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a²x² +4ax +4 = 0

D = b² - 4ac

= (4a)² -4 x a² x 4

= 16a² - 16a²

= 0

D = 0

If D = 0, then the quadratic equation has 1 solution.

Correct option  (C)  3x + 2y = 0

Explanation :

the equation of  the line is

-3(x + 2) -2(y - 3) = 0

3x + 2y = 0

Correct option  (D)  x +  6y + 16 = 0

Explanation :

Slope of the line joining the points (4, 1) and (-2,2) is

2 - 1/-2 - 4(= -1/6)

Hence, by Theorem 2.2, the equation of the given line is

y + 3 = 1/6(x - 2)

x + 6y + 16 = 0

Correct option (A)

Explanation :

Since the line passes through ( 1, 4),

we have

3a(-1) + 5(4) + a - 2 = 0

-2a + 18 = 0

 Hence, a = 9 and the line is 27x + 5y + 7  = 0.

Answer is (c)  P(A'B') = P(A').P(B')

(B) With increase in temperature, the value of K for endothermic reaction increases because unfavourable change in entropy of the surroundings decreases 

(D) With increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surrounding decreases

Answer is (b) P(A ∩ B) = 0

(A) Adsorption is accompanied by decrease in enthalpy and decrease in entropy of the system 

(B) The critical temperatures of ethane and nitrogen are 563 K and 126 K, respectively. The adsorption of ethane will be more than that of nitrogen on same amount of activated charcoal at a given temperature 

In adsorption process both ΔH & ΔS is – ve. Higher the critical temperature of a gas higher the extent of adsorption. 

Cloud is not an emulsion. 

Brownian motion depends on the size of the particles.

Answer is (b) 1/2

Answer is (b) A ⊂ B

Answer is (b) 2 log(1 + √x) + c

Answer is (d) 1

Given, the function f : R → R is defined as f(x) = x³, x ∈ R

Let x₁, x₂ ∈ R 

f(x₁) = f(x₂)

x³₁ = x³₂

x₁ = x₂

Thus the given function is one-one or injective.

Let Polar coordinate if point (1,2) is (r,θ)

We know the relation of cartesian coordinates (x, y) with polar coordinates (r,θ) is

x = rcosθ  \(\Rightarrow\) rcosθ = 1 .............(1)

y = rsinθ \(\Rightarrow\) rsinθ = 2 .............(2)

dividing equation (2) by equation (1), we get

tanθ = 2 \(\Rightarrow\) θ = tan^-12 = 1.107

Now, adding square of equation (1) and square of equation (2)

we get,

r²cos²θ + r²sin²θ = 1+4

r² = 5 \(\Rightarrow\) r = \(\sqrt5\)

therefore, polar coordinates of (1,2) is (\(\sqrt5\),1.107) or (\(\sqrt5\),tan^-12)

Let \(C(k,7) = (x_1,y_1), A(-5,1) = (x_2,y_2), B(3, 5) = (x_3, y_3)\)

Since points are equidistant, so

AC = BC

\(AC^2 = BC^2\)

Using distance formula,

\((x_3-x_1)^2 + (y_3-y_1)^2 = (x_2-x_1)^2+(y_2-y_1)^2\)

\(\Rightarrow (3-k)^2+(5-7)^2 = (-5-k)^2+(1-7)^2\)

\(\Rightarrow 9 - 6k + k^2 + (-2)^2 = 25 + 10k + k^2 + (-6)^2\)

\(\Rightarrow 9-6k+k^2+4=25+10k+k^2+36\)

\(\Rightarrow 13-6k=61+10k\)

\(\Rightarrow -6k-10k =61-13\)

\(\Rightarrow -16k = 48\)

\(\Rightarrow k = \cfrac{48}{-16}\)

\(\Rightarrow k = -3\)

dy/dx = 5a^x.loga - 1/x = 3/2√x

y = √(x + y)y² - 2y = x

dy/dx = 1/(2y - 1)

 \(S_n = 1 + 4 + 1 +...+ 3n - 2\)    \(\begin{pmatrix}\because \text{nth term of sequence is}\\a_n = a+ (n - 1)d\\= 1+(n-1)3\\= 3n- 2\end{pmatrix}\) 

\(= \sum (3n - 2)\)

\(= 3\sum n - 2\sum 1\)

\(= \frac{3(n(n+1))}{2}-2n\)

\(= \frac{3n^2+ 3n - 4n}{2}\)

\(= \frac{3n^2 - n}{2}\)

\(= \frac{n(3n -1)}{2}\)

Correct option is (d) \(5x^2 + 5y^2 + 60x + 7=0\)

\(\frac{\sqrt{{x_1}^2+ {y_1}^2+ 4x_1 +3}}{\sqrt{{x_1}^2 + {y_1}^2 - 6x_1 + 5}} = \frac23\)

⇒ \(\frac{{x_1}^2+ {y_1}^2+ 4x_1 +3}{{x_1}^2 + {y_1}^2 - 6x_1 + 5} = \frac49\)

⇒ \(9{x_1}^2 + 9{y_1}^2 + 36x_1 + 27= 4{x_1}^2 + 4{y_1}^2-24x_1 + 20\)

⇒ \(5{x_1}^2+ 5{y_1}^2+ 60x_1 + 7 = 0\)

\(\therefore\) Locus of point is \(5x^2 + 5y^2 + 60x + 7=0\)

Let initial radius of the circle is r then, area of circle is πr²
Now, area of circle is decreased by 36% .
e.g., area of new circle = πr² - 36% of πr² = 0.64 πr²
area of new circle = π(0.8r)²
Hence, raidus of new circle = 0.8r
% decreased in radius = change in radius/intial radius × 100
= (r - 0.8r)/r × 100
= 0.2r/r × 100
= 20 %

Hence, radius of circle is decreased by 20%

Let f(x₁) = f(x₂) for some x₁, x₂ ∈

(x₁)³ = (x₂)³  

x₁ = x₂,

Hence f(x) is one − one

Since √a is not defined for a ∈ (−∞, 0) 

∴ √a = b is not a function.

The element(s) of relation R be removed to make R an equivalence relation is (1,2)

Geomorphology is the science dealing with the study and interpretation of the origin and development of landforms on the earth's surface. Geomorphology is an aid to resource evolution, engineering construction, and planning. It includes the study of the landforms and of the processes operating on them.

• The primary areas covered are weathering, hillslope and mass movement, fluvial features, glacial features, aeolian features, coastal and karst landforms.
• Though it is a branch of physical geography, it is nevertheless, closely related to surface geology, hydrology, and meteorology, various exogenic and endogenic processes account for the development of different types of landform features.
• Therefore, it involves the study of specific landforms and an interpretation of how each landform is related to others in space and time. 
• However, the forms of the land surfaces are the only records, which help to describe the history of the earth’s surfaces.
• The study of landforms gives meaningful information to physical and even to human geographers.

correct option:

(B) (x + 5)/3

(d) Mc kinley is the highest mountain peak of the USA. Mount Mc Kinley or Denali is the highest mountain peak in North America, with a summit elevation of 20,237 feet above sea level. At some 18,000 feet, the base to peak rise is considered the largest of any mountain situated entirely above sea level.

2⁶ reflexive relations

We know that total number of reflexive relations on set A having n element is 

2^{\(n^2-n\).}

 Given that = {a₁, a₂, a₃, a₄, a₅ }.

Therefore, A has 5 elements.

Therefore, total number of reflexive relations defined on the set A = 2^{\(5^2-5\)}. = 2²⁰.

We have relation R : A ⟶ A, R = {(x, y): x and y are students of same set}.

⇒ R = {(x, y) : x and y ∈ A}.

Reflexivity : Let x ∈ A.

⇒ (x, x) ∈ R.

Therefore, relation R is reflexive.

Symmetricity : Let (x, y) ∈ R 

⇒ x and y belongs to same set A.

⇒ y and x belongs to same set A

⇒ (y, x) ∈ R.

Therefore, relation R is symmetric.

Transitivity : Let (x, y) ∈ R and (y, z) ∈ R

⇒ x and y, y and z ∈ A

⇒ x and z ∈ A

⇒ (x, z) ∈ R.

Therefore, relation R is transitive.

Since, relation R is reflexive, symmetric and transitive.

Therefore, relation R is an equivalence relation.