Mathematics Test 163

Result

Mathematics Test 163

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Weekly Quiz Competition

Question 1
4 / -1

The value of

is equal to

A
0

B
1

C
2

D
irrational

Solution
Question 2
4 / -1

The distance between the line

A
10/9

B

C
3/10

D
10/3

Solution
Question 3
4 / -1

If θ ∈ R and the equation sinx + cos (θ + x) + cos(θ – x) = 2 has a solution then which of the following is true

A
sin² θ > 3/4

B
cos² θ  1/4

C
sin² θ  1/4

D
cos² θ  3/4

Solution
Question 4
4 / -1

(where [.] denotes the greatest integer function) is

A
0

B
1

C
cot 1

D
tan 1

Solution

As denominator is not converging
The limit is of the form

∴ Value of limit = 0
Question 5
4 / -1

A
1

B
3

C
9

D
2

Solution
Question 6
4 / -1

Minimum length of tangent drawn from a point Q on 3x + 4y = 20 to the circle x²+ y² = 1 is

A
15

B

C
0

D
3

Solution

PT² = OP² – r²

= 16 – 1 = 15
Question 7
4 / -1

One side of square PQRS is on the line y = 2x – 17 and other 2 vertices are on parabola y = x², then minimum area of square is:

A
80

B
4 √5

C
1280

D
16 √5

Solution
Question 8
4 / -1

If the variable line y = kx + 2h is tangent to an ellipse 2x² + 3y² = 6, then locus of P(h, k) is a conic C whose eccentricity equals

A
√5 / 2

B
√7 / 3

C
√7 / 2

D

Solution

By using condition of tangency, we get 4h² = 3k² + 2

∴ Locus of P(h, k) is 4x² – 3y² = 2 (which is hyperbola.)

Hence
Question 9
4 / -1

Let x and y be real numbers satisfying the equation 4y² + 4xy + x + 6 = 0. The complete set of values for x, is

A
{x : – 2 ≤ x ≤ 3}

B

C
{x : – 2 ≤ x ≤ 2}

D

Solution

We have 4y² + (4x)y + (x + 6) = 0

As y ∈ R, so discriminant ≥ 0.

⇒ (4x)² – 4 × 4 (x + 6) ≥ 0

⇒ x² – (x + 6) ≥ 0 ⇒ x² – x – 6 ≥ 0

⇒ (x – 3) (x + 2) ≥ 0

Hence x ∈ (– ∞, – 2] ∪ [3, ∞)
Question 10
4 / -1

The solution set of inequality (cot^–1x) (tan^–1x) + cot^–1x – 3tan^–1x – 3 > 0 is

A
x ∈ (tan 2, tan 3)

B
x ∈ (cot 3, cot 2)

C
x ∈ (– ∞ , tan 2) ∪ (tan 3, ∞)

D
x ∈ (– ∞ , cot 3) ∪ (cot 2, ∞)

Solution

⇒ (cot^–1x – 3) (2 – cot^–1x) > 0

⇒ (cot^–1x – 3) (cot^–1x – 2) < 0

⇒ 2 < cot^–1x < 3

⇒ cot3 < x < cot2 (As cot^–1x is a decreasing function.)

Hence x ∈ (cot3, cot2)