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SRMJEEE Maths Mock Test - 9 - JEE MCQ


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30 Questions MCQ Test - SRMJEEE Maths Mock Test - 9

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SRMJEEE Maths Mock Test - 9 - Question 1

Consider the given expression:

The negation of the above expression is

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 1

Here,

Hence, this is the required solution.

SRMJEEE Maths Mock Test - 9 - Question 2

The locus of the point of intersection of two normals to the parabola x2=8y, which are at right angles to each other,is

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 2

Let P(4t1, 2t12) and Q(4t2, 2t22) be the points on x2 = 8y. Equation of the normals at P and Q are 
y - 2t12 = -1/t1(x - 4t12)  ....(1) 
y - 2t22 = -1/t2 (x - 4t22)   ....(2)
Since the normals are at right angles, we have
(-1/t1)(-1/t2) = -1
⇒ t1t2 = -1   ....(3)
Solving Eqs. (1) and (2) and from Eq. (3), we have

⇒ 2y = x2 + 12  which is the required locus.

SRMJEEE Maths Mock Test - 9 - Question 3

If is a unit vector, then is

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 3

We know from the standard result of vector triple product that for any 3 vectors 

Now note that 

SRMJEEE Maths Mock Test - 9 - Question 4

If A and B are any 2 x 2 matrics, then |A + B| = 0 implies

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 4

|A+B| = 0
⇒ A + B = O, where O is a zero matrix.
⇒ A = O − B = −B
Take determinant on both sides
⇒ |A| = −|B|
⇒ |A| + |B| = 0

SRMJEEE Maths Mock Test - 9 - Question 5

The period of the function sin(πx/2)+cos(πx/2) is

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 5

t = T/|a|
  = 2π/π/2
  = 4

SRMJEEE Maths Mock Test - 9 - Question 6

The area common to the parabola y = 2x2 and y = x2 + 4 is

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 6

SRMJEEE Maths Mock Test - 9 - Question 7

If  and f (0) = 300 , then find f(x).

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 7

Given equation is

We know that,
Y = f(x)
So,

So, this is linear differential equation.
Where,

So,
I.F

So, complete equation is

Given that

Putting the value of c in equation (i), we
get,

Thus,

Hence, this is required solution.

SRMJEEE Maths Mock Test - 9 - Question 8

Let f(x) = x3 + 3x2 + 3x + 2. Then, at x = -1

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 8

f(x) = (x+1)3 + 1      
∴ f'(x) = 3(x+1)2
f'(x) = 0  ⇒  x = -1
Now, f" (-1 - ∈) = 3(-∈)2 > 0, f'(-1 + ∈)2 = 3∈2 > 0
∴ f(x) has neither a maximum nor a minimum at x = -1
Let f'(x) = φ ′ (x) = 3(x+1)2    
∴ φ′(x) = 6(x+1).
φ′(x) = 0  ⇒  x = -1
φ ′ (-1-∈) = 6(-∈) < 0, φ ′ (-1-∈) = 6∈ > 0
∴ φ (x) has a minimum at x = -1

SRMJEEE Maths Mock Test - 9 - Question 9

The length of the shadow of a rod inclined at 10o to the vertical towards the sun is 2.05 metres when the elevation of the sun is 38o.The length of the rod is

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 9

SRMJEEE Maths Mock Test - 9 - Question 10

The area of the region bounded by the curve y = x - x2 between x = 0 and x = 1 is

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 10

SRMJEEE Maths Mock Test - 9 - Question 11

A and B are square matrices of order n x n, then (A - B)2 is equal to

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 11

(A - B)2
⇒ (A - B) x (A - B)
⇒ (A - B) x A - (A - B) x B
⇒  A2 - AB - BA + B2

SRMJEEE Maths Mock Test - 9 - Question 12

The equation of the line parallel to the tangent to the circle x2 + y2 = r2 at the point (x₁, y₁) and passing thro' origin is

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 12

Correct Answer :- D

Explanation : mm1 = -1

(y1/x1)*m1 = -1

m1 = - x1/y1

So, equation of line passing through (x₁, y₁) 

y = m1x

y = (-x1/y1)*x

yy1+ xx1 = 0

SRMJEEE Maths Mock Test - 9 - Question 13

If x=a[(cost)+(log tan(t/2))], y=a sint, (dy/dx)=

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 13

SRMJEEE Maths Mock Test - 9 - Question 14

If equation λx+ 2y- 5xy + 5x - 7y + 3 = 0, represents two straight lines, the value of λ is

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SRMJEEE Maths Mock Test - 9 - Question 15

3nC₀-8nC₁+13nC₂-18nC₃+...+ to (n+1) terms=

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 15

We need to evaluate the expression:

  • The expression involves a series of binomial coefficients: 3nC₀, 8nC₁, 13nC₂, and so on.
  • The pattern alternates in sign, beginning with a positive term.
  • Each term is derived from the binomial coefficient multiplied by a coefficient that follows the pattern of increasing integers.
  • Upon analysis, the series can be represented as a polynomial of degree n.
  • Using the identity for binomial coefficients, the sum evaluates to zero when summed over the complete set of coefficients.

Conclusion: The entire sum results in 0.

SRMJEEE Maths Mock Test - 9 - Question 16
The lines 2x-3y=5 and 3x-4y=7 are diameters of a circle with an area of 154 square units. What is the equation of this circle?
Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 16

To find the equation of the circle:

  • The given lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of the circle.
  • First, find the intersection point of these lines, which is the centre of the circle.
  • Solve the equations:
    • 2x - 3y = 5
    • 3x - 4y = 7
  • Using simultaneous equations, solve to find the intersection point.
  • The centre is (1, -1).

The area of the circle is 154 square units.

  • Area formula: πr² = 154
  • Calculate the radius: r² = 154/π ≈ 49
  • The radius r is 7.

Thus, the equation of the circle with centre (1, -1) and radius 7 is:

  • (x - 1)² + (y + 1)² = 49

Convert to standard form:

  • Expand: x² - 2x + 1 + y² + 2y + 1 = 49
  • Simplify: x² + y² - 2x + 2y = 47

The correct option is C.

SRMJEEE Maths Mock Test - 9 - Question 17

The number of ways in which we can put letters of the word PERSON in the squares of the Fig so that no row remains empty is

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 17

R3 cannot remain empty.
Total ways = 8C6×6!=20160
When R1 is empty =6!=720
When R2 is empty =6!=720
∴ When no is empty =20160−720×2=18,720

SRMJEEE Maths Mock Test - 9 - Question 18
What is the value of the series 1/2! - 1/3! + 1/4! - 1/5! + .......?
Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 18

The series given is an alternating series:

  • 1/2! - 1/3! + 1/4! - 1/5! + ...

This resembles the expansion of ex when x = -1, which is:

  • e-1 = 1 - 1/1! + 1/2! - 1/3! + 1/4! - ...

By comparing both, the given series is part of the expansion for e-1:

  • It starts from the 1/2! term, skipping the initial terms.

Thus, the series converges to e-1.

SRMJEEE Maths Mock Test - 9 - Question 19

How many 5 digit telephone numbers can be constructed using the digits 0 to 9, if each number starts with 67 and no digit appears more than once?

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 19

Since, telephone number start with 67, so two digits is already fixed
Now, we have to arrangement of three digits from remaining eight digits (i.e., 0,1,2,3,4,5,8,9)

= 8 × 7 × 6
= 336 ways

SRMJEEE Maths Mock Test - 9 - Question 20

If the co-ordinates of the points A,B,C be (−1, 3, 2), (2, 3, 5) and (3, 5,−2)  respectively, then ∠A=

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 20

Equation of AB is 
and that of AC is 
Hence 

SRMJEEE Maths Mock Test - 9 - Question 21

The orthocentre of the triangle with vertices (−2,−6), (−2,4) and (1,3) is

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 21

Let A (−2,−6), B(–2,4) and C(1,3) are the vertices of the △ABC
Now using the distance formula, we get 

(AB)2 = (BC)2 + (CA)2
so the △ABC is a right angle triangle, right angle at C and we know that in right-angle triangle orthocenter is the point where right angle formed,
Therefore orthocenter is C(1, 3)

SRMJEEE Maths Mock Test - 9 - Question 22

Find the value of 

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 22

Given,

Using by parts rule,

Then, we get,

Thus,

Hence, this is required solution.

SRMJEEE Maths Mock Test - 9 - Question 23


then  = f '(1)

SRMJEEE Maths Mock Test - 9 - Question 24

If two vectors are parallel, then find the value of λ.

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 24

Given that,
 that are parallel to each other.
If and if these vectors are parallel to each other, then they must follow, 

SRMJEEE Maths Mock Test - 9 - Question 25

If then angle between the vectors andis

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 25


⇒ sinθ = cosθ
⇒ θ = 45°

SRMJEEE Maths Mock Test - 9 - Question 26

Solution of the differential equation 

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 26

SRMJEEE Maths Mock Test - 9 - Question 27

Evaluate: 

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 27

Let


Thus,

Hence, this is required solution

SRMJEEE Maths Mock Test - 9 - Question 28

The value of b such that the scalar product of the vector î+ĵ+k̂ with the unit vector parallel to the sum of the vectors 2î + 4ĵ - 5k̂ and bî+ 2ĵ + 3k̂ is one is

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 28

 Parallel vector =(2+b)i+6j−2k

Unit vector = (2+b)i+6j−2k / (b2+4b+44)1/2

According to the question, 

1 = (2+b)+6−2/b2+4b+44

⇒b2+4b+44=b2+12b+36

⇒8b=8⇒b=1

SRMJEEE Maths Mock Test - 9 - Question 29

Four normal dice are rolled once. The number of possible outcomes in which at least one die shows up 2 is -

Detailed Solution for SRMJEEE Maths Mock Test - 9 - Question 29

Total number of outcomes when four normal dice are rolled
= 6 × 6 × 6 × 6 = 6= 1296
Total number of ways in which no dice shows up 2 i.e.
Each of the four dice shows up 1,3,4,5 or  6 as outcomes
=5 × 5 × 5 × 5 = 5= 625
Hence total number of possible outcomes when no dice shows up 2
=1296 − 625 = 671

SRMJEEE Maths Mock Test - 9 - Question 30


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