**Gaurav Sharma**, Director of Genius Classes Of Mathematics (GCM), Karnal ( Haryana ), completed his graduation from Kurukshetra University in Electronics. Being inclined towards Mathematics, he pursued his Masters in Mathematics and also completed his L.L.B. He believes that, when a man quits learning he will be more senile onwards. He is a mentor right from his college days and loves teaching and inspiring students. Team Genius Classes, along with the regular classes, organizes workshops, events and seminars time to time. You can find lot of good stuffs in his Facebook group - CAT Mausi

Two sides of a triangle are given by the roots of the equation x^2 – 2 root(3)x + 2 = 0. The angle between the sides is Pi/3. The perimeter of the triangle is

- 6 + root(3)
- 2 root(3) + root(6)
- 2 root(3) + root(10)
- None of these

Let ABC be the triangle such that it’s sides a = BC and b = CA are the roots of the equation x^2 – 2 root(3)x + 2 = 0

a + b = 2 root(3) and ab = 2

it is also given that angle C = Pi/3 so Cos C = ½

(a^2 + b^2 – c^2)/2ab = ½

a^2 + b^2 – c^2 = ab

(a + b)^2 – c^2 = 3ab

(2 root(3))^2 – c^2 = 3 x 2 = c^2 = 6

c = root(6)

perimeter of triangle ABC = a + b + c = 2 root(3) + root(6)

Find the range of x for | |2x – 4| + 3| < 7

- (0, 8)
- ( -4, 0)
- [-4, 0]
- (0, 4)

We have | |2x – 4| + 3| < 7

-7 < |2x – 4| + 3 < 7

-10 < |2x – 4| < 4

|2x – 4| is always greater than – 10, so we need to look only one part of the inequality

|2x – 4| < 4

-4 < 2x – 4 < 4

0 < 2x < 8

0 < x < 4

x e (0, 4)

If a, b, c, d are four consecutive terms of an increasing AP then the roots of the equation (x – a) (x – c) + 2( x – b) (x – d) = 0 are

- Non real complex
- Real and equal
- Integers
- Real and distinct

Given than a < b < c < d

f(x) = (x – a) (x – c) + 2 (x – b) (x – d)

f(b) = (b – a) (b – c) < 0

And f(d) = (d – a) (d – c) > 0

f(x) = 0 has one root in (b, d). Also, f(a) f(c) < 0

So the other root lies in (a, c)

Hence roots of the equation are real and distinct

If four squares are chosen at random on a chess board, find the probability that they lie on a diagonal line

- 364/64C4
- 182/64C4
- 364/64C2
- 91/64C4

Total number of ways to select 4 squares out of 64 squares 64C4

Clearly only diagonals 4, 5, 6 … 12 contain 4 or more squares

Number of ways in which we can select 4 squares from the diagonals in one direction =

4C4 + 5C4 + 6C4 + 7C4 + 8C4 + 7C4 + 6C4 + 5C4 + 4C4

= 1 + 5 + 15 + 35 + 70 + 35 + 15 + 5 + 1

= 182

Similarly we can select 182 squares from the diagonals in the other direction

Total number of ways = 182 + 182 = 364

So, required probability = 364/64C4

If ABCD is a parallelogram and Q, R are the circum-centers of triangles ABC and ADC respectively. Then AQCR is always

- Rectangle
- Trapezium
- Rhombus
- None of these

Since the triangles ABC and ADC are congruent their circumcircles will be congruent.

Therefore, AQ, QC, AR and CR will be equal as they are all radii of the circumcircles.

So AQCR is a rhombus.

An Infinite GP has its first term a and sum = 8, what is the range of values a can take

- (0, 16)
- [0, 16]
- (1, 15)
- (0, 8)

Sum of infinite GP = a/ ( 1 – r) = 8

-1 < r < 1 , r e (-1, 1)

So, 1 – r e (0, 2)

a = 8 ( 1 – r)

Now if -1 < r < 1

So, 0 < 1 – r < 2

8 x 0 < 8 ( 1 – r) < 8 x 2

0 < 8 ( 1 – r) < 16

0 < a < 16

N! is expressed in base 12 ends in exactly 7 trailing zeroes. How many values can N take

- 7
- 5
- 0
- 11

N! is a multiple of 12^7 but not 12^8

N! is a multiple of (2^2 x 3)^7 but not ( 2^2 x 3)^8

2^14 x 3^7 divide N! but 2^16 x 3^8 does not.

Now by trial and error method, Let us start with N = 15

Highest power of 2 that divides N! = 7 + 3 + 1 = 11

Highest power of 3 that divide N! = 5 + 1 = 6

15! = 2^11 x 3^6 x p

16! = 2^15 x 3^6 x p

17! = 2^15 x 3^6 x q

18! = 2^16 x 3^8 x q

18! Is a multiple of 12^8 and 17! Is not a multiple of 12^7

So, there are no number N such that N! is a multiple of 3^7 but not 3^8

There is no number N such that N! is a multiple of 12^7 but not 12^8

0 values of N satisfy the above condition.

Triangle ABC has a perimeter 18. What is the highest value that the inradius can take?

- 3
- Root(2)
- Root(3)
- 3 root(3)

Perimeter = 18

Area = A

rs = A

semi perimeter = 9

r = A/s = A/9

the higher the area, the higher the radius

For a given perimeter, the maximum area will for an equilateral triangle

In radius = a / 2root(3) = 6/2root(3) = 3/root(3) = root(3)

The number of values of k for which [x^2 – (k – 2)x + k^2] * [x^2 + kx + (2k – 1)] is a perfect square is

- 2
- 1
- 0
- None of these

For the given equation [x^2 – (k – 2)x + k^2] * [x^2 + kx + (2k – 1)] should have roots common or each should have equal roots. If both roots are common, then

1/1 = - ( k – 2)/k = k^2/(2k – 1)

k = - k + 2 and 2k – 1 = k^2 => k = 1

If both the equations have equal roots, then

( k – 2)^2 – 4k^2 = 0 and k^2 – 4( 2k – 1) = 0

(3k – 2) (-k – 2) = 0 and k^2 – 8k + 4 = 0 ( no common value )

k = 1 is the only possible value.

If a is an integer lying in [-5, 30], then the probability that the graph of y = x^2 + 2(a + 4)x – 5a + 64 is strictly above the x-axis.

- 1/6
- 7/36
- 2/9
- 3/5

x^2 + 2(a + 4)x – 5a + 64 ≥ 0

if D ≤ 0 then

(a + 4)^2 – ( -5a + 64) < 0

Or a^2 + 13a – 48 < 0

(a + 16) (a – 3) < 0

-16 < a < 3

-5 ≤ a ≤ 2

Then the favorable cases is equal to the number of integers in the interval [-5, 2] i.e 8

Total number of cases is equal to the number of integers in the interval [ -5, 30] i.e 36

Required probability = 8/36 = 2/9