Mathematics Test 134

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Mathematics Test 134

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Question 1
4 / -1

There are two balls in an urn whose colours are not known (each ball can be either white or black). A white ball is put into the urn. A ball is drawn from the urn. The probability that it is white is

A
1/4

B
1/3

C
2/3

D
1/6

Solution

Let E_i(0 < i < 2) denote the event that urn contains 'i' white and '(2 – i)' black balls.
Let A denote the event that a white ball is drawn from the urn.
We have P(E_i) = 1/3 for i = 0, 1, 2
P (A|E₁) = 1/3, P(A|E₂) = 2/3, P(A|E₃) = 1.
By the total probability rule,
P(A)= P(E₁)P(A|E₁) + P(E₂)P(A|E₂) + P(E₃)P(A|E₃)
Question 2
4 / -1

The plane passing through the point (−2, −2, 2) and containing the line joining the points
(1, 1, 1) and (1, −1, 2) makes intercepts on the coordinates axes then sum of the lengths of intercepts is

A
3

B
4

C
6

D
12

Solution
Question 3
4 / -1

If a, b, c are in GP and are in AP, then a, b, c are the lengths of the sides of a triangle which is

A
acute angled

B
obtuse angled with ∠A being obtuse

C
right angled with ∠A being 90 °

D
equilateral

Solution

a, b, c are in GP b² = ac
are in AP.
2(log 2b – log 3c) = log a – log 2b + log 3c – log a
log 2b = log 3c 2b = 3c
∴ b² = ac & 2b = 3c
b = 2a/3 and c = 4a/9
Since (a+b) = 5a/3 > c, (b+c) = 10a/9 > a and
(c + a) = 13a/9 > b, therefore, a, b, c are the sides
of a triangle. Also, as ‘a’ is the greatest side, let us
find angle A of ΔABC

Hence ΔABC is an obtuse angled triangle.
Question 4
4 / -1

Sum of the series
1 + 3 + 6 + 10 + 15 + ……………………….n terms is

A

B

C

D

Solution
Question 5
4 / -1

The solution of the equation (2x + y + 1) dx + (4x + 2y – 1) dy = 0 is

A
log (2x + y – 1) = C + x + y

B
log (4x + 2y – 1) = C + 2x + y

C
log (2x + y + 1) + x + y = C

D
log (2x + y – 1) + x + 2y = C

Solution

Put 2x + y = X ⇒ Therefore, the given equation is reduced to

2(2x + y) + log (2x + y – 1) = 3x + constant
x + 2y + log (2x + y – 1) = C
Question 6
4 / -1

The area bounded by curves y = f(x), the x-axis and the ordinates x = 1 and x = b is (b - 1) sin (3b + 4). Then f(x) is-

A
(x-1) cos (3x + 4)

B
sin (3x + 4)

C
sin (3x + 4) + 3 (x – 1) cos (3x + 4)

D
sin (3x + 4) + (x – 1) cos (3x + 4)

Solution

1. Area bounded by curve y = f(x), x =1 and x = b is

Now differentiating both sides with respect to b we get
fB. = sin (3b + 4) + 3(b – 1) cos (3b + 4)
Question 7
4 / -1

If sin 5x + sin 3 x + sin x = 0, then the value of x other than zero, lying between 0 < x < π/2 is

A
π/6

B
π/12

C
π/3

D
π/4

Solution
Question 8
4 / -1

For n > 2 the product where is equal to

A

B

C

D
none of these

Solution
Question 9
4 / -1

The value of K, for which the equation (K–2)x² + 8x + K + 4 = 0 has both the roots real distinct and negative is:

A
0

B
2

C
3

D
-4

Solution

If the equation f(x) = 0, has real distinct and negative roots then:

–6 < K < 4 & (K < –4 or K > 2) & K > 2
–6 < K < 4 & _{K < –4 & K > 2) or (K > 2 & K > 2}
– 6 < K < 4 &
2 < K < 4
K∈(2, 4)
Among the given options
3 lies in the given range,
Hence K = 3
Question 10
4 / -1

The equation of the common tangent touching the circle (x-3)² + y² = 9 and the parabola y² = 4x above the x-axis, is

A
y√3 = 3x + 1

B
y√3 = - (x+3)

C
y√3 = (x+3)

D
y√3 = -(3x + 1)

Solution

Equation of any tangent to parabola

If the tangent to parabola is also tangent to the circle (x-3)² + y² = 9 then:

From the figure it is evident that for the tangent above x-axis Hence equation of required common tangent is: