2 nd batch Admission Test Math Solution

41. The value of Ф (4500) is

(a) 150 (b) 1200 (c) 320 (d) 1500

Solution

Theory:

In number theory, Euler's totient or phi function, φ(n) is an arithmetic function that counts the totatives of n, that is, the positive integers less than or equal to n that are relatively prime to n. Thus if n is a positive integer, then φ(n) is the number of integers k in the range 1 ≤ k ≤ n for which gcd(n, k) = 1. The totient function is a multiplicative function, meaning that if two numbers m and n are relatively prime (to each other), then φ(mn) = φ(m)φ(n).[3][4]

For example let n = 9. Then gcd(9, 3) = gcd(9, 6) = 3 and gcd(9, 9) = 9. The other six numbers in the range 1 ≤ k ≤ 9, that is, 1, 2, 4, 5, 7 and 8, are relatively prime to 9. Therefore, φ(9) = 6. As another example, φ(1) = 1 since gcd(1, 1) = 1.

The totient function is important mainly because it gives the order of the multiplicative group

of integers modulo n (the group of units of the ring ).

Solve:

Ф (4500) = Ф (125*9*4) = Ф (125) Ф (9) Ф (4) = (125−25)*(9−3)*(4−2) = 1200

Answer: (b) 1200

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42. If 7x = 15(mod 40), then the value of x is

(a) 25 (b) 20 (c) 35 (d) 15

Solution

7x = 15(mod 40) is equivalent to solving the equation: 7x = 15 + 40q

Now check with the answer choices:

If x = 25, q = (7*25-15)/40 = 4

If x = 20, q = (7*20-15)/40 = 3.125

If x = 35, q = (7*35-15)/40 = 5.75

If x = 15, q = (7*15-15)/40 = 2.25

Answer: (a) 25

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43. The solution of the matrix equation AX = b, where A = and b =

is

(a) (b) (c) (d)

Solution

AX = b [by calculator]

X = A-1 b

=

Answer: (c)

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44. The dimension of the vector space Pn is

(a) n (b) (n+1) (c) n+2 (d) n/2

Answer: (b) (n+1)

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45. If A = , then A-5 is

(a) (b)

(c) (d)

Solution

[using scientific calculator]

Answer: (a)

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46. The pair of lines ax2+2hxy+by2 = 0 are parallel if

a) a = -b (b) h2 > ab (c) h2 = ab (d) h2<ab

Solution

Theory:

If h2 ≥ ab, then ax2 + 2hxy + by2 = 0 represent a pair of straight lines passing through the origin.

If h2 < ab, then ax2 + 2hxy + by2 = 0 represents two imaginary lines having real point of intersection, the origin.

If h2 = ab, then ax2 + 2hxy + by2 = 0 represents coincident lines. If h2 > ab, then ax2 + 2hxy + by2 = 0 represents two real and different lines. If θ is an angle between the lines represented by ax2 + 2hxy + by2 = 0, then

Solve:

Putting θ = 0 in any of the above equation yields h2 = ab

Answer: (c) h2 = ab

47. If a, b, c are vectors, then a x (b x c) + b x (c x a) + c x (a x b) is equal to

a) 1 (b) 0 (c) 2 (d) -1

Solution

Answer: (b) 0

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48. Integration of is

a) (b) In

(c) (d) 2 In

Solution: Try differentiating the answer choices

(a)

=

=

Answer: (a) 2 In

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49. Г(m+1) Г(1-m) is equal to

a) π / (sin mπ) (b) mπ / (sin π) (c) mπ / (sin mπ) (d) mπ / [sin (π-m]

Solution

Theory:

Basic Properties of Г(x):

One of the most important formulas satisfied by the Gamma function is

For any x > 0, we know

Solution

Г(m+1) Г(1-m) = m * Гm * Г(1-m) = m * π / (sin mπ) = mπ / (sin mπ)

Answer: (c) mπ / (sin mπ)

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50. If x = 1 is a three times repeated root of the equation x4 – 6x3 +12x2 -10x + 3 = 0, then the fourth root is

a) 3 (b) -3 (c) 31/2 (d) -31/2

Solution

Now, solve with calculator

we get, x=1,1,3.

Answer: (a) 3

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51. If 3/2, 3 + are the two roots of the equation, 2x3 – 15x2 +46x - 42 = 0, then the third root is

a) -3 + (b) 3 - (c) 3 - (d) 3 +

Solution

Complex roots come in pairs.

If 3 + is one root, then the other root is 3 -

Answer: (d) 3-

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52. Three vectors u, v, w are collinear if they are related by au + by + cw = 0 and

(a) a – b +c = 0 (b) a - b – c = 0 (c) a + b + c = 0 (d) a + b + c = 0

Answer: (d) a + b + c = 0

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53. The tangent of the circle x2 + y2 + 2x - 4y – 8 = 0 at (1, -1) is

a) 3x - 2y – 5 = 0 (b) 3x + 2y – 5 = 0 (c) 2x + 3y – 5 = 0 (d) 2x - 3y – 5 = 0

Solution

Center (-1,2), given point (1,-1)

Equation of radius:

Equation of tangent, 2x-3y+k=0

putting (1,-1) in the above equation, we get: 2(1) - 3 (-1) + k = 0

Hence, equation of tangent is:

Answer: (d) 2x-3y-5=0

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54. The probability that A can solve a problem is ¼ and B can solve it is ¾, then the probability that both can solve it together is

a) 1 (b) ½ (c) 3/16 (d) 13/16

Solution

=

=

Answer: (d) 13/16

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55. If y = sce x0, then dy/dx =

a) π/180 sec x0 tan x0 (b) π/180 sec x tan x (c) sec x tan x (d) sec x0 tan x0

Solution

y = sec x0 = sec (πx/180) [in radian]

dy/dx = sec(πx/180) tan(πx/180) * d/dx (πx/180)

dy/dx = sec(πx/180) tan(πx/180) * (π/180)

dy/dx = π/180 sec x tan x

Answer: (b) π/180 sec x tan x

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56. The solution of the differential equation (D2 + 8D + 16)y = 0 is,

a) (A + B)e-4x (b) (A + Bx)e-4x (c) (A + Bx)e4x (d) Acos4e + Bsin4x

Solution

Auxiliary equation:

Answer: (b) (A + Bx)e-4x

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57. Which of the following is incorrect?

(a) (b)

(c) (d)

Solution

Options (a), (c) and (d) are different formulas of integration. Option (b) is not correct.

Answer: (b)

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58. (ax – bx) / x is

a) ln (b/a) (b) ln (a/b) (c) 0 (d) non existent

Solution

Use L’Hospital rule. The given limit is in form.

(differentiating nominator & denominator)

=

=

=

Answer: (b) ln (a/b)

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59. Г(½)

a) (b) (c) (d)

Answer: (b)

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60. cos8x sin6x dx =

a) 5 / 4096 (b) 5π / 4096 (c) π / 4096 (d) 5π / 2048

Solution

Using scientific calculator the given integration yields: 0.003835

Value of option (a) = 0.001221 Value of option (b) = 0.003835

Value of option (c) = 0.000767 Value of option (d) = 0.007669

Answer: (b) 5π /4096

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1 st batch Admission Test Math Solution

Question No. 26

Solution:

Geometrical Progression series:

here, a = 1st term

r = common ratio

n-th term=

4th term =

9th term=

So, and

= = 243

Ans. c) 3

Question No. 27

The coefficient of x2 in the expansion of 10 is

a) 480 b) 360 c) 1024 d) 720 e) 120

Solution:

The general term in the expansion of

(a+x)n is [that is (r+1)th term]

tr+1 = nCr an-r xr

term for expansion of is, tr+1 = 10Cr

= 10Cr

if this term contains r = 2

hence t2+1 = 10C2

coefficient of x2 is 720

Ans. d) 720

Question No. 28

Solution:

Please refer to the van-diagram shown:

here,

b)

Ok. hence it is the answer. You can verify c), d) & e) in similar way.

Ans. .

Question No. 29

Solution:

if

and

if is the angle between

then, =

and =

here, u =

v = v = =

u . v =

again,

Ans. c) 600

Question No. 30

Solution:

General equation of circle is,

Center = (-g, -f)

Radius =

General equation of a line passing through the points (x1, y1) is

Where m is the slope

Perpendicular distance of a line from the point (x’, y’) is =

Equation of circle = 0

……………. (1)

Radius = =

= =

Equation of the line passing through the points (-4, -2) is

……………. (2)

Perpendicular distance of the line (2) from the center of the circle is

line (2) will be tangent to the circle if the perpendicular distance from the center of the circle is equal radius.

equation of tangent (2) is

- -

Not consistent with answer. So the question is wrong.

Question No. 31

Solution:

This is the case of combination. if 2 books are executed always then we have to select 5 books from 10 books.

We know r no. of items can be selectwd from n no. of items

5 books be selected from 10 books in = 252 ways

Ans. C) 252.

Question No. 32

Solution:

[for acute angle, sin A is positive]

Question No. 33

Solution:

Each dice contains the numbers1, 2, 3, 4, 5, and 6.

When two dice are rolled, number of possible outcomes are

the sum of the numbers will be 7 when two dice will give following values.

1 st dice 2 nd dice

1 66 12 55 23 44 3

total 6 results are possible

Ans:

Question No. 34

Solution:

when (5-2x) is positive

when (5-2x) is negative

hence solution is and

Ans. B.

Question No. 35

Solution:

the differentiation of with respect to is

Given,

Ans. b)

Question No. 36

Solution:

Ans: d)

Question No. 37

Solution:

for is called the auxiliary equation

Roots of Auxiliary equation Complete solution

case 1: All roots are real and different

case 2: but other roots are real & different

case 3: (Imaginary roots) or,

1. is a pair of imaginary roots. or,

2. repeated twice

Solve: for choice (b) auxiliary equation is

here

Ans: b)

Question No. 38

Solution:

refer to the theory of previous question.

Auxiliary equation:

Solution is

Ans: c)

Question No. 39

Solution:

Solution: Options (a), (b), (c) and (e) are different formulas of integration.

Option (e) is applicable for f(-x) = f(x)

Option (d) is not correct.

Ans: (d)

Question No. 40

Solution:

if f is a real valued function,

f (x1, x2, x3, …….., xn)

Hessian of f , H

Ans: d) hessian of u

41. The point of intersection of the straight lines represented by the equation

a) (3,-1) b) (-1,2) c) (1,-3) d) (4,2) e) (-3,-1)

Solution:

The point of intersection of two lines are represented in the equation

--------- (i)

The point of intersection will satisfy equation (1)

For point (3, -1) ie. X = 3, y = 1

Left side of equation (1) is

= 8

For (-1,2), LHS = -27 RHS

For (1, -3) LHS = 0 RHS *

For (4,2), LHS = 8 RHS

For (-3,-1), LHS= 20 RHS

Ans : c) (1,-3)

42. the limit of

a) 1 b) -1 c) 2 d) e) does not exist

Solution:

for x = tanx = tan

Putting x = the fraction attains from.

so we can apply L’ Hospital rule

which is applicable for forms.

Ans: a) 1

43. If

a) (1+logx) x b) c)

d) e)

Solution:

let, z = xx

now,

Ans: e)

44. If x =

a) b) c) d) e)

Solution:

=

=

= 2tan

Ans: b)

45. The minimum value of is

a) b) c) d) e)

Solution:

let,

for maximum or minimum value,

Which is greater than 0

Hence, minimum value is obtained by putting

= =

Ans: b)

46. The value of is

a) b) c)

d) + c e)

Solution:

I =

let, x = tan

=

=

=

= -

= -

=

Ans : c)

47. The derived set of the set 1 is

a) 1 b) 1 c) d) e)

Solution:

Answer choices are wrong.

48. [by using calculator]

now,

Answer is

Ans: d)

49. sin5x cos6x dx equals

a) b) c) d) e)

Solution:

by using calculator (keep it in Radian mode).

The solution is c)

Ans: c)

50. The question of the circle which cuts the circles and orthogonally is

a) b)

c) d)

e)

Solution:

Two circles are said to cut orthogonally if angle of intersection of these circles at a point of intersection is a right angle ie. if the tangents to these circles at common point are perpendicular to each other.

Two circles S: and : cut orthogonally if

Let the equation of required circle be …………..(i)

Given,

………….(ii)

and

………….(iii)

if (i) & (ii) cuts orthogonally,

………….(iv)

if (i) & (iii) cuts orthogonally,

…………..(v)

from (iv) & (v) 2g+2f-6g-8=7

………….(vi)

also (iv)

From choice a)

from choice b)

from choice c)

from d)

from e)

now, check equations (iv) and (vi) for the choices.

from choice a) g =

f =

=

hence, choice (a) is the right choice

Ans : a)