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A prime number is a positive integer that has exactly two distinct whole number factors (or divisors), namely 1 and the number itself.

\(\therefore\) The number of divisors of any prime number is 2.

We can say the LCM of two prime numbers is their product. Hence, the statement in the question is TRUE.

∵ 10 = 1 x 2 x 5

Positive divisor of 10 are 1, 2 and 5.

∴ A = {1, 2, 5}.

2, 3, 5 and 7 are prime numbers less than 10.

∴ B = {2, 3, 5, 7}.

Let x be constant function from set A to set B.

i.e. x(a) = b, where a ∈ A and b ∈ B and b is a fixed constant.

Then numbers of constant functions = n(B) = 4.

First 5 whole numbers are 0,1,2,3 and 4.

Since, no number is divisible by 0.

Therefore, there is no number which is exactly divisible by the first 5 whole numbers.

The correct option is: (B) 25 cm

The correct option is :(A) 1

In ∆ PQR 12 cm > 10 cm > 8 cm 

∴ QR > PQ > PR 

∴ ∠ P > ∠ R > ∠ Q

The biggest angle is ∠P and the smallest angle is ∠Q.

The correct option is : (C) 5

The correct option is : (A) One 

\(\frac{sin^3x}{1 + cos x} + \frac{cos^3x}{1 + sin \,x}\)

\(= \frac{sin^3x}{1 + cos \,x} \times \frac{1- cos\,x}{1 - cos \,x} + \frac{cos^3x}{1 + sin\,x} \times \frac{1 - sin\,x}{1 - sin\, x}\)

\(= \frac{(sin^3x)(1 - cos\,x)}{1 - cos^2x} + \frac{cos^3x}{1 - sin^2x} (1 - sin\,x)\)

\(= \frac{(sin^3x)(1 - cos\,x)}{sin^2x} + \frac{cos^3x(1 - sin\,x)}{cos^2x}\)

\(= sin\,x (1 - cos\,x) + cos\, x(1 - sin\, x)\)

\(= sin\,x + cos\,x - 2sin\,x\;cos\,x\)

\(= \sqrt 2\left(\frac1{\sqrt2} sin\,x + \frac1{\sqrt2} cos\,x\right) - sin(2x)\)

\(= \sqrt 2 (sin\,x \;cos(\frac\pi4) + cos\,x \;sin(\frac\pi4)) - sin (2x)\)

\(= \sqrt 2\; sin(x + \frac\pi4) - sin(2x)\)

or \(\sqrt 2\, cos(x - \frac\pi4) - sin(2x)\)

⇒ \(\sqrt 2 \,cos(\frac\pi4 - x) - sin(2x)\)

a) At t= 6 the position of ant will be y= 2, Since the path will repeat after t=4

b) Speed of ant is s = sq.rt. of 13

Distance d = sq.rt(6² + 4²) = (sq.rt of 13)*2

Time t = d/s

t = sq.rt(13)*2/sq.rt(13) = 2 sec

The remainder will be zero.

a^n + b^n is divisible by a + b (provided n is odd).

So, 1^2017+2^2017+3^2017+4^2017 will be divisible by (1+2+3+4)

So, it is divisible by 2017 as well.

Debtors’ Ledger Adjustment Account in General Ledger 

Creditors’ Ledger Adjustment Account in General Ledger 

Correct option is (1) \(1:1:1\)

\(x^2 + x + 1=0\)

\(x = \frac{-1 \pm \sqrt{-3}}2 = \frac{-1 \pm\sqrt 3 i}2\)

Since, roots are imaginary.

Given that one root of equation ax² + bx + c = 0 and x² + x +1 = 0 is common.

Since, root is imaginary so other root is its reciprocal. So, if one root is common then other root must be common.

Hence, both given quadratic equations have two common roots.

\(\therefore\) Both equation are congruent to each other 

\(\therefore \frac a1 = \frac b1 = \frac c1\)

Hence, \(a = b= c\)

\(\therefore a:b:c = 1:1:1\)

(i)Heaven Life Insurance Co. Valuation balance Sheet as at 31st march, 2017

(ii) Statement showing Net Profit for the valuation period 

(iii) Amount due to Policyholders

clearly the given triangle is a right angle triangle, right angle at (6,6) centroid (4,4)  and circumcentre of a triangle lies on the midpoint of hypt.   

circumcentre {(6+0)/2,(0+6)/2} =(3,3)  

now distance ={(4-3)² +(4-3)²}^1/2                           

 =√2

(i) Under Single column In the Books of H. O. Delhi Branch Stock Account 

(ii) Under Double Column In the Books of H. O. Delhi Branch Stock Accoun

Let direction ratio's of the required line be a, b, c since it is perpendicular to the two given lines 

2a - 3b - 2c = 0 ...(i)

and -a + 2b + 3c = 0 ...(ii)

Solving (i) and (ii) by cons multiplication,let us

a/-5 = b/-4 = c/1 = k (let)

Thus the required line passes through (-1,2,3) and has direction ratio propersonal to -5, 4, 1. so it eqn. in

(x + 1)/-5 = (y - 2)/-4 = (x - 3)/1

Answer is (b) - 6/7

Given:

Efficiency of Anil = 2 × Efficiency of Binay

Time difference between Anil and Binay to finish a work = 20 days

Concept used:

Efficiency = Total Work/Time

Calculation:

According to the question

Efficiency ratio of Anil and Binay = 2 ∶ 1

The ratio of time taken by Anil and Binay = 1 ∶ 2

Time difference between Anil and Binay = 1 unit

⇒ 20 days = 1 unit

So,

Time taken by Anil = 20 × 1 = 20 days

Time taken by Binay = 20 × 2 = 40 days

Now, 

Total work = LCM of 20 and 40

⇒ 40 units

Time taken by Anil and Binay working together = Total Work/Efficiency

⇒ 40/(1 + 2)

⇒ 40/3

⇒ 13.33 days

∴ Time taken by Anil and Binay to finish work together is 13.33 days.

∣PA−PB∣<AB

Answer is (b) sec² x

Answer is (d) 2

Concept:

If A is the arithmetic mean of numbers a and b and is given by ⇔ A =  (a + b)/2

If G is the geometric mean of the numbers a and b and is given by  ⇔ G = \(\rm \sqrt{ab}\)

Calculation:

Let the numbers be a and b 

According to given condition,

\(\rm \frac{a+b}{2\sqrt {ab}}=\frac m n\)

Applying componendo and dividendo, we get 

\(\begin{array}{l} \rm \frac{a+b+2 \sqrt{a b}}{a+b-2 \sqrt{a b}}=\frac{m+n}{m-n} \\ \rm \frac{(\sqrt{a}+\sqrt{b})^{2}}{(\sqrt{a}-\sqrt{b})^{2}}=\frac{m+n}{m-n} \\ \rm \frac{(\sqrt{a}+\sqrt{b})}{(\sqrt{a}-\sqrt{b})}=\frac{\sqrt{m+n}}{\sqrt{m-n}} \end{array}\)

Applying componendo and dividendo again, we get 

\( \rm \frac{(\sqrt{a}+\sqrt{b})+(\sqrt{a}-\sqrt{b})}{(\sqrt{a}+\sqrt{b})-(\sqrt{a}-\sqrt{b})}=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}} \)

Squaring both sides, we get

\(\rm \frac{a}{b}=\left(\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}\right)^{2} \)

\(\rm\frac{a}{b}=\frac{2 m+\sqrt{m^{2}-n^{2}}}{2 m-2 \sqrt{m^{2}-n^{2}}} \\ \rm \frac{a}{b}=\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}\)

Concept:

Let a and b two numbers.

AM = \(\rm (a+b)\over 2\) and GM = \(\rm \sqrt{ab}\)

Calculation:

Let, two numbers be a and b,

We know, AM = \(\rm (a+b)\over 2\) and GM =\(\rm \sqrt{ab}\)

Given that, GM : AM = n : m 

⇒ AM : GM = m : n

\(⇒ \frac{\rm a+b}{2\rm \sqrt{ab}}= \frac{\rm m}{\rm n}\)

\(⇒ \frac{(\rm a+b)^{2}}{4\rm a b}= \frac{\rm m^{2}}{\rm n^{2}}\;\;\;\;\; \ldots \ldots \ldots . . \text { (i) } \\ ⇒\frac{(\rm a+b)^{2}-4 a b}{4\rm a b}=\frac{\rm m^{2}- \rm n^{2}}{ \rm n^{2}} \\ ⇒ \frac{(\rm a-b)^{2}}{4 \rm a b}=\frac{\rm m^{2}-\rm n^{2}}{\rm n^{2}} \ldots \ldots \ldots . . \text { (ii) }\)

Since, on dividing eqn (i) and (ii), we get 

\( \rm \frac{(a+b)^{2}}{(a-b)^{2}}=\frac{m^{2}}{m^{2}-n^{2}}\\ \rm ⇒ \frac{a+b}{a-b}=\frac{m}{\sqrt{m^{2}-n^{2}}} \\ \rm ⇒ \frac{(a+b)+(a-b)}{(a+b)-(a-b)}=\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}\)

\(\rm \Rightarrow \frac{2a}{2b}=\frac{a}{b}=\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}\)

Hence, option (2) is correct.

The correct option is (B) 15.

Explanation:

Dividend on Rs. 20 = Rs. (9/100)x 20 = Rs.9/5.  

Rs. 12 is an income on Rs. 100.  

Rs. 9/5 is an income on Rs.[ (100/12) x (9/5)] = Rs. 15.

Answer is (a) -1

Answer is (b) (x = 2,y = - 8)