2af(a)−a²f′(a)

Given system of equations can be written as
3x - y - 2z = 2,
2y - z = -1 and
3x - 5y = 3. and its matrix form is AX = B , where
\(A=\left[\begin{array}{lll} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array}\right], X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } B=\left[\begin{array}{c} 2 \\ -1 \\ 3 \end{array}\right]\)
\(\text {Now, }|A|=3(0-5)-0+3(1+4)=-15+15=0\)
\(\therefore A \) is a singular matrix.
\((\text { adjA })=\left[\begin{array}{lll} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{array}\right] \)
\(\therefore(a d j A) B=\left[\begin{array}{lll} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{array}\right]\left[\begin{array}{c} 2 \\ -1 \\ 3 \end{array}\right]=\left[\begin{array}{l} -10-10+15 \\ -6-6+9 \\ -12-12+18 \end{array}\right]=\left[\begin{array}{l} -5 \\ -3 \\ -6 \end{array}\right] \neq O\)
Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent

\(\left.\begin{array}{l} \text { Deluxe } & \text { Premium } & \text { Standard } \\ A & 6 & 4 & 5 \\ B & 8 & 3 & 4 \\ \text { C } & 4 & 2 & 3 \end{array}\right] \text { . } \)

Since 1*2 =1.2 ³   = 8 but 2*1 = 2.1³ = 2 
So 1*2 is not equal to 2* 1, so the given Binary Operation is not Commutative

Let y = tan⁻¹(−1)
⇒ tan y = −1
We know that the range of the principal value branch of tan⁻¹x is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \text {. }\)
\( \Rightarrow \tan y=\tan \left(\frac{-\pi}{4}\right) \)
\(\therefore y=\frac{-\pi}{4} \)
Hence, the principal value of tan⁻¹(−1) is \(\frac{-\pi}{4} \).

\(\text { [Ans. } f(x) \text { is increasing in }(2,3], f(x) \text { is decreasing is } [3, \infty)]\)
 

Given  
 \(\tan ^{-1}\left(\frac{n}{\pi}\right)>\frac{\pi}{4} \)
\(\Rightarrow \frac{n}{\pi}>\tan \frac{\pi}{4} \quad\left[\because \tan ^{-1} \theta>\phi \Rightarrow \theta>\tan \phi\right] \)
\(\Rightarrow \frac{n}{\pi}>1 \Rightarrow n>\pi \)
\(\Rightarrow n>3.14\) 
Hence, the minimum value of n is 4 .

We shall prove the result by using principle of mathematical induction
\(\mathrm{P}(n): \text { If } \mathrm{A}=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \text {, then } \mathrm{A}^n=\left[\begin{array}{cc} \cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta \end{array}\right], n \in \mathbf{N}\)
\(P(1): A=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \text {, so } A^1=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\)
Therefore, the result is true for n = 1.
Let the result be true for n = k. So
\(\mathrm{P}(k): \mathrm{A}=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \text {, then } \mathrm{A}^k=\left[\begin{array}{cc} \cos k \theta & \sin k \theta \\ -\sin k \theta & \cos k \theta \end{array}\right]\)
Now, we prove that the result holds for n = k +1
\( \mathrm{A}^{k+1} =\mathrm{A} \cdot \mathrm{A}^k=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc} \cos k \theta & \sin k \theta \\ -\sin k \theta & \cos k \theta \end{array}\right] \)
\(=\left[\begin{array}{cc} \cos \theta \cos k \theta-\sin \theta \sin k \theta & \cos \theta \sin k \theta+\sin \theta \cos k \theta \\ -\sin \theta \cos k \theta+\cos \theta \sin k \theta & -\sin \theta \sin k \theta+\cos \theta \cos k \theta \end{array}\right]\)
\(=\left[\begin{array}{cc} \cos (\theta+k \theta) & \sin (\theta+k \theta) \\ -\sin (\theta+k \theta) & \cos (\theta+k \theta) \end{array}\right]=\left[\begin{array}{cc} \cos (k+1) \theta & \sin (k+1) \theta \\ -\sin (k+1) \theta & \cos (k+1) \theta \end{array}\right]\)
Therefore, the result is true for n = k + 1. Thus by principle of mathematical induction, we have \(A^n=\left[\begin{array}{rr} \cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta \end{array}\right]\), holds for all natural numbers.

Here \(\Delta\)is a skew-symmetric determinant of 3rd (odd) order
\(\Delta\)=0

\(We\quad have:\quad y={ x }^{ 3 }+tan\quad x\)
\(\frac { dy }{ dx } ={ 3x }^{ 2 }+{ sec }^{ 2 }x\)
\(and\quad \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =6x+2secx.\frac { d }{ dx } (sec\quad x)\)
\(=6x+2secx.secxtanx\)
\(=6x+2{ sec }^{ 2 }xtanx\)

We have, \(f(x)=\left\{\begin{array}{ll} \frac{1}{2}-x, & \text { if } 0 \leq x<\frac{1}{2} \\ 1, & \text { if } x=\frac{1}{2} \\ \frac{3}{2}-x, & \text { if } \frac{1}{2} 
Here, given function is a polynomial function, so it is continuous in the given intervals.
Now, we have to check the continuity at \(x=\frac{1}{2}\) .
\( \mathrm{LHL} =\lim _{x \rightarrow \frac{1}{2}} f(x)=\lim _{x \rightarrow \frac{1}{2}}\left(\frac{1}{2}-x\right) \)
\(=\lim _{h \rightarrow 0}\left[\frac{1}{2}-\left(\frac{1}{2}-h\right)\right] \) 
 \(=\lim _{h \rightarrow 0} h=0\) 
\(\left[\text { put } x=\frac{1}{2}-h ; \text { when } x \rightarrow \frac{1^{-}}{2}, \text { then } h \rightarrow 0\right]\) 
and 
\( \mathrm{RHL} =\lim _{x \rightarrow \frac{1^{+}}{2}} f(x)=\lim _{x \rightarrow \frac{1^{+}}{2}}\left(\frac{3}{2}-x\right) \)
\(=\lim _{h \rightarrow 0}\left[\frac{3}{2}-\left(\frac{1}{2}+h\right)\right] \) 
\(=\lim _{h \rightarrow 0}\left[\frac{3}{2}-\frac{1}{2}-h\right]=\lim _{h \rightarrow 0}[1-h]=1\) 
\(\because \mathrm{LHL} \neq \mathrm{RHL}\) 
Hence, f(x) is not continuous at \(x=\frac{1}{2}\) 

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 
\( \Rightarrow\left[\begin{array}{l} p \\ q \\ r \end{array}\right]=\frac{1}{1200}\left[\begin{array}{ccc} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{array}\right]\left[\begin{array}{c} 2 \\ 5 \\ -4 \end{array}\right] \)
\(=\frac{1}{1200}\left[\begin{array}{c} 150+750-300 \\ 220-500-120 \\ 144+0+96 \end{array}\right] \)
\(=\frac{1}{1200}\left[\begin{array}{c} 900-300 \\ 220-620 \\ 144+96 \end{array}\right]=\frac{1}{1200}\left[\begin{array}{c} 600 \\ -400 \\ 240 \end{array}\right]=\left[\begin{array}{c} 1 / 2 \\ -1 / 3 \\ 1 / 5 \end{array}\right]\)
\(\Rightarrow \quad p=\frac{1}{2}, q=-\frac{1}{3}, r=\frac{1}{5}\)
\(\therefore \quad x  =2, y=-3 \text { and } z=5\)

\( \text { Let } A=\left[\begin{array}{lll} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{array}\right] \text { and } B=\left[\begin{array}{rrr} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{array}\right]\)
Here, number of  columns of A = number of rows of B = 3, so AB is defined.   
\(\text { Now, } \quad A B=\left[\begin{array}{lll} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{array}\right]\left[\begin{array}{rrr} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{array}\right]\) 
\(c_{11}=R_{1} C_{1}=\left[\begin{array}{lll} 2 & 3 & 4 \end{array}\right]\left[\begin{array}{l} 0 \\ 3 \end{array}\right]=2+0+12=14 \)
\(c_{12}=R_{1} C_{2}=\left[\begin{array}{lll} 2 & 3 & 4 \end{array}\right]\left[\begin{array}{c} -3 \\ 2 \\ 0 \end{array}\right]=-6+6+0=0 \)
\(c_{21}=R_{2} C_{1}=\left[\begin{array}{lll} 3 & 4 & 5 \end{array}\right]\left[\begin{array}{l} 1 \\ 0 \\ 3 \end{array}\right]=3+0+15=18 \) 
\(c_{22}=R_{2} C_{2}=\left[\begin{array}{lll} 3 & 4 & 5 \end{array}\right]\left[\begin{array}{r} -3 \\ 2 \\ 0 \end{array}\right]=-9+8+0=-1 \)
\(c_{23}=R_{2} C_{3}=\left[\begin{array}{lll} 3 & 4 & 5 \end{array}\right]\left[\begin{array}{l} 5 \\ 4 \\ 5 \end{array}\right]=15+16+25=56 \) 
\(c_{31}=R_{3} C_{1}=\left[\begin{array}{lll} 4 & 5 & 6 \end{array}\right]\left[\begin{array}{l} 1 \\ 0 \\ 3 \end{array}\right]=4+0+18=22 \)
\(c_{32}=R_{3} C_{2}=\left[\begin{array}{lll} 4 & 5 & 6 \end{array}\right]\left[\begin{array}{r} -3 \\ 2 \\ 0 \end{array}\right]=-12+10+0=-2 \)
\( \text { and } c_{33}=R_{3} C_{3}=\left[\begin{array}{lll} 4 & 5 & 6 \end{array}\right]\left[\begin{array}{l} 5 \\ 4 \\ 5 \end{array}\right]=20+20+30=70 \)
\(\text { Hence, } C=A B=\left[\begin{array}{lll} c_{11} & c_{12} & c_{13} \\ c_{21} & c_{22} & c_{23} \\ c_{31} & c_{32} & c_{33} \end{array}\right]=\left[\begin{array}{ccc} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \end{array}\right] \)

The binary operation * on N is defined as a * b = L.C.M. of a and b.
(i) 5 * 7 = L.C.M. of 5 and 7 = 35
20 * 16 = L.C.M of 20 and 16 = 80
(ii) It is known that:
L.C.M of a and b = L.C.M of b and a &mn For E; a, b ∈ N.
∴ a * b = b * a
Thus, the operation * is commutative.
(iii) For a, b, c ∈ N, we have:
(a * b) * c = (L.C.M of a and b) * c = LCM of a, b, and c
a * (b * c) = a * (LCM of b and c) = L.C.M of a, b, and c
∴ (a * b) * c = a * (b * c)
Thus, the operation * is associative.
(iv) It is known that:
L.C.M. of a and 1 = a = L.C.M. 1 and a &mnForE; a ∈ N
⇒ a * 1 = a = 1 * a &mnForE; a ∈ N
Thus, 1 is the identity of * in N.
(v) An element a in N is invertible with respect to the operation * if there exists an element b in N, such that a * b = e = b * a.
Here, e = 1
This means that:
L.C.M of a and b = 1 = L.C.M of b and a
This case is possible only when a and b are equal to 1.
Thus, 1 is the only invertible element of N with respect to the operation *.

the required trapezium is as given in figure. Draw perpendiculars DP and CQ on AB. Let AP = x cm. Note that \(\Delta \)APD \(\cong \) \(\Delta \) BQC.
Therefore, QB = x cm. Also, by pythagoras theorem DP = QC = \(\sqrt { 100-{ x }^{ 2 } } \)
Let A be the area of the trapezium.

Then, A \(\equiv \) A(x)
\(=\frac { 1 }{ 2 } (sum\ of\ parallel\ sides)\times (height)\)
\(=\frac { 1 }{ 2 } (2x+10+10)\sqrt { 100-{ x }^{ 2 } } \)
= (x + 10)\(\sqrt { 100-{ x }^{ 2 } } \)
\(or\ A'(x)=(x+10)\frac { (-2x) }{ 2\sqrt { 100-{ x }^{ 2 } } } +(\sqrt { 100-{ x }^{ 2 }) } \)
\(=\frac { -2{ x }^{ 2 }-10x+100 }{ \sqrt { 100-{ x }^{ 2 } } } \)
Now, A'(x) = 0 gives 2x² + 10x - 100 = 0,
i.e., x = 5 and x = -10
So, x = 5.
Now, A''(x) = \(\frac { \sqrt { 100-{ x }^{ 2 } } (-4x-10)-(-2x^{ 2 }-10x+100)\frac { (-2x) }{ 2\sqrt { 100-{ x }^{ 2 } } } }{ 100-{ x }^{ 2 } } \)
\(=\frac { { 2x }^{ 3 }-300x-1000 }{ { (100-{ x }^{ 2 }) }^{ \frac { 1 }{ 2 } } } \)
(on simplification)
or \(A''(5)=\frac { { 2(5) }^{ 3 }-300(5)-1,000 }{ (100-(5{ ) }^{ 2 })^{ \frac { 3 }{ 2 } } } \)
\(=\frac { -2,250 }{ 75\sqrt { 75 } } =\frac { -30 }{ \sqrt { 75 } } <0\)
Thus, area of trapezium is maximum at x = 5 and the maximum area is given by
A(5) = (5 + 10)\(\sqrt { 100-{ (5) }^{ 2 } } \)
= 15\(\sqrt { 75 } =75\sqrt { 3 } { cm }^{ 2 }\)

Putting x = cos \(\theta \) in L.H.S., we get
\(LHS={ tan }^{ -1 }\left[ \frac { \sqrt { 1+cos\theta } -\sqrt { 1-cos\theta } }{ \sqrt { 1+cos\theta } +\sqrt { 1-cos\theta } } \right] \)
\(={ tan }^{ -1 }\left[ \frac { \sqrt { 2 } cos\frac { \theta }{ 2 } -\sqrt { 2 } sin\frac { \theta }{ 2 } }{ \sqrt { 2 } cos\frac { \theta }{ 2 } +\sqrt { 2 } sin\frac { \theta }{ 2 } } \right] \)
\(={ tan }^{ -1 }\left[ \frac { 1-tan\frac { \theta }{ 2 } }{ 1+tan\frac { \theta }{ 2 } } \right] \)
\(={ tan }^{ -1 }\left[ tan\left( \frac { \pi }{ 4 } -\frac { \theta }{ 2 } \right) \right] \)
\(=\frac { \pi }{ 4 } -\frac { \theta }{ 2 } \)
\(=\frac { \pi }{ 4 } -\frac { 1 }{ 2 } { cos }^{ -1 }x=RHS\)

In the interval \(\left(0, \frac{\pi}{2}\right), f(x)=\cos x\) 
\(\Rightarrow \quad f^{\prime}(x)=-\sin x\) 
which gives \(f^{\prime}(x)<0 \text { in }\left(0, \frac{\pi}{2}\right)\) 
Hence, f(x) = cos x is decreasing in \(\left(0, \frac{\pi}{2}\right)\) 

\(A^{-1} \text {exist iff }|A| \neq 0\)

We know that if a matrix is of order m x n, then it has mn elements. Thus, to find all possible orders of a matrix with 8 elements, we will find all ordered pairs of natural numbers, whose product is 8. Thus, all possible ordered pair are (1,8), (8, I), (2, 4), (4, 2).

T₁ and T₃ are similar as their sides are proportional.

(i) (b) \(\ln \triangle A B D\)
\(\tan \alpha  =\frac{B D}{A D}\)
\( =\frac{30}{30 \sqrt{3}}=\frac{1}{\sqrt{3}} \)
\(\Rightarrow  \alpha  =30^{\circ}\)
\(\therefore  \sin \alpha =\sin 30^{\circ}=\frac{1}{2}\)
\(\Rightarrow  \alpha  =\sin ^{-1}\left(\frac{1}{2}\right)\)
(ii) (c) \(\cos \alpha=\cos 30^{\circ}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \alpha =\cos ^{-1}\left(\frac{\sqrt{3}}{2})\right.\)
(iii) (d) \(\ln \triangle B DC\)
\(\tan \beta  =\frac{(BD)}{(C D)}\)
\(=\frac{30}{10 \sqrt{3}}=\sqrt{3} \)
\(\Rightarrow \beta =\tan ^{-1}(\sqrt{3})\)
(iv) \( \text { (c) Since, } \alpha=\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6} \text {, } \)
\(\beta=\tan ^{-1}(\sqrt{3})=\frac{\pi}{3} \)
\(\therefore \quad \angle A B C=\pi-\left(\frac{\pi}{6}+\frac{\pi}{3}\right)\)
\( =\pi-\frac{\pi}{2}=\frac{\pi}{2}\)
(v) (c) We know that domain and range of \(\cos ^{-1} x\) are [-1,1] and \([0, \pi]\) respectively.

(i) (d) Given, R = {(V₁,V₂) :  V₁,V₂ ∈ l}and both use their voting right in general election-2019. Since, X,Y∈ I⋅X exercised his voting right while Y did not cast her vote in general election-2019.
∴ Clearly, (X,Y)∉R
(ii) (a) Relation is symmetric.
∴(X,W)∈R
⇒(W,X)∈R
(iii) (a) Since, (F,F)∈R,F ∈ I and F use their voting right.
⇒R is reflexive.
⇒(F1,F2)∈R
⇒(F2,F1)∈R
R is symmetric.
and (F₁,F₂) ∈ R 
and  (F₂,F₃) ∈ R
⇒ (F₁,F₃) ∈ R 
(By transitive property)
(iv) (c) Given, relation RR is reflexive, symmetric and transitive.
∴R is equivalence relation.
(v) (a) Clearly, Mr. Shyam exercised his voting right in general election-2019, then Mr. Shyam is related to all those eligible voters who cast their votes.

(i) (b) :To create a garden using 200 ft fencing, we
need to maximise its area.
(ii) (c) : Required relation is given by 2x + y = 200.
(iii) (c) : Area of garden as a function of x can be represented as
\(A(x)=x \cdot y=x(200-2 x)=200 x-2 x^{2}\)
(iv) (a) : \(\begin{equation} A(x)=200 x-2 x^{2} \Rightarrow A^{\prime}(x)=200-4 x \end{equation}\)
For the area to be maximum A'(x) = 0
\(\begin{equation} \Rightarrow 200-4 x=0 \Rightarrow x=50 \mathrm{ft} \end{equation}\)
(v) (c) : Maximum-area of the garden
= 200(50) - 2(50)2 = 10000 - 5000 = 5000 sq. ft

 (i) (d)
(ii) (a)
(iii) (b) ,
(iv) (a)
(v) (b)

(i) (a) : Let \(\begin{equation} \Delta=\left|\begin{array}{cc} 1 & -2 \\ 4 & 3 \end{array}\right| \end{equation}\) 
Cofactor of 1 = 3, cofactor of -2 =-4
Cofactor of 4 = 2, cofactor of 3 = 1
\(\therefore\) Required sum = 3 - 4 + 2 + 1 = 2
(ii) (b) : Let \(\begin{equation} \Delta=\left|\begin{array}{ccc} 5 & 6 & -3 \\ -4 & 3 & 2 \\ -4 & -7 & 3 \end{array}\right| \end{equation}\) 
Minor of \(\begin{equation} a_{21}=\left|\begin{array}{cc} 6 & -3 \\ -7 & 3 \end{array}\right|=18-21=-3 \end{equation}\) 
(iii) (c) : Let \(\begin{equation} \Delta=\left|\begin{array}{ccc} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{array}\right| \end{equation}\) 
Clearly, a₃₂ = 5
and A₃₂ = cofactor of a₃₂ in \(\begin{equation} \Delta=(-1)^{3+2}\left|\begin{array}{ll} 2 & 5 \\ 6 & 4 \end{array}\right| \end{equation}\) 
= (-1)(8-30) = 22
\(\begin{equation} \therefore \end{equation}\) a₃₂·A₃₂ = 5 x 22 = 110
(iv) (d) : Here, \(\begin{equation} \Delta=\left|\begin{array}{lll} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array}\right| \end{equation}\) 
\(\therefore\) Minor of  \(\begin{equation} a_{23}=\left|\begin{array}{ll} 5 & 3 \\ 1 & 2 \end{array}\right|=10-3=7 \end{equation}\) 
(v) (b) : Here,\(\begin{equation} \Delta=\left|\begin{array}{ccc} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{array}\right| \end{equation}\) 
 \(\begin{equation} A_{11}=(-1)^{1+1}\left|\begin{array}{cc} 0 & 4 \\ 5 & -7 \end{array}\right|=1(0-20)=-20 \end{equation}\) 
\(\begin{equation} A_{12}=(-1)^{1+2}\left|\begin{array}{cc} 6 & 4 \\ 1 & -7 \end{array}\right|=-1(-42-4)=46 \end{equation}\) 
\(\begin{equation} A_{13}=(-1)^{1+3}\left|\begin{array}{ll} 6 & 0 \\ 1 & 5 \end{array}\right|=1(30-0)=30 \end{equation}\) 
\(\begin{equation} \therefore \quad \Delta=a_{11} A_{11}+a_{12} a_{12}+a_{13} A_{13} \end{equation}\) 
= 2(-20) -3(46) + 5(30) = -28
\(\begin{equation} \Rightarrow|\Delta|=28 \end{equation}\)

(a) e^xy + log (xy) + cos(xy) + 4 = 0
\(\begin{array}{ll} \Rightarrow & e^{x y} \frac{d}{d x}(x y)+\frac{1}{x y} \frac{d}{d x}(x y)-\sin (x y) \frac{d}{d x}(x y)=0 \end{array}\)
\(\begin{array}{ll} \Rightarrow & \frac{d}{d x}(x y)\left\{e^{x y}+\frac{1}{x y}-\sin (x y)\right\}=0 \end{array}\)
\(\begin{array}{ll} \because & e^{x y}+\frac{1}{x y}-\sin (x y) \neq 0 \end{array}\)
\(\begin{array}{ll} \therefore & \frac{d}{d x}(x y)=0 \end{array}\)
 

(c) Assertion is correct, reason is incorrect.

(a) Assertion is correct, reason is correct;reason is a correct explanation for assertion.

(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.

(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.