In \(\triangle PQR, \angle 1=\angle 2\)  [given]
\(\Rightarrow \) PR = PQ  [since, sides opposite to equal angles of a triangle are not equal]
Given that, \(\frac { QR }{ QS } =\frac { QT }{ PR } \Rightarrow \quad \frac { QR }{ QS } =\frac { QT }{ PQ } \)
[\(\because \)  PQ = PR, proved above]
\(\Rightarrow \frac { QS }{ QR } =\frac { PQ }{ QT } \)  
[on taking of the terms reciprocals]  ... (i)
In \(\triangle PQS\) and \(\triangle TQR\), we have
\(\angle PQS=\angle TQR\)     [common]
and \(\Rightarrow \frac { QS }{ QR } =\frac { QP }{ QT } \)  [from Eq.(i)]
\(\therefore \triangle PQS\sim \triangle TQR\)    [by SAS similarity criterion]
Hence proved.

Given pair of linear equations is
41x+53y=135   ..(i)
and 53x+41y=147  ...(ii)
On adding Eqs. (i) and (ii), we get
94x+94y=282
\(\Rightarrow\) x+y=3 [dividingboth sides by 94] ...(iii)
On subtracting Eq. (i) from Eq. (ii), we get
12x-12y=12
\(\Rightarrow\) x-y=1 [dividing both sides by 12]   ...(iv)
Now, on adding Eqs.(iii) and (iv), we get
2x=4 \(\Rightarrow\) x=2
On substituting x=2 in Eq. (iii), we get
y=3-2=1
Hence, x=2 and  y=1 is the required solution.

Given, a and β are the zeroes of the polynomial
p(x)=ax²+bx+c.
Sum of zeroes, a+β=\(-b\over a\) and product of zeroes, aβ=\(c\over a\)
Now, a²β+aβ²=aβ(a+β)
\(=\frac { c }{ a } \times \frac { (b) }{ a } =\frac { -bc }{ a^{ 2 } } \)

Let C be the position of peacock and A be the position of serpent.

Given, ∠DCA = 30°
\(\Rightarrow \angle B A C=\angle D C A=30^{\circ}\) [alternate angles]
∵ Distance = Speed x Time
∴ \(A C=300 \ \frac{12}{60}\left[\because 12 \mathrm{~s}=\frac{12}{60} \mathrm{~min}\right]\)
\(\Rightarrow A C=60 \mathrm{~m}\)
In right angled ΔABC,
\(\sin 30^{\circ}=\frac{P}{H}=\frac{B C}{A C}=\frac{h}{60}\)
30 m.

The list of two-digit numbers divisible by 3 is :
12, 15, 18, . . . , 99
Is this an AP? Yes it is. Here, a = 12, d = 3, a_n = 99.
As   a_n = a + (n – 1) d,
we have   99 = 12 + (n – 1) x 3
i.e., 87 = (n – 1) x 3
i.e.,  \(n-1=\frac{87}{3}=29\)
i.e., n = 29 + 1 = 30
So, there are 30 two-digit numbers divisible by 3.

Given that coordinates of one end point of the diameter is (4, -1) and centre of the circle is (1,- 3).

Let coordinates of the other end of the diameter be (x, y). 
We know that the centre of the circle (1,-3) is the mid-point of diameter.
\(\Rightarrow \frac { 4+x }{ 2 } =1\) and \( \frac { (-1+y) }{ 2 } =-3\)
\(\Rightarrow 4+x=2\) and \( -1+y=-6\Rightarrow x=-2\) and \( y=-6+1=-5\)
Thus, coordinates of the other end of the diameter are (-2, -5).

Let base be x cm, then height be (x-7) cm
By Pythagoras Theorem, (base)² + (height)²=(hypotenuse)²
\(\Rightarrow\) x²+(x-7)2=(13)² \(\Rightarrow\) x²+x²+49-14x=169
\(\Rightarrow\) 2x²-14x-120=0 \(\Rightarrow\) x²-7x-60=0
\(\Rightarrow\) x²-12x+5x-60=0 \(\Rightarrow\) x(x-12)+5(x-12)=0
\(\Rightarrow\) (x+5)(x-12)=0 \(\Rightarrow\) x-12=0 or x+5=0
\(\Rightarrow\) x=12 or x=-5 (-5 is rejected as sides can never be negative)
\(\Rightarrow\) Base=12 cm and altitude=12-7=5 cm

\(L H S=\frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\frac{\cos A}{\sin A}-\cos A}{\frac{\cos A}{\sin A}+\cos A}\)
\(=\frac{\cos A\left(\frac{1}{\sin A}-1\right)}{\cos A\left(\frac{1}{\sin A}+1\right)}=\frac{\left(\frac{1}{\sin A}-1\right)}{\left(\frac{1}{\sin A}+1\right)}=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}=\operatorname{RHS}\)

We have
x² + 7x + 10 = (x + 2)(x + 5)
So, the value of x² + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0, i.e., when x = – 2 or x = –5. Therefore, the zeroes of x² + 7x + 10 are – 2 and – 5. Now,
sum of zeroes = \(-2+(-5)=-(7)=\frac{-(7)}{1}=\frac{-(\text { Coefficient of } x)}{\text { Coefficient of } x^{2}}\) 
product of zeroes = \((-2) \times(-5)=10=\frac{10}{1}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)

First, solve the given equations similar to question 1 to get the value of x and y and then put these value in y=mx+7 or y=mx+3 to get required value of m.
x=2, y=-3, m=-5

Rational

Let point on y-axis be (0,a)
Now distance of this point from (5,-2) equal to distance from point (-3,2)
i.e.,  \(\Rightarrow \sqrt { 5^{ 2 }+\left( -2-a \right) ^{ 2 } } =\sqrt { \left( 3 \right) ^{ 2 }+\left( a-2 \right) ^{ 2 } } \)
Squaring and simplifying, we get
25+4+a²+4a = 9+a²+4-4a ⇒ 8a=-16  ⇒ a=-2

\(4x^2+4\sqrt3+3=0\)
\(\Rightarrow \quad { x }^{ 2 }+\frac { 4 }{ 4 } \sqrt { 3 } x+\frac { 3 }{ 4 } =0\quad \Rightarrow \quad { x }^{ 2 }+\sqrt { 3 } x=-\frac { 3 }{ 4 } \)
Now adding square of (a coefficient of x) on both sides, we get
\(\Rightarrow \quad { x }^{ 2 }+\sqrt { 3 } x+{ \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }=-\frac { 3 }{ 4 } +{ \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }\quad \Rightarrow \quad { \left( x+\frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }=-\frac { 3 }{ 4 } +\frac { 3 }{ 4 } \)
\(\Rightarrow \quad { \left( x+\frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }=0\quad \Rightarrow \quad x+\frac { \sqrt { 3 } }{ 2 } =0\)
\(\Rightarrow \quad x=\frac { \sqrt { 3 } }{ 2 } \)
Hence, the roots are\(\frac { -\sqrt { 3 } }{ 2 } \) and\(\frac { -\sqrt { 3 } }{ 2 } \).

(d) If Assertion is incorrect but Reason is correct.

(a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

(i) (c) (6, 7)
(ii) (b) \(\mathrm{AB}=\sqrt{(8-0)^{2}+(3+1)^{2}}\) 
(iii) (b) BD = 10 km
(iv) (b) \(\mathrm{AC}=\sqrt{(6-0)^{2}+(7+1)^{2}}\) 
\(=\sqrt{36+64}=\sqrt{100}=10\) 
(v) (a) From (ii)  \(\mathrm{AB}=4 \sqrt{5}\) 
\(\mathrm{DC}=\sqrt{(6+2)^{2}+(7-3)^{2}}\) 
= \(\sqrt{64+16}=\sqrt{80}=4 \sqrt{5}\) 
\(\mathrm{AD}=\sqrt{(-2-0)^{2}+(3+1)^{2}}\) 
\(=\sqrt{4+16}=\sqrt{20}=2 \sqrt{5}\) 
and \(\mathrm{BC}=\sqrt{(6-8)^{2}+(7-3)^{2}}\) 
 = \(=\sqrt{4+16}=\sqrt{20}=2 \sqrt{5}\) 
\(\therefore \quad \mathrm{AB}=\mathrm{DC} \text { and } \mathrm{AD}=\mathrm{BC}\) 
Thus, opposite sides are equal.
From (iii) and (iv) we have
BD = 10 and AC = 10
Since opposite sides are equal and diagonals are equal therefore ABCD is a rectangle.

(i) (b): \(\angle X A C=45^{\circ}\) 
\(\therefore \quad \angle A C D=45^{\circ}\) [Alternate interior angles]
(ii) (b)
(iii) (c) : \(\text { In } \Delta A C D\)
\(\frac{A D}{D C}=\tan 45^{\circ} \)
\(\Rightarrow \frac{100}{D C}=1 \Rightarrow D C=100 \mathrm{~m}\)
(iv) (d): \(\text { In } \Delta A B D, \frac{A D}{B D}=\tan 30^{\circ}\)
\(\Rightarrow \quad \frac{100}{B D}=\frac{1}{\sqrt{3}} \)
\(\Rightarrow \quad B D=100 \sqrt{3} \mathrm{~m}\)
(v) (a): \(\text { In } \Delta A D C\)
\(\frac{A D}{A C}=\sin 45^{\circ} \Rightarrow \frac{100}{A C}=\frac{1}{\sqrt{2}} \Rightarrow A C=100 \sqrt{2} \mathrm{~m}\)

Here the smallest 3-digit number divisible by 7 is 105. So, the number of bacteria taken into consideration is 105, 112, 119, .... ,994 So, first term (a) = 105, d = 7 and last term = 994
(i) (c): t₅ = a + 4d = 105 + 28 = 133
(ii) (b): Let n samples be taken under consideration.
\(\because\) Last term = 994
\(\Rightarrow\) a + (n - 1)d = 994 \(\Rightarrow\) 105 + (n - 1)7 = 994 \(\Rightarrow\) n = 128
(iii) (a): Total number of bacteria in first 10 samples
\(=S_{10}=\frac{10}{2}[2(105)+9(7)]=1365\)
(iv) (a): t₇ from end = (128 - 7 + 1)^th term from beginning = 122^th term = 105 + 121(7) = 952
(v) (c): t₅₀ = 105 + 49 x 7 = 448