JEE Advanced Math Test 15

Result

JEE Advanced Math Test 15

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Weekly Quiz Competition

Question 1
4 / -1

If then

A
g(x) represent an equation of parabola, whose concave side is upward

B
g(x) have point of extrema at x = π/4

C
g(x) represent an equation of parabola, whose concave side is downward

D
Option B and option C are correct

Solution
Question 2
4 / -1

If α,β are the roots of equation, x²- 2x a² + 1 = 0 and γ, δ are the roots of the equation, x² - 2(a + 1)x + x(a - 1) = 0 such that α,β ∈ γ,δ then the interval in which ‘a’ lies,

A

B

C

D

Solution
Question 3
4 / -1

Directions For Questions

Consider the curves,

c₃ ≡ x = 0

...view full instructions

Maximum area bounded by all of these three curve is,

A

B

C
2

D
1

Solution

The value of K₁& K₂ decide the opening of parabola
Question 4
4 / -1

Directions For Questions

Consider the curves,

c₃ ≡ x= 0

...view full instructions

Minimum area bounded by all of these three curve is,

A

B

C
2

D
1

Solution
Question 5
4 / -1

Directions For Questions

Triangle ABC has circumcentre ‘O’ and orthocentre ‘H’. Points E and F are chosen on sides AC and ABrespectively such that AE = AO and AF = AH. If ‘R’ be radius of circle circumscribing the triangle ABC.Then,

...view full instructions

The area of quadrilateral BFEC is equal to,

A

B

C

D

Solution
Question 6
4 / -1

Directions For Questions

Triangle ABC has circumcentre ‘O’ and orthocentre ‘H’. Points E and F are chosen on sides AC and ABrespectively such that AE = AO and AF = AH. If ‘R’ be radius of circle circumscribing the triangle ABC.Then,

...view full instructions

The length of segment EF is equal to,

A
RcosA

B
R

C
Rsin2A

D
2RcosA

Solution

AE = R,AF = 2RcosA

⇒ EF = R
Question 7
4 / -1

Directions For Questions

Triangle ABC has circumcentre ‘O’ and orthocentre ‘H’. Points E and F are chosen on sides AC and ABrespectively such that AE = AO and AF = AH. If ‘R’ be radius of circle circumscribing the triangle ABC.Then,

...view full instructions

Let, then sum of squares of three altitudes of triangle AFE is equal to,

A

B
2R²

C

D

Solution
Question 8
4 / -1

If the equation x⁴ – 4x³ + ax² + bx + 1 = 0 has four positive roots then

A
a = 6, b = –4

B
a = 5, b = –4

C
a = 6, b = –3

D
a = 6, b ∈ R

Solution