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WBJEE Maths Test - 6 - JEE MCQ


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30 Questions MCQ Test - WBJEE Maths Test - 6

WBJEE Maths Test - 6 for JEE 2025 is part of JEE preparation. The WBJEE Maths Test - 6 questions and answers have been prepared according to the JEE exam syllabus.The WBJEE Maths Test - 6 MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for WBJEE Maths Test - 6 below.
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WBJEE Maths Test - 6 - Question 1

The approx value of (7.995)13 correct to four decimal places is

WBJEE Maths Test - 6 - Question 2

The value of (7.995)1/3 correct to four decimal places is

WBJEE Maths Test - 6 - Question 3

If in the expansion of (1 + x)20, the coefficients of rth and (r + 4)th terms are equal, then r is

WBJEE Maths Test - 6 - Question 4

The coordinates of the point from which equal tangents are drawn to the circle x2+y2=1, x2+y2+8x+15=0 and x2+y2+10y+24=0 are

WBJEE Maths Test - 6 - Question 5

(1 − ω + ω2)7 + (1 + ω − ω2)7 =

WBJEE Maths Test - 6 - Question 6

Circles x2+y2+2gx+2fy=0 and x2+y2+2g'x+2f'y=0 touch externally, if

WBJEE Maths Test - 6 - Question 7
The reciprocal of 3+√7i is
WBJEE Maths Test - 6 - Question 8
The area bounded by the curve y=sinx, y=0, x=0 and x=(π/2) is
WBJEE Maths Test - 6 - Question 9

is equal to

WBJEE Maths Test - 6 - Question 10

WBJEE Maths Test - 6 - Question 11

If a ≠ 6, b, c satisfy = 0 then abc =

WBJEE Maths Test - 6 - Question 12

WBJEE Maths Test - 6 - Question 13

The differential equation dy/dx+sin 2y=x3cos2y is reduced to linear from dv/dx+Pv=Q where P and Q are function of x alone, by changing the variable as :

WBJEE Maths Test - 6 - Question 14

The derivative of is

WBJEE Maths Test - 6 - Question 15

If f (x)   = (x − x0) φ (x) and φ (x) is continuous at x = x0 , then f ′(x) 0 is equal to

WBJEE Maths Test - 6 - Question 16

The eccentricity of the ellipse 9x2 + 5y2 − 30y = 0 is

WBJEE Maths Test - 6 - Question 17

The roots of the equation = 0 are

Detailed Solution for WBJEE Maths Test - 6 - Question 17


WBJEE Maths Test - 6 - Question 18

The solution of differential equation

WBJEE Maths Test - 6 - Question 19

The sum of distances of any point on the ellipse 3x2 + 4y2 = 24 from its foci is

WBJEE Maths Test - 6 - Question 20

The locus of the middle points of chords of the hyperbola 3x2 − 2y2 + 4x − 6y = 0 parallel to y = 2x is

WBJEE Maths Test - 6 - Question 21

The eccentricity of the conic 9x2 - 16y= 144 is

WBJEE Maths Test - 6 - Question 22

If f(x) = x3 + 4x2 + λx + 1 is a monotonically decreasing function of x in the largest possible interval (- 2, -2/3) then

Detailed Solution for WBJEE Maths Test - 6 - Question 22

WBJEE Maths Test - 6 - Question 23

If cos⁻1x+cos⁻1y+cos⁻1z = 3π then xy + yz + zx is equal to

WBJEE Maths Test - 6 - Question 24

The maximum area of the rectangle that can be inscribed in a circle of radius r is

WBJEE Maths Test - 6 - Question 25

If |z₁|=|z₂| and amp. z₁+amp.z₂=0, then

WBJEE Maths Test - 6 - Question 26

If A and B are any 2 x 2 matrics, then det (A + B) = 0 implies

WBJEE Maths Test - 6 - Question 27

Detailed Solution for WBJEE Maths Test - 6 - Question 27

L=lim(n->∞) n(13+23+…….n3)2/12+22+…..n2)3=?

Prerequisite r. rΣn =13+23+…..n3=n2(n+1)2/4

                       r=1Σn r2=12+22+…..n2=n[(n+1)(2n+1)]/6

L=lim(n->∞) n[(n2(n+1)2)]/[(n+1)(2n+1)/6]3

L=lim(n->∞)[n2(n+1)/2]/[(2n+1)3/27]

=(1+1/∞)/(1+1/2n∞)3 x (27/16)

=27/16

lim(n->∞) n(13+23+…….n3)2/12+22+…..n2)3=27/16

WBJEE Maths Test - 6 - Question 28

The parabola with directrix x + 2y − 1 = 0 and focus (1, 0) is

WBJEE Maths Test - 6 - Question 29

The line y=mx+c touches the parabola x2=4ay if

WBJEE Maths Test - 6 - Question 30

How many total words can be formed from the letters of the word INSURANCE in which vowels are always together?

Detailed Solution for WBJEE Maths Test - 6 - Question 30
  1. Identify the letters
    The word “INSURANCE” consists of 9 letters in total: I, N, S, U, R, A, N, C, E.

  2. Identify the vowels and consonants
    The vowels in “INSURANCE” are I, U, A, E (4 vowels). The consonants are N, S, R, N, C (5 consonants).

  3. Treat vowels as a single unit
    To ensure that the vowels are never separated, we treat the group of vowels (I, U, A, E) as one single unit. Thus we have the “vowel‐block” plus N, S, R, N, C.

  4. Count the arrangements of the units
    We have 6 units in all: (vowel‐block), N, S, R, N, C.
    Since N is repeated twice, the number of ways to arrange these 6 units is 6! divided by 2!:
    Arrangements = 6! ÷ 2! = 720 ÷ 2 = 360

  5. Count the arrangements of the vowels
    Inside the vowel‐block, the 4 vowels can be arranged in 4! ways = 24

  6. Total arrangements
    Multiply the two results: 360 × 24 = 8640

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