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SRMJEEE Maths Mock Test - 8

35 Questions MCQ Test SRMJEEE Subject Wise & Full Length Mock Tests | SRMJEEE Maths Mock Test - 8

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Attempt SRMJEEE Maths Mock Test - 8 | 35 questions in 50 minutes | Mock test for JEE preparation | Free important questions MCQ to study SRMJEEE Subject Wise & Full Length Mock Tests for JEE Exam | Download free PDF with solutions

QUESTION: 1

y = {x(x - 3)}² increases for all values of x lying in the interval

A.
0 < x < 3/2
B.
0 < x < ∞
C.
- ∞ < x < 0
D.
1 < x < 3

Solution:

Since y = {x(x - 3)}²
∴ dy dx = 2x (x - 3) (2x - 3)
For increasing function dy/dx > 0
or 2x (x - 3) (2x - 3) > 0
or x(x - 3) (x - 3/2) > 0
∴ x∈(0, 3/2) ∪ (3, ∞)

QUESTION: 2

The area bounded by the x-axis and the curve y = 4x - x² - 3, is

A.
4/3
B.
3/4
C.
7
D.
3/2

Solution:

QUESTION: 3

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

A.
356
B.
252
C.
210
D.
120

Solution:

QUESTION: 4

If the equation 3x² + xy - y² - 3x + 6y + K = 0 represents pair of lines, then the value of K is

A.
9
B.
1
C.
- 9
D.
0

Solution:

QUESTION: 5

A.
√2
B.
2√2
C.
3√2
D.
3√3

Solution:

QUESTION: 6

The equation of a circle with centre (4, 1) and having 3x + 4y -1 = 0 as tangent is

A.
x² + y² - 8x - 2y - 8 = 0
B.
x² + y² - 8x - 2y + 8 = 0
C.
x² + y² - 8x + 2y + 8 = 0
D.
x² + y² - 8x -2y + 4 = 0

Solution:

QUESTION: 7

A.
- 2abc
B.
abc
C.
0
D.
a² + b² + c²

Solution:

QUESTION: 8

If |(z-1)/(z-i)| = 1, then the locus of z is

A.
a circle
B.
an ellipse
C.
a straight line
D.
a parabola

Solution:

QUESTION: 9

The focus of the parabola y²-x-2y+2=0 is

A.
[(1/4),0]
B.
(1,2)
C.
[(3/4),1]
D.
[(5/4),1]

Solution:

QUESTION: 10

If y=a cos (log x) + b sin (log x) where a,b are parametres, then x²y+xy'

A.
y
B.
-y
C.
2y
D.
-2y

Solution:

QUESTION: 11

The derivative of

Solution:

QUESTION: 12

∫[(1)/(x-x²)]dx=

A.
logx-log(1-x)+c
B.
log(1-x²)+c
C.
-logx+log(1-x)+c
D.
log(x-x²)+c

Solution:

QUESTION: 13

Solution:

QUESTION: 14

4tan⁻¹(1/5)-tan⁻¹(1/239)=

A.
π
B.
π/2
C.
π/3
D.
π/4

Solution:

QUESTION: 15

What is the value of

A.
1
B.
-1
C.
∞
D.
limit does not exist

Solution:

QUESTION: 16

If f : ℝ → ℝ is defined by

then f is continuous on the set

A.
ℝ
B.
ℝ − { − 2 }
C.
ℝ − { − 1 }
D.
ℝ − { − 1, − 2 }

Solution:

QUESTION: 17

If A is a non-singular, then adj A is

A.
Non-singular
B.
Singular
C.
Symmetric
D.
Undefined

Solution:

QUESTION: 18

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

A.
f(x) has a maximum
B.
f(x) has a minimum
C.
f'(x) has a maximum
D.
f'(x) has a minimum

Solution:

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

QUESTION: 19

Can a quartile, a decile and a percentile be the median?

A.
Only quartile but not decile and percentile
B.
Quartile and decile but not percentile
C.
Decile and percentile but not quartile
D.
Quartile, decile and percentile, all the three

Solution:

QUESTION: 20

The gap between the highest and the lowest score is called

A.
range
B.
class interval
C.
length of class interval
D.
class mark

Solution:

QUESTION: 21

The number of 4-digit even numbers that can be formed using 0, 1, 2, 3, 4, 5, 6 without repetition is

A.
120
B.
300
C.
420
D.
20

Solution:

Each even number must have 0, 2, 4 or 6 in is units place
Here total number of digits = 7

When 0 occurs at units place there is no restriction on other places and when 2 or 4 occurs at units place there is restriction on thousands' place as 0 can not be put at thousands' place
Case I When 0 occurs at units place:

∴ The number of numbers formed in this case

Case II. When 0 does not occur at units place:

The units' place can be filled up by any one of the three digits 2, 4 and 6 in 3 ways
∴ The number of numbers formed in this case

∴ The required number = 120 + 300 = 420

QUESTION: 22

The base of triangle is passing through a fixed point (a,b) and its sides are bisected at right angles by lines, y² - 4xy - 5x² = 0, then the locus of the vertex of triangle is

A.
2x² + 2y² + (3a + 2b)x + (2a - 3b)y = 0
B.
2x² + 2y² + (3a + 2b)x + (2a + 3b)y = 0
C.
2x² - 2h² + (3a + 2b)x + (2a + 3b)y = 0
D.
None of these

Solution:

y² - 4xy - 5x² = 0
⇒ (y - 5x) (y + x) = 0
y - 5x, cuts AB at M at right angle
and y + x = 0 cuts AC at N at right angle

By collinearity of (x₁, y₁), (x₂, y₂), (a, b) we get h = x, k = y
2x² + 2y² + (3a + 2b)x + (2a - 3b)y = 0