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MCQ Practice Test & Solutions: WBJEE Maths Test - 10 (75 Questions)

You can prepare effectively for JEE WBJEE Sample Papers, Section Wise & Full Mock Tests 2026 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "WBJEE Maths Test - 10". These 75 questions have been designed by the experts with the latest curriculum of JEE 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 120 minutes
  • - Number of Questions: 75

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WBJEE Maths Test - 10 - Question 1

In the expansion of [7 1 3 + 11 1 9 ]6561, the number of terms free from radicals is

Detailed Solution: Question 1

General term,
tr + 1 = 6564Cr (7 1 3 )6561-r (11 1 9 )r
⇒ tr + 1 = 6561Cr . 7( 6561-r 3 ) . 11( r 9 )
tr + 1
will be free from radical signs, if
( 6561-r 3 ) and ( r 9 )
are integers for
0 ≤ r ≤ 6561
∴ r = 0, 9, 18, 27, ...6561
0, 9, 18, ...6561 is in A.P.
First term, a = 0
Common - difference = 18 - 9 = 9
Last term, t = 6561
l = a + (n - 1)d, where n = number of terms
⇒ 6561 = 0 + (n - 1)9
⇒ 9(n - 1) = 6561
⇒ n - 1 = 729
⇒ n = 729 + 1
⇒ n = 730

WBJEE Maths Test - 10 - Question 2

If the sum of the coefficients in the expansion of (x+y)n is 1024, then the value of the greatest coefficient in the the expansion is

WBJEE Maths Test - 10 - Question 3

The equation of the normal to the circle x2 + y2 + 6x + 4y - 3 = 0 at (1, -2) is

WBJEE Maths Test - 10 - Question 4

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

WBJEE Maths Test - 10 - Question 5

WBJEE Maths Test - 10 - Question 6

Two circles x2 + y2 - 2x + 6y + 6 = 0 and x2 + y2 - 5x + 6y + 15 = 0

WBJEE Maths Test - 10 - Question 7

If |z₁ + z₂| = |z₁ - z₂|, then the difference of arguments of z₁ and z₂ is

WBJEE Maths Test - 10 - Question 8

(1 − ω + ω 2 ) (1 − ω 2 + ω 4 ) (1 − ω 4 + ω 8 ) (1 − ω 8 + ω 16)    =

WBJEE Maths Test - 10 - Question 9

The area of the curve |x|+|y|=4 is

WBJEE Maths Test - 10 - Question 10

WBJEE Maths Test - 10 - Question 11

WBJEE Maths Test - 10 - Question 12

WBJEE Maths Test - 10 - Question 13

The singular solution of the differential equation y=px+p3, (p=dy/dx) is :

WBJEE Maths Test - 10 - Question 14

WBJEE Maths Test - 10 - Question 15

The solution of the differential equation

Detailed Solution: Question 15

WBJEE Maths Test - 10 - Question 16

( d dx )(x2 + cos x)4 =

Detailed Solution: Question 16

d dx

(x2 + cos x)4
= 4.(x2 + cos x)3 x d dx (x2 + cos x)
= 4(x2 + cos x)3 x (2x - sin x)

WBJEE Maths Test - 10 - Question 17

(d/dx)[tan⁻1(secx+tanx)]=

WBJEE Maths Test - 10 - Question 18

The radius of the circle passing through the foci of the ellipse ((x2/16) + (y2/9) = 1), and having its centre (0,3) is

WBJEE Maths Test - 10 - Question 19

The equation of the ellipse in the form of ((x2/a2) + (y2/b2) = 1), given the eccentricity to be 2/3 and latus rectum 2/3, is

WBJEE Maths Test - 10 - Question 20

The tangents to the hyperbola x2 - y2 = 3 are parallel to the st. line 2x + y + 8 = 0 at the following points

WBJEE Maths Test - 10 - Question 21

The st. line lx + my + n = 0 touches the hyperbola x2/a2- y2/b2 = 1 if

WBJEE Maths Test - 10 - Question 22

The function f(x) = sin 4x + cos 4x increases if

Detailed Solution: Question 22

f ′ x > ⇒ 4 sin 3 x . cos x − 4 cos 3 x . sin x > 0


⇒ sin x . cos x sin 2 x − cos 2 x > 0
o r sin 2 x . cos 2 x < 0 o r sin 4 x < 0
∴ π < 4 x < 2 π o r 3 π < 4 x < 4 π
∴ π 4 < x < π 2 o r 3 π 4 < x < π .

WBJEE Maths Test - 10 - Question 23

WBJEE Maths Test - 10 - Question 24

tan⁻1(x/y) - tan⁻1 (x - y/x + y) is

WBJEE Maths Test - 10 - Question 25

If A and B are two square matrices such that B = -A⁻1 BA, then (A + B)2 =

WBJEE Maths Test - 10 - Question 26

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

Detailed Solution: Question 26

WBJEE Maths Test - 10 - Question 27

|(1)/(2 + i)2 −(1)/(2 − i)2 | =

WBJEE Maths Test - 10 - Question 28

The focus of the parabola x2-8x+2y+7=0 is

WBJEE Maths Test - 10 - Question 29

Number of divisors of n = 38808 (except 1 and n) is

Detailed Solution: Question 29

Factorizing the given number, we have
38808 = 23 . 32 . 72 . 11
Therefore the total number of divisors
= (3 + 1) (2 + 1) (1 + 1) - 1 = 71
But this includes the division by the number itself
Hence, the required number of divisors
= 71 - 1 = 70

WBJEE Maths Test - 10 - Question 30

If nP4:nP3=1:2, then n=

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