Mathematics Test 200

Result

Mathematics Test 200

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Weekly Quiz Competition

Question 1
4 / -1

Let f(x) be a one-to-one function such that f(1) = 3, f(3) = 1, f '(1) = – 4 and f '(3) = 2. If g = f ^–1, then the slope of the tangent line to 1/g at x = 1 is

A
1/√2

B
-1/9

C
-1/18

D
1/32

Solution
Question 2
4 / -1

A
6

B
12

C
18

D
24

Solution
Question 3
4 / -1

If g (x³ + 1) = x⁶ + x³ + 2, then the value of g(x² – 1) is

A
x⁴ – 3x² + 3

B
x⁴ + x² + 4

C
x⁴ – 3x² + 4

D
x⁴ + x² + 2

Solution

g(x³ + 1) = x⁶ + x³ + 2 = (x³ + 1)2 – x³ + 1

= (x³+ 1)²– (x³+ 1 – 1) + 1 = (x³+ 1)²– (x³+ 1) + 2

Put x³ + 1 = t

So, g(t) = t² – t + 2

⇒ g(x² – 1) = (x² – 1)² – (x² – 1) + 2

= x⁴ – 3x² + 4.
Question 4
4 / -1

Suppose that f (0) = 0 and f ' (0) = 2, and let g (x) = f (- x + f (f (x))). The value of g ' (0) is equal to

A
0

B
1

C
6

D
8

Solution

g (x) = f (- x + f (f (x))) ;

f (0) = 0; f ' (0) = 2

g ' (x) = f ' (- x + f (f ( x )))· [- 1 + f ' (f (x))· f ' (x )]

g ' (0) = f ' (f (0))· [- 1 + f '(0) · f '(0)]

= f ' (0) [- 1 + (2)(2)]

= (2) (3) = 6
Question 5
4 / -1

The value of the definite integral,

A

B

C

D

Solution
Question 6
4 / -1

A line L is perpendicular to the curve at its point P and passes through (10, –1). The coordinates of the point P are

A
(2, –1)

B
(6, 7)

C
(0, –2)

D
(4, 2)

Solution
Question 7
4 / -1

then the sum of the square of reciprocal of all the values of x where f(x) is non-differentiable, is equal to

A
1

B
81

C
82

D
82/81

Solution

Clearly f(x) is non differentiable at x = 1/9, 1

∴ sum of squares of reciprocals

= 9² + 1 = 82 Ans.
Question 8
4 / -1

h(x) = {x},k(x) = 5^{log₂(x + 3)} then in [0, 1], Lagranges Mean Value Theorem is NOT applicable to

[Note : where [x] and {x} denote the greatest integer and fractional part function of x respectively]

A
f, g, h

B
h, k

C
f, g

D
g, h, k

Solution

f is not differentiable at x = 1/2

g is not continuous in [0, 1] at x = 0 & 1 h is not continuous in [0, 1] at x = 1

k (x) = (x + 3)^ln₂5= (x + 3)^p where 2 < p < 3
Question 9
4 / -1

If the function f (x) = ax e^–bx has a local maximum at the point (2, 10), then

A
a = 5; b = 0

B
a = 5e, b = 1/2

C
a = 5e², b = 1

D
none

Solution

f (2) = 10, hence 2ae^–2b = 10

⇒ ae^–2b = 5 ....(1)

f ' (x) = a [e^–bx – bx e^–bx] = 0

f ' (2) = 0

a(e^–2b – 2be^–2b) = 0

ae^–2b (1 – 2b) = 0

⇒ b = 1/2 or a = 0 (rejected)

from (1) if b = 1/2; a = 5e

∴ a = 5e and b = 1/2 Ans.]
Question 10
4 / -1

then the value of x satisfying the equation f (x, 10) = f (x, 11), is

A
9

B
10

C
11

D
none

Solution