\(f(x)=\left\{\begin{array}{ll} x+1, & \text { if } x \geq 1 \\ x^{2}+1, & \text { if } x<1 \end{array}\right.\)
The given function f is defined at all the points of the real line.
Let c be a point on the real line.
Case I:\(\text { If } c<1, \text { then } f(c)=c^{2}+1 \text { and } \lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^{2}+1\right)=c^{2}+1\)
\(\therefore \lim _{x \rightarrow c} f(x)=f(c) \)
Therefore, f is continuous at all points x, such that x < 1
Case II: \(\text { If } c=1, \text { then } f(c)=f(1)=1+1=2\)
The left hand limit of f at x = 1 is,
\(\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}\left(x^{2}+1\right)=1^{2}+1=2\)
The right hand limit of f at x = 1 is,
\(\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(x+1)=1+1=2 \)
\(\therefore \lim _{x \rightarrow 1} f(x)=f(1) \)
Therefore, f is continuous at x = 1
Case III: \(\text { If } c>1, \text { then } f(c)=c+1 \)
\(\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(x+1)=c+1 \)
\(\therefore \lim _{x \rightarrow c} f(x)=f(c) \)
Therefore, f is continuous at all points x, such that x > 1
Hence, the given function f has no point of discontinuity

Let \(\frac{1}{x}=p, \frac{1}{y}=q\) and \(\frac{1}{z}=r\)
Then, the given equations becomes \(2 p+3 q+10 r=2\)
\(4 p-6 q+5 r=5 \)
6 p+9 q-20 r=-4
This system can be written as AX = B, where
\(A=\left[\begin{array}{ccc} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{array}\right], X=\left[\begin{array}{l} p \\ q \\ r \end{array}\right], B=\left[\begin{array}{c} 2 \\ 5 \\ -4 \end{array}\right]\)
Here, \(|A|=\left|\begin{array}{ccc}2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20\end{array}\right|\)
=2(120-45)-3(-80-30)+10(36+36) 
\( =150+330+720 =1200 \neq 0\)
Thus, A is non-singular, therefore its inverse exists.
Therefore, the above system has a unique solution given by X=A^-1 B
Cofactors of A are
\(C_{11}  =75, \quad C_{21}=150, \quad C_{31}=75\)
\(C_{12}  =110, C_{22}=-100, \quad C_{32}=30\)
\(C_{13}  =72, C_{23}=0 \text { and } C_{33}=-24 \)
\(\therefore \operatorname{adj}(A)  =\left[\begin{array}{ccc} 75 & 110 & 72 \\ 150 & -100 & 0 \\ 75 & 30 & -24 \end{array}\right] \)
\(=\left[\begin{array}{ccc} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{array}\right]\)
\(\therefore \quad A^{-1}  =\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{1200}\left[\begin{array}{ccc} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{array}\right]\)
Now, X=A^-1B 
\( \Rightarrow\left[\begin{array}{l} p \\ q \\ r \end{array}\right]=\frac{1}{1200}\left[\begin{array}{ccc} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{array}\right]\left[\begin{array}{c} 2 \\ 5 \\ -4 \end{array}\right] \)
\(=\frac{1}{1200}\left[\begin{array}{c} 150+750-300 \\ 220-500-120 \\ 144+0+96 \end{array}\right] \)
\(=\frac{1}{1200}\left[\begin{array}{c} 900-300 \\ 220-620 \\ 144+96 \end{array}\right]=\frac{1}{1200}\left[\begin{array}{c} 600 \\ -400 \\ 240 \end{array}\right]=\left[\begin{array}{c} 1 / 2 \\ -1 / 3 \\ 1 / 5 \end{array}\right]\)
\(\Rightarrow \quad p=\frac{1}{2}, q=-\frac{1}{3}, r=\frac{1}{5}\)
\(\therefore \quad x  =2, y=-3 \text { and } z=5\)

Let ABCD be a rectangle inscribed in a given circle with centre at 0 and radius a.
Let AB = 2x and BC = 2y

Then, OA² = OM² + AM² 
\(\Rightarrow a^2=y^2+x^2\)
\(\Rightarrow y=\sqrt{a^2-x^2}\)
Let A be the area of the rectangle.
\(\therefore A=4xy=4x\sqrt{x^2-x^2}\)
\(\Rightarrow \ \ \frac{dA}{dx}=4\{\frac{a^2-2x^2}{\sqrt{a^2-x^2}}\}\)
For maximum or minimum value of A,
\(\frac{dA}{dx}=0\)
\(=4\{\frac{a^2-2x^2}{\sqrt{a^2-x^2}}\}=0\Rightarrow\ \ x=\frac{a}{\sqrt{2}}\)
Now, \(\frac{d^2A}{dx^2}=4\frac{d}{dx}\{(a^2-2x^2)(a^2-x^2)^{-1/2}\}\)
\(\Rightarrow \frac{d^2A}{dx^2}=4[-4x(a^2-x^2)^{-1/2}+(a^2-2x^2)\times(-1/2)(a^2-x^2)^(-3/2)(-2x)]\)
\(=[\frac{-4x}{\sqrt{a^2-x^2}}+\frac{x(a^2-2x^2)}{(a^2-x^2)^{3/2}}]\)
\(\therefore\ (\frac{d^2A}{dx^2})_{x=\frac{a}{\sqrt2}}=-16<0\)
Thus A is maximum when \(x=\frac{a}{\sqrt2}\)
putting \(x=\frac{a}{\sqrt2}\) in (i) \(y=\frac{a}{\sqrt2}\)
Therefore \(x=y=\frac{a}{\sqrt2}\)
Hence area is maximum when x = y ⇒ 2x = 2y
i.e., the rectangle is a square.

\( \text { We have, } \tan \frac{1}{2}\left(\sin ^{-1} \frac{2 x}{1+x^{2}}+\cos ^{-1} \frac{1-y^{2}}{1+y^{2}}\right) \)
\(=\tan \frac{1}{2}\left(2 \tan ^{-1} x+2 \tan ^{-1} y\right) \) 
 \(\left[\because 2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right] \)
\(= \tan \frac{1}{2} \times 2\left(\tan ^{-1} x+\tan ^{-1} y\right) \)
\(= \tan \left[\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right] \)
\(= \frac{x+y}{1-x y} \because \tan ^{-1} x+\tan ^{-1} y=\mid \tan ^{-1}\left(\frac{x+y}{1-x y}\right) \)
\(\text { and } \tan \left(\tan ^{-1} \theta\right)=\theta \)

R is reflexive, since (1, 1), (2, 2) and (3, 3) lie in R. Also, R is not symmetric, as (1, 2) \(\in\) R but (2, 1) \(\notin\) R. Similarly, R is not transitive, as (1, 2) \(\in\) R and (2, 3) \(\in\) R but (1, 3) \(\notin\) R.

Note that
f '(x) = 3x² – 6x + 4
= 3(x² – 2x + 1) + 1
= 3(x – 1)² + 1 > 0, in every interval of R
Therefore, the function f is increasing on R.

The given system of equation is
x + 3y = 5 and
2x + 6y = 8.
The given system of equations can be written in the form of  A X = B, where 
\(A=\left[\begin{array}{ll} 1 & 3 \\ 2 & 6 \end{array}\right], X=\left[\begin{array}{l} x \\ y \end{array}\right] \text { and } B=\left[\begin{array}{l} 5 \\ 8 \end{array}\right] \).
Now, |A| = 1(6) - 3(2) = 6 - 6 = 0
∴ A  is a singular matrix.
\(\text { Now, }(\operatorname{adj} A)=\left[\begin{array}{cc} 6 & -3 \\ -2 & 1 \end{array}\right]\)
\((a d j A) B=\left[\begin{array}{cc} 6 & -3 \\ -2 & 1 \end{array}\right]\left[\begin{array}{l} 5 \\ 8 \end{array}\right]=\left[\begin{array}{c} 30-24 \\ -10+8 \end{array}\right]=\left[\begin{array}{l} 6 \\ -2 \end{array}\right] \neq O\)
Thus, the solution of the given system of equations does not exist. Hence, the system of equation is inconsistent 

Let \(\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\theta \Rightarrow \cos \theta=\frac{\sqrt{3}}{2}\)
We know that the principal value branch of cos^-1 is \( [0, \pi] \)
\(\therefore \quad \cos \theta=\frac{\sqrt{3}}{2}=\cos \frac{\pi}{6} \Rightarrow \theta=\frac{\pi}{6} \)
\(\Rightarrow \cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6} \in[0, \pi] \)
Hence, the principal value of \(\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right) \text { is } \frac{\pi}{6}\)

For x = 0,
f(x) is continuous and
at x = 1, LHL = 1 and RHL = 5,
hence discontinuous and
at x = 2;
f(x) = 5 which is continuous

The given function is \( f(x)=(x-2)^{4}(x+1)^{3} \text { . }\)
\(\therefore f^{\prime}(x) =4(x-2)^{3}(x+1)^{3}+3(x+1)^{2}(x-2)^{4} \)
\(=(x-2)^{3}(x+1)^{2}[4(x+1)+3(x-2)] \)
\(=(x-2)^{3}(x+1)^{2}(7 x-2) \)
\(\text { Now, } f^{\prime}(x)=0 \Rightarrow x=-1 \text { and } x=\frac{2}{7} \text { or } x=2\)
Now, for values of x close to \( \frac{2}{7} \)and to the left of \(\frac{2}{7}, f^{\prime}(x)>0\). Also, for values of x close to \(\frac{2}{7}\)and to
\(\text { the right of } \frac{2}{7}, f^{\prime}(x)<0\)
\(\text {Thus, } x=\frac{2}{7} \)is the point of local maxima
 Now, for values of x close to 2 and to the left of 2,f′(x)<0. Also, for values of x close to 2 and to the " id="MathJax-Element-1863-Frame" role="presentation" style="position: relative;" tabindex="0">Now, for values of x close to 2 and to the left of 2,f′(x)<0. Also, for values of x close to 2 and to the  right of 2,f′(x)>0" id="MathJax-Element-1864-Frame" role="presentation" style="position: relative;" tabindex="0">right of 2,f′(x)>0
 Thus, x=2 is the point of local minima. " id="MathJax-Element-1865-Frame" role="presentation" style="position: relative;" tabindex="0"> Now, for values of x close to 2 and to the left of 2,f′(x)<0. Also, for values of x close to 2 and to the " role="presentation" style="position: relative;" tabindex="0">
 right of 2,f′(x)>0" id="MathJax-Element-1864-Frame" role="presentation" style="position: relative;" tabindex="0"> 
 Now, as the value of x varies through −1,f′(x) does not changes its sign. " id="MathJax-Element-1866-Frame" role="presentation" style="position: relative;" tabindex="0"> Now, as the value of x varies through −1,f′(x) does not changes its sign. 
 Thus, x=−1 is the point of inflexion. " id="MathJax-Element-1867-Frame" role="presentation" style="position: relative;" tabindex="0"> Thus, x=−1 is the point of inflexion.

(i) We have
\(\mathrm{A}=\left[\begin{array}{lll} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{array}\right] \Rightarrow \mathrm{A}^{\prime}=\left[\begin{array}{cc} 3 & 4 \\ \sqrt{3} & 2 \\ 2 & 0 \end{array}\right] \Rightarrow\left(\mathrm{A}^{\prime}\right)^{\prime}=\left[\begin{array}{lll} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{array}\right]=\mathrm{A}\)
Thus \(\left(\mathrm{A}^{\prime}\right)^{\prime}=\mathrm{A}\)
(ii) We have
\(\mathrm{A}=\left[\begin{array}{lll} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{array}\right], \mathrm{B}=\left[\begin{array}{rrr} 2 & -1 & 2 \\ 1 & 2 & 4 \end{array}\right] \Rightarrow \mathrm{A}+\mathrm{B}=\left[\begin{array}{ccc} 5 & \sqrt{3}-1 & 4 \\ 5 & 4 & 4 \end{array}\right]\)
Therefore \((A+B)^{\prime}=\left[\begin{array}{cc} 5 & 5 \\ \sqrt{3}-1 & 4 \\ 4 & 4 \end{array}\right]\)
Now \(\mathrm{A}^{\prime}=\left[\begin{array}{cc} 3 & 4 \\ \sqrt{3} & 2 \\ 2 & 0 \end{array}\right], \mathrm{B}^{\prime}=\left[\begin{array}{rr} 2 & 1 \\ -1 & 2 \\ 2 & 4 \end{array}\right] \text {, }\)
So \(A^{\prime}+B^{\prime}=\left[\begin{array}{rr} 5 & 5 \\ \sqrt{3}-1 & 4 \\ 4 & 4 \end{array}\right]\)

Thus \((\mathrm{A}+\mathrm{B})^{\prime}=\mathrm{A}^{\prime}+\mathrm{B}^{\prime} \)

(iii) We have
\(k B=k\left[\begin{array}{rrr} 2 & -1 & 2 \\ 1 & 2 & 4 \end{array}\right]=\left[\begin{array}{ccc} 2 k & -k & 2 k \\ k & 2 k & 4 k \end{array}\right]\) 
and 
\((k B)^{\prime} =\left[\begin{array}{ccc} 2 k & -k & 2 k \\ k & 2 k & 4 k \end{array}\right]^{\prime}=\left[\begin{array}{cc} 2 k & k \\ -k & 2 k \\ 2 k & 4 k \end{array}\right].\)
\(=k\left[\begin{array}{rr} 2 & 1 \\ -1 & 2 \\ 2 & 4 \end{array}\right]=k B^{\prime} \) 
Thus, (kB)' = kB'

\(We\quad have\quad :\quad xy={ e }^{ (x-y) }\)
\(Taking\quad logs.,log(xy)=log{ e }^{ (x-y) }\)
\(logx+logy=(x-y)log\quad e\)
\(logx+logy=(x-y)\)
\(Diff.w.r.t.x,\quad \frac { 1 }{ x } +\frac { 1 }{ y } \frac { dy }{ dx } =1-\frac { dy }{ dx } \)
\(\Rightarrow \left( \frac { 1 }{ y } +1 \right) \frac { dy }{ dx } =1-\frac { 1 }{ x } \)
\(\Rightarrow \frac { 1+y }{ y } \frac { dy }{ dx } =\frac { x-1 }{ x } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { y(x-1) }{ x(y+1) } \)

The given system of equation is:
2x - y = -2
3x + 4y = 3
these can be written as A X = B
= X = A^-1B
where  \(A=\begin{bmatrix} 2&-1\\3&4\end{bmatrix}, X=\begin{bmatrix} x\\y\end{bmatrix}\ and\ B=\begin{bmatrix}-2\\3 \end{bmatrix}\)
\(|A|=\begin{bmatrix} 2&-1\\3&4\end{bmatrix}=8+3=11\neq0\Rightarrow A^{-1}\)exists.
Now \(adj\ A=\begin{bmatrix} 4&-3\\1&2\end{bmatrix}=\begin{bmatrix}4&1\\-3&2 \end{bmatrix}\)
\(A^{-1}={1\over |A|}(adj\ A)={1\over11}\begin{bmatrix}4&1\\3&2 \end{bmatrix}\)
\(X={1\over11}\begin{bmatrix}4&1\\-3&2 \end{bmatrix}\begin{bmatrix} -2\\3\end{bmatrix}\)
\(={1\over11}\begin{bmatrix} -8+.3\\6+6\end{bmatrix}={1\over11}\begin{bmatrix} -5\\12\end{bmatrix}\)
\(\Rightarrow \begin{bmatrix} x\\y\end{bmatrix}=\begin{bmatrix} -5/11\\12/11\end{bmatrix}\)
\(x={-5\over11},y={12\over11}\)|

Let's consider \(\tan ^{-1}(1)=x\). Then, \(\tan x=1=\tan \left(\frac{\pi}{4}\right)\). \(\therefore \tan ^{-1}(1)=\frac{\pi}{4}\)
Let's assume,\(\cos ^{-1}\left(-\frac{1}{2}\right)=y\).
Then, \(\cos y=-\frac{1}{2}=-\cos \left(\frac{\pi}{3}\right)=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \left(\frac{2 \pi}{3}\right)\)
\(\therefore \cos ^{-1}\left(-\frac{1}{2}\right)=\frac{2 \pi}{3}\)
Let's again assume that \(\sin ^{-1}\left(-\frac{1}{2}\right)=z\).
Then, \(\sin z=-\frac{1}{2}=-\sin \left(\frac{\pi}{6}\right)=\sin \left(-\frac{\pi}{6}\right)\).
\(\therefore \sin ^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\)
\(\therefore \tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)\)
\(=\frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6} \)
\(=\frac{3 \pi+8 \pi-2 \pi}{12}=\frac{9 \pi}{12}=\frac{3 \pi}{4}\)

No two different students of the class can have same roll number. Therefore, f must be one-one. We can assume without any loss of generality that roll numbers of students are from 1 to 50. This implies that 51 in N is not roll number of any student of the class, so that 51 can not be image of any element of X under f. Hence, f is not onto.

We have,\(f(x)=x^{3}-3 x, x \in[0, \sqrt{3}]\) 
For (x), Rolle's theorem is satisfied
\(f^{\prime}(c)=0\)  \(\left[\because f^{\prime}(x)=3 x^{2}-3\right]\) 
\( \Rightarrow 3 c^{2}-3=0\)
\(\Rightarrow c^{2}=\frac{3}{3}=1\)
\(\Rightarrow c=\pm 1, \text { where } 1 \in(0, \sqrt{3})\)
\(\therefore c=1 \)

We have
\(f(x)=x^{2}+2 x-8, x \in[-4,2]\) 
Rolle'stheorem is satisfied
\(\therefore f^{\prime}(c)=0 \)
\(\Rightarrow f^{\prime}(c)=2 c+2=0 \)
\(c+1=0 \Rightarrow c=-1 \)

As cos(-680°) = cos 680°
= cos(720° – 40°) = cos 40°
∴ cos^-1[cos(-680°)J = cos^-1 (cos 40°)
= 40° = \(\frac{2\pi}{9}\).

(i) (c) : Let C(x) be the maintenance cost function, then C(x) = 5000000 + 160x - 0.04x²
(ii) (b) : We have, C(x) = 5000000 + 160x - 0.04x²
Now, C(x) = 160 - 0.08x
For maxima/minima, put C'(x) = 0
\(\Rightarrow\) 160 = 0.08x
\(\Rightarrow\) x = 2000
(iii) (b) : Clearly, from the given condition we can see that we only want critical points that are in the interval [0,4500].
Now, we have C(0) = 5000000
C(2000) = 5160000
and C(4500) = 4910000
\(\therefore\) Maximum value of C(x)would be Rs.5160000
(iv) (a) : The complex must have 4500 apartments to minimise the maintenance cost.
(v) (a) : The minimum maintenance cost for each apartment woud be Rs.1091.11

we have,\(\begin{equation} f(x)=\left\{\begin{array}{ll} x-3 & , x \geq 3 \\ 3-x & , 1 \leq x<3 \\ \frac{x^{2}}{4}-\frac{3 x}{2}+\frac{13}{4} & , x<1 \end{array}\right. \end{equation}\) 
(i) (b) : \(\begin{equation} \mathrm{R} f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \end{equation}\) 
\(\begin{equation} =\lim _{h \rightarrow 0} \frac{3-(1+h)-2}{h}=\lim _{h \rightarrow 0}-\frac{h}{h}=-1 \end{equation}\) 
(ii) (b) : \(\begin{equation} \mathrm{L}_{\mathrm{s}}^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h} \end{equation}\) 
\(\begin{equation} =\lim _{h \rightarrow 0} \frac{-1}{h}\left[\frac{(1-h)^{2}}{4}-\frac{3(1-h)}{2}+\frac{13}{4}-2\right] \end{equation}\) 
\(\begin{equation} =\lim _{h \rightarrow 0}\left(\frac{1+h^{2}-2 h-6+6 h+13-8}{-4 h}\right) \end{equation}\) 
\(\begin{equation} =\lim _{h \rightarrow 0}\left(\frac{h^{2}+4 h}{-4 h}\right)=-1 \end{equation}\) 
(iii) (c) : Since, R.H.D. at x = 3 is 1 and L.H.D. at x = 3 is-1
\(\therefore\)  f(x) is non-differentiable at x = 3.
(iv) (d)
(v) (c) : From above, we have
\(\begin{equation} f^{\prime}(x)=\frac{x}{2}-\frac{3}{2}, x<1 \end{equation}\) 
\(\begin{equation} \therefore f^{\prime}(-1)=\frac{-1}{2}-\frac{3}{2}=-2 \end{equation}\)
 

(i) (a) : Let F be the matrix representing the number of family members and R be the matrix representing the requirement of calories and proteins for each person. Then
 
 
(ii) (b) : The requirement of calories and proteins for each of the two- families is given by the product matrix FR.
\(F R=\left[\begin{array}{lll} 4 & 4 & 4 \\ 2 & 2 & 2 \end{array}\right]\left[\begin{array}{ll} 2400 & 45 \\ 1900 & 55 \\ 1800 & 33 \end{array}\right]\) 
\(=\left[\begin{array}{ll} 4(2400+1900+1800) & 4(45+55+33) \\ 2(2400+1900+1800) & 2(45+55+33) \end{array}\right]\) 
 
(iii) (c)
(iv) (c) : Since, AB = B ...(i) and BA = A ...(ii)
\(\therefore\) A² + B² = A·A + B·B 
= A(BA) + B(AB) [using (i) and (ii)]
= (AB)A + (BA)B [Associative law]
= BA +AB [using (i) and (ii]
= A+B