JEE mathematics

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Question
CBSEENMA11015694

For each t ∈R, let [t] be the greatest integer less than or equal to t. Then

limx0+ x1x+2x+......+15x

  • does not exist (in R)

  • is equal to 0

  • is equal to 15

  • is equal to 120

Solution

D.

is equal to 120

limx0+ x1x+2x +.......+15x Now 1x - 1 < 1x1x2x-1<2x2x15x-1 <15x15xlimx0+ x   15x -15 < limx0+ x1x + 2x + ..... + 15xlimx0+ x   15 xlimx0+ x  15-5x < L limx0+ 15L 120

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Question
CBSEENMA11015695

If the tangent at (1, 7) to the curve x2 = y – 6 touches the circle x2 + y2 + 16x + 12y + c = 0 then the value of c is

  • 95

  • 195

  • 185

  • 85

Solution

A.

95

Equation of tangent at (1,7) to curve x2 = y -6 is

x - 1 =12(y +7) -6

2x - y +5 = 0 ... (i)

Centre of circle  = (-8,-6)

Radius of circle = 64 + 36-c = 100-c

 Line (i) touches the circle 2(-8) -(-6) +54 + 1 = 100-C5 = 100-c c = 95

Question
CBSEENMA11015696

If α, β ∈ C are the distinct roots, of the equation x2 -x + 1 = 0, then α101 + β107 is equal to

  • 2

  • -1

  • 0

  • 1

Solution

D.

1

x2-x + 1 = 0

Roots are -ω, -ω2

Let α = -ω, β = -ω2

α101 + β107 = (-ω)101 + (-ω2)107

= -( ω101214)
= - (ω2 + ω)
= 1

Question
CBSEENMA11015697

PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the angles of elevation of the top of the tower at P, Q and R are respectively 45o, 30o and 30o, then the height of the tower (in m) is

  • 502

  • 100

  • 50

  • 1003

Solution

B.

100

Let ST = h (height of tower)

PT = ST = h

STQT = tan 30°QT = h3Now PT2 + QT2 = 20024h2 = 2002h = 100

Question
CBSEENMA11015698

The sum of the coefficients of all odd degree terms in the expansion of x + x3 -15 + x - x3-15, (x>1) is

  • 2

  • -1

  • 0

  • 1

Solution

A.

2

x  + x3 -15 + x  -x3 -15 =2 [C02x5 + C25 x3 (x3-1) + C45 x (x3-1)2] = 2 [x5 + 10 (x6 - x3) + 5x(x6-2x3 +1)] = 2[ x5 + 10x6 - 10x3 + 5x7 -10x4 + 5x]= 2[5x7 + 10x6 + x5 - 10x4-10x3 + 5x]

Sum of odd degree terms coefficients
= 2(5 + 1 – 10 + 5)
= 2