Mathematics Test 272

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Mathematics Test 272

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Question 1
4 / -1

If the coefficients of (r +1)th term and (r + 3)th term in the expansion of (1+x)²ⁿ be equal then

A
n + r – 1 = 0

B
n = r – 1

C
n = r + 1

D
none of these.

Solution

T_r+1 = ²ⁿCr(x)^r (1)^2n-r

T_r+3 = ²ⁿC_r+2 (x)^r+2 (1)^2n-r-2

T_r+1 : T_r+3

= ²ⁿC_r = ²ⁿC_r+2

=> 2n!/(2n-r)!r! = 2n!/(r+2)!(2n-r-2)!

=> 1/(2n-r)(2n-r-1) = 1/(r+2)(r+1)

=> (r+2)(r+1) = (2n-r)(2n-r-1)

=> r2 + 3r + 2 = 4n2 - 2nr - 2n - 2nr + r^2 + r

=> 0 = 4n2 - 4nr - 2n + r - 3r - 2

=> 0 = 4n2 - 4nr - 2n + r - 3r - 2 - 2 + 2

=> 0 = 4n2 - 4nr - 4 - [2n + 2r - 2]

=> 4n(n-r-1) -2(n-r-1)

Therefore n - r - 1 = 0

=> n = r+1
Question 2
4 / -1

In the expansion of (1+x)⁶⁰, the sum of coefficients of odd powers of x is

A
2⁶¹

B
2⁶⁰

C
2⁵⁹

D
none of these.

Solution

(1+x)ⁿ= nC0+ nC1x+ nC2x²+...+ nC0xⁿ ....(1)

(1−x)ⁿ= nC0 − nC1x+ nC2x²−...+ nC0 xⁿ....(2)

⇒(1+x)ⁿ−(1−x)ⁿ = nC0+ nC1x+ nC2x²+...+ nC0xⁿ− nC0+ nC1x− nC2x²+...+ nC0xⁿ

=2(nC1x+nC3x³+....+xⁿ)

Given: n=60 and put x=1

⇒(1+1)⁶⁰−(1−1)⁶⁰

=2⁶⁰

⇒2⁶⁰=2(nC1x+nC3x³+....+xⁿ)

∴Sum of odd powers of x in the expansion is = nC1x+nC3x³+....+xⁿ=2⁶⁰−1 =2⁵⁹
Question 3
4 / -1

The middle term in the expansion of (1+x)²ⁿ is

A

B
2ⁿC_n−1xⁿ⁻¹

C
2ⁿC_n+1xⁿ⁺¹

D
2ⁿC_n

Solution

Middle term in the expansion of (1+x)²ⁿ; is

=tn+1 = 2nCn.1⁽²ⁿ⁻ⁿ⁾.xⁿ

= {(2n)!.xⁿ}/(2n−n)!n!

= {{2n(2n−1)(2n−2)(2n−3)....4×3×2×1}/n! n!}/xⁿ

= {{2ⁿ[n(n−1)(n−2)....×2×1][(2n−1)(2n−3)....3×1]}/n! n!}xⁿ

= [[(2n−1)(2n−3)....3×1]/n!] 2ⁿxⁿ
Question 4
4 / -1

A
a negative integer

B
an irrational number

C
a rational number

D
none of these

Solution
Question 5
4 / -1

The circumcentre of the triangle with vertices (0, 0), (3, 0) and (0, 4) is

A
(1, 1)

B
(2, 3/2)

C
(3/2, 2)

D
None of these

Solution

Step-by-step explanation:

The given points when plotted show a right angled triangle

Since we know that the circumcentre of a right angled triangle lies on the midpoint of the hypotenuse, using section formula:

C=(3/2,4/2) => (3/2,2)
Question 6
4 / -1

The mid points of the sides of a triangle are (5, 0), (5, 12) and (0, 12), then orthocentre of this triangle is

A
(0, 0)

B
(0, 24)

C
(10, 0)

D

Solution
Question 7
4 / -1

Area of a triangle whose vertices are (a cos q, b sinq), (–a sin q, b cos q) and (–a cos q, –b sin q) is

A
Ab sin q cos q

B
A cos q sin q

C
1/2 ab

D
Ab

Solution
Question 8
4 / -1

The point A divides the join of the points (–5, 1) and (3, 5) in the ratio k : 1 and coordinates of points B and C are (1, 5) and (7, –2) respectively. If the area of ΔABC be 2 units, then k equals

A
k = 7 or k = 9

B
k = 6 or k = 7

C
k = 7 or k = 31/9

D
k = 9 or k = 31/9

Solution
Question 9
4 / -1

If A(cosa, sina), B(sina, – cosa), C(1, 2) are the vertices of a ΔABC, then as a varies, the locus of its centroid is

A
x² + y² – 2x – 4y+3=0

B
x²+y² – 2x–4y+1 = 0

C
3(x² + y²) – 2x – 4y+1=0

D
None of these

Solution
Question 10
4 / -1

Two perpendicular tangents to the circle x²+y² = r² meet at P. The locus of P is

A
x + y = r

B

C
x²+y² = 4r²

D
x²+ y² = 2r²

Solution