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SRMJEEE Maths Mock Test - 2 - JEE MCQ


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30 Questions MCQ Test - SRMJEEE Maths Mock Test - 2

SRMJEEE Maths Mock Test - 2 for JEE 2025 is part of JEE preparation. The SRMJEEE Maths Mock Test - 2 questions and answers have been prepared according to the JEE exam syllabus.The SRMJEEE Maths Mock Test - 2 MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for SRMJEEE Maths Mock Test - 2 below.
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SRMJEEE Maths Mock Test - 2 - Question 1
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If the area of a Δ A B C be λ then a2 sin 2B + b2 sin 2A is equal to

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SRMJEEE Maths Mock Test - 2 - Question 2
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The radius of the circle in which the sphere x2 + y2 + z2 + 2x - 2y - 4z - 19 = 0 is cut by the plane x + 2y + 2z + 7 = 0, is

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Radius of sphere is=5 then perpen dicular dice Frome plane=4 now radius of circle r,sqr=25_16=9so r=3
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SRMJEEE Maths Mock Test - 2 - Question 3
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The pole of the line lx+my+n=0 w.r.t. the parabloa y2 =4ax

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SRMJEEE Maths Mock Test - 2 - Question 4
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The difference of an integer and its cube is divisible by
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Solution:


  • Let the integer be x.

  • The difference of the integer and its cube is x - x^3.

  • For this difference to be divisible by a number, the number should divide the difference without leaving a remainder.

  • We need to find a number which divides x - x^3 without leaving a remainder.

  • Factoring out x from x - x^3 gives x(1 - x^2).

  • Further factoring 1 - x^2 gives (1 - x)(1 + x).

  • Therefore, the difference x - x^3 = x(1 - x)(1 + x).

  • The number that divides x - x^3 without leaving a remainder is the common factor of x, 1 - x, and 1 + x.

  • The common factor of x, 1 - x, and 1 + x is 1.

  • Therefore, the difference of an integer and its cube is divisible by 1, which means it is divisible by any integer including 4, 6, 10, and 9.

  • However, the smallest number among the given options that divides x - x^3 without leaving a remainder is 6.

  • Hence, the correct answer is B: 6.


    •  
SRMJEEE Maths Mock Test - 2 - Question 5
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If A+B+C=180o, then [(tanA+tanB+tanC)/(tanA tanB tanC)]=
Detailed Solution for SRMJEEE Maths Mock Test - 2 - Question 5

Given, A + B + C = 180
So, A + B = 180 - C
Taking tan on both sides we get,
⇒ tan(A+B) = tan(180-C)
⇒ 
⇒ tanA + tanB = -tanC(1 - tanA tanB)
⇒ tanA + tanB = - tanC + tanA tanB tanC
⇒ tanA + tanB + tanC = tanA tanB tanC

SRMJEEE Maths Mock Test - 2 - Question 6
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The area (in square units) of the region enclosed by the curves y = x2 and y = x3 is
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By solving equations 1and 2 we can get 0,1as values for x which can be the limits and by subtracting y=x^2 from x^3 and integrating the result within the limits 0 to 1 we can get area = 1\12
SRMJEEE Maths Mock Test - 2 - Question 7
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In a Δ A B C , a = 1 and the perimeter is six times the AM of the sines of the angles. The measure of ∠ A is
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SRMJEEE Maths Mock Test - 2 - Question 8
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if A is a 3 x 3 matrix and B is its adjoint matrix. If ∣B∣ = 64, then ∣A∣ =
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SRMJEEE Maths Mock Test - 2 - Question 9
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If and are two square matrices such that , then

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SRMJEEE Maths Mock Test - 2 - Question 10
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If f(x) = {2x - 3, x ≤ 2} then f(1) is equal to
Detailed Solution for SRMJEEE Maths Mock Test - 2 - Question 10

To find f(1) from the function f(x) = 2x - 3 when x ≤ 2, follow these steps:

  • Substitute x = 1 into the function: f(1) = 2(1) - 3.
  • Calculate the value: 2 - 3 = -1.

Now, calculate f(2):

  • Use the same function with x = 2: f(2) = 2(2) - 3.
  • Compute the result: 4 - 3 = 1.

We found f(1) = -1 and f(2) = 1.

Therefore, f(1) is equal to -f(2).

SRMJEEE Maths Mock Test - 2 - Question 11
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If x dy = y(dx + y dy), y > 0 and y (1) = 1, then y (-3) is equal to
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SRMJEEE Maths Mock Test - 2 - Question 12
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Equation 2x2+7xy+3y2+8x+14y+λ=0 represents a pair of straight lines, value of λ is
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SRMJEEE Maths Mock Test - 2 - Question 13
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In the expansion of (y1/6 - y-1/3)9 the term independent of y is :
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SRMJEEE Maths Mock Test - 2 - Question 14
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If then a and b are
SRMJEEE Maths Mock Test - 2 - Question 15
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The points with position vectors 10î + 3ĵ, 12î - 5ĵ and aî + 11ĵ are collinear if a =
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Given, A = (10i + 3j​)
B = (12i - 5j)​
C = (ai + 11j​)
AB = 2i - 8j​
AC = (a - 10)i + 8j​
AB and AC are collinear
⇒ 2/(a - 10) = -8/8
⇒ 2 = 10 - a
⇒ a = 8

SRMJEEE Maths Mock Test - 2 - Question 16
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SRMJEEE Maths Mock Test - 2 - Question 17
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If  then the value of a + c is equal to
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Comparing the elements we get c + 1 = 0 ⇒ c = −1 and a−1=0⇒a=1

SRMJEEE Maths Mock Test - 2 - Question 18
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A committee consists of 9 experts from three institutions A, B and C, of which 2 are from A, 3 from B and 4 from C. If three experts resign, then the probability that they belong to different institutions is
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SRMJEEE Maths Mock Test - 2 - Question 19
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The number of vectors of unit length perpendicular to vectors and is
SRMJEEE Maths Mock Test - 2 - Question 20
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If then is
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Since for all i.
for all
SRMJEEE Maths Mock Test - 2 - Question 21
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The sum of an infinite G.P. is 3. The sum of the series formed by squaring its terms is also 3. The series is
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SRMJEEE Maths Mock Test - 2 - Question 22
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If the two circles 2x2 + 2y2 -3x + 6y + k = 0 and x2 + y2 - 4x + 10y + 16 = 0 cut orthogonally, then the value of k is
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SRMJEEE Maths Mock Test - 2 - Question 23
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The equation line passing through the point P(1,2) whose portion cut by axes is bisected at P, is
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SRMJEEE Maths Mock Test - 2 - Question 24
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The strength of a beam varies as the product of its breadth b and square of its depth d. A beam cut out of a circular log of radius r would be strong when
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SRMJEEE Maths Mock Test - 2 - Question 25
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The equation of the common tangent to the curves y2=8x and xy=-1 is :
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SRMJEEE Maths Mock Test - 2 - Question 26
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If a matrix is such that
Then what is equal to?
Detailed Solution for SRMJEEE Maths Mock Test - 2 - Question 26
Let A be a matrix such that
Post multiply by on both the sides, we get

SRMJEEE Maths Mock Test - 2 - Question 27
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If α, β are the roots of the equation x2- 2x + 2 = 0, then the value of α2 + β2 is
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α + β = 2
αβ = 2
(α + β)2 = 22
α2 + β2 +2αβ = 4
α2 + β2 +2(2) = 4
α2 + β2 + 4 = 4
α2 + β2 = 0

SRMJEEE Maths Mock Test - 2 - Question 28
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The order and degree of differential equation √(dy/dx) - 4 (dy/ dx) - 7x = 0 are
Detailed Solution for SRMJEEE Maths Mock Test - 2 - Question 28

√(dy/ dx)- 4 (dy/dx) -7x = 0
or √(dy/ dx) = 4 (dy/dx) + 7x
Squaring both side
dy/ dx = [4 (dy/dx ) + 7x]2
or dy dx = 16 ( dy/dx )2 + 49x2 + 56 dy/ dx x
∴ order = 1
degree = 2

SRMJEEE Maths Mock Test - 2 - Question 29
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How many total words can be formed from the letters of the word INSURANCE in which vowels are always together?
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SRMJEEE Maths Mock Test - 2 - Question 30
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The 5th term of a G.P. is 2, then the product of its first 9 term is
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