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# JEE Mathematics Solved Question Paper 2005

#### Multiple Choice Questions

1.

If the coefficients of rth, (r+ 1)th and (r + 2)th terms in the binomial expansion of (1 + y)m are in A.P., then m and r satisfy the equation

• m2 – m(4r – 1) + 4r2 – 2 = 0

• m2 – m(4r+1) + 4r2 + 2 = 0

• m2 – m(4r + 1) + 4r2 – 2 = 0

• m2 – m(4r – 1) + 4r2 + 2 = 0

C.

m2 – m(4r + 1) + 4r2 – 2 = 0 197 Views

2.

Let P be the point (1, 0) and Q a point on the locus y2 = 8x. The locus of mid point of PQ is

• y2 – 4x + 2 = 0

• y2 + 4x + 2 = 0

• x2 + 4y + 2 =

• x2 – 4y + 2 = 0

A.

y2 – 4x + 2 = 0

P = (1, 0) Q = (h, k) such that
k2 = 8h
Let (α, β) be the midpoint of PQ 442 Views

3.

The value of α for which the sum of the squares of the roots of the equation x2 – (a – 2)x – a – 1 = 0 assume the least value is

• 1

• 0

• 3

• 2

A.

1

x2 – (a – 2)x – a – 1 = 0
⇒ α + β = a – 2
α β = –(a + 1)
α2 + β2 = (α + β)2 - 2αβ
= a2 – 2a + 6 = (a – 1)2 + 5
⇒ a = 1

152 Views

4.

If roots of the equation x2 – bx + c = 0 be two consectutive integers, then b2 – 4c equals

• – 2

• 3

• 2

• 1

D.

1

Let α, α + 1 be roots
α + α + 1 = b
α(α + 1) = c
∴ b2 – 4c = (2α + 1)2 - 4α(α + 1) = 1.

170 Views

5.

In a triangle PQR, ∠R =π/2. If (P/2) and tan (Q/2) are the roots of ax2 +bx+ c = 0, a ≠ 0 then

• a = b + c

• c = a + b

• b = c

• b = a + c

B.

c = a + b 239 Views

6.

The system of equations
αx + y + z = α - 1,
x + αy + z = α - 1,
x + y + αz = α - 1

has no solution, if α is

• -2

• either-2 or 1

• not -2

• 1

A.

-2

αx + y + z = α - 1,
x + αy + z = α - 1,
x + y + αz = α - 1 = α(α2 – 1) – 1(α - 1) + 1(1 - α)
= α (α - 1) (α + 1) – 1(α - 1) – 1(α - 1)
⇒ (α - 1)[α2 + α - 1 – 1] = 0
⇒ (α - 1)[α2 + α - 2]= 0 [α2 + 2α - α - 2] = 0
(α - 1) [α(α + 2) – 1(α + 2)]= 0
(α - 1) = 0,
α + 2 = 0 ⇒ α = –2, 1; but α ≠ 1

275 Views

7.

If the cube roots of unity are 1, ω, ω2 then the roots of the equation (x – 1)+ 8 = 0, are

• -1 , - 1 + 2ω, - 1 - 2ω2

• -1 , -1, - 1

• -1 , 1 - 2ω, 1 - 2ω2

• -1 , 1 + 2ω, 1 + 2ω2

C.

-1 , 1 - 2ω, 1 - 2ω2

(x – 1)3 + 8 = 0
⇒ (x – 1) = (-2) (1)1/3
⇒ x – 1 = -2 or -2ω or -2ω2 or
n = -1 or 1 – 2ω or 1 – 2ω2 .

446 Views

8. • • • tan 1

• D.  is equals  107 Views

9.

Area of the greatest rectangle that can be inscribed in the ellipse • 2ab

• ab

• • a/b

A.

2ab Area of rectangle ABCD = (2acosθ)
(2bsinθ) = 2absin2θ
⇒ Area of greatest rectangle is equal to 2ab when sin2θ = 1.
178 Views

10.

If in a frequently distribution, the mean and median are 21 and 22 respectively, then its mode is approximately

• 22.0

• 20.5

• 25.5

• 24.0

D.

24.0

Mode + 2Mean = 3 Median
⇒ Mode = 3 × 22 – 2 × 21= 66 – 42= 24.

154 Views