\(\therefore \mathrm{HM} =\frac{2 a b}{a+b} \)
\(=\frac{2(64)}{32}= 4 \)

\(x =\frac{1}{\cos 80^{\circ}}-\frac{\sqrt{3}}{\sin 80^{\circ}} \)
\(=\frac{\sin 80^{\circ}-\sqrt{3} \cos 80^{\circ}}{\sin 80^{\circ} \cos 80^{\circ}} \)
\(\frac{x}{2} =\frac{\frac{1}{2} \sin 80^{\circ}-\frac{\sqrt{3}}{2} \cos 80^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}} \)
\(=\frac{\sin 80^{\circ} \cos 60^{\circ}-\cos 80^{\circ} \sin 60^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}} \)
\(=\frac{\sin 20^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}} \)
\(=\frac{\sin 160^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}} \)
\(=\frac{2 \sin 80^{\circ} \cos 80^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}}=4 \)

f(0) = 0, f(-3) = 9 and f(3) = 9
.'. S is [0,9]

Given points are A(1, 2, -3) and B(-1, -2, 1).
Then \(\overrightarrow { AB } =\overrightarrow { OB } -\overrightarrow { OA } =(-\hat { i } -2\hat { j } +\hat { k } )-(\hat { i } +2\hat { j } -3\hat { k } 0=-2\hat { i } -4\hat { j } +4\hat { k } \)
\(|\overrightarrow { AB } |=\sqrt { { (-2) }^{ 2 }+{ (-4) }^{ 2 }+{ 4 }^{ 2 } } =\sqrt { 4+16+16 } =\sqrt { 36 } =6\)
Now, \(l=\frac { x }{ |\overrightarrow { AB } | } =\frac { -2 }{ 6 } =\frac { -1 }{ 3 } \)
\(m=\frac { y }{ |\overrightarrow { AB } | } =\frac { -4 }{ 6 } =\frac { -2 }{ 3 } \)
\(n=\frac { z }{ |\overrightarrow { AB } | } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)
Thus, the direction cosines of \(\overrightarrow { AB } \) are \(\left( -\frac { 1 }{ 3 } ,-\frac { 2 }{ 3 } ,\frac { 2 }{ 3 } \right) \)

Take logarithm :
log y = \(\sqrt{x} \log x\)
Differentiating implicitly,
\({y'\over y}=\sqrt{x}.{1\over x}+{1\over 2\sqrt{x}}.log\ x\)
\(={log \ x+2\over 2\sqrt{x}}\)
Therefore, \({d\over dx}(x^{\sqrt{x}})=y'=x^{{\sqrt{x}}}({log \ x+2\over 2\sqrt{x}})\) .

In the letters of the word, ARTICLE, there are three vowels namely A, I, E.
There are 3 even places.
3 vowels can occupy the even places in 3P₃ = 3! ways.
Remaining 4 letters can occupy 4 places in 4! ways.
Hence, total number of ways of arrangement = 4! \(\times\) 3!
=  \(4\times 3\times 2\times 3\times 2\)
=144