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In which quadrant are sin, cos and tan positive?a)IInd quadrantb)IVth quadrantc)IIIrd quadrantd)Ist quadrantCorrect answer is 'D'. Can you explain this answer?
|
Anshu Joshi answered |
- All three of them are positive in Quadrant I
- Sine only is positive in Quadrant II
- Tangent only is positive in Quadrant III
- Cosine only is positive in Quadrant IV
SinA = 1/√10 , SinB= 1/√5 If A and B are both acute angles,then , A+B=?- a)300
- b)750
- c)600
- d)450
Correct answer is option 'D'. Can you explain this answer?
SinA = 1/√10 , SinB= 1/√5 If A and B are both acute angles,then , A+B=?
a)
300
b)
750
c)
600
d)
450
|
Riya Banerjee answered |
We know that:
Sin θ = Opposite / Hypotenuse
Sin θ = Opposite / Hypotenuse
∴ SinA = 1/√10
CosA= 3/√10
similarly, SinB = 1/√5
CosB= 2/√5
CosA= 3/√10
similarly, SinB = 1/√5
CosB= 2/√5
⇒ Multiply:
Cos(A+B)= CosA x CosB - SinA x SinB
Cos(A+B)= CosA x CosB - SinA x SinB
Substituting the value in above equation we get:
= 3/√10 x 2/√5 - 1/√10 x 1/√5
= 6/√50 - 1/√50
= 6-1/5√2. ........(√50=5√2)
= 1/ √2
= 6/√50 - 1/√50
= 6-1/5√2. ........(√50=5√2)
= 1/ √2
we know that, sin 45 = 1/ √ 2 therefore
sinθ / cosθ = 45
sinθ / cosθ = 45
Can you explain the answer of this question below:tan x = - 5/12, x lies in the second quadrant. So sinx=?
- A:
5/13
- B:
-5/13
- C:
-12/13
- D:
12/13
The answer is a.
tan x = - 5/12, x lies in the second quadrant. So sinx=?
5/13
-5/13
-12/13
12/13
|
Krishna Iyer answered |
tanx = -5/12
Therefore perpendicular = -5, base = 12
Applying pythagoras theorem,
(hyp)2 = (per)2 + (base)2
⇒ (-5)2 + (12)2
hyp = [25+144]1/2
hyp = (169)1/2
hyp = 13
sinx = perpendicular/hypotenous
= -5/13
In second quadrant, only sin x, cosec x are positive
So. sinx = 5/13
Therefore perpendicular = -5, base = 12
Applying pythagoras theorem,
(hyp)2 = (per)2 + (base)2
⇒ (-5)2 + (12)2
hyp = [25+144]1/2
hyp = (169)1/2
hyp = 13
sinx = perpendicular/hypotenous
= -5/13
In second quadrant, only sin x, cosec x are positive
So. sinx = 5/13
What is the value of sin 7π ?
- a)1
- b)-1
- c)-1/2
- d)0
Correct answer is option 'D'. Can you explain this answer?
What is the value of sin 7π ?
a)
1
b)
-1
c)
-1/2
d)
0
|
Om Desai answered |
Sin 7π = Sin 7*180 = Sin 2π * 7 = 0
# Remember Sin nπ =0
# Remember Sin nπ =0
The next term of the sequence 1, 2, 4, 7,11,…. Is- a)18
- b)17
- c)16
- d)15
Correct answer is option 'C'. Can you explain this answer?
The next term of the sequence 1, 2, 4, 7,11,…. Is
a)
18
b)
17
c)
16
d)
15
|
Gaurav Kumar answered |
The given series is: 1,2,4,7,11,...
Difference between second and first term = 2 - 1 = 1
Difference between third and second term = 4 - 2 = 2
Difference between fourth and third term = 7 - 4 = 3
Difference between fifth and fourth term = 11 - 7 = 4
Difference between sixth and fifth term = 16 - 11 = 5
Difference between second and first term = 2 - 1 = 1
Difference between third and second term = 4 - 2 = 2
Difference between fourth and third term = 7 - 4 = 3
Difference between fifth and fourth term = 11 - 7 = 4
Difference between sixth and fifth term = 16 - 11 = 5
cos 68° cos 8° + sin 68° sin 8° = ?
- a)1/2
- b)1/4
- c)1
- d)0
Correct answer is option 'A'. Can you explain this answer?
cos 68° cos 8° + sin 68° sin 8° = ?
a)
1/2
b)
1/4
c)
1
d)
0
|
Lavanya Menon answered |
We know,
cosA cosB + sinA sinB = cos(A-B)
cosA cosB + sinA sinB = cos(A-B)
cos 68° cos 8° + sin 68° sin 8° = Cos (68-8) = Cos60°
=1/2
=1/2
Can you explain the answer of this question below:What is the value of 
- A:√3/2
- B:1/2
- C:1
- D:1/√2
The answer is d.
What is the value of
A:
√3/2
B:
1/2
C:
1
D:
1/√2
|
Harshitha Mehta answered |
We know ,π = 180deg
So cos 41π/4 = Cos( 41*180/4)
= Cos (1845deg)
= Cos (1800 + 45)
= Cos (10π + π/4)
= Cos (π/4)
= 1/√2
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
a)
b)
c)
d)
|
Yogesh Singla answered |
D is right sinx= -cox and x*n=1/n x*n+1
In which quadrant are sin, cos and tan positive?- a)IInd quadrant
- b)IVth quadrant
- c)IIIrd quadrant
- d)Ist quadrant
Correct answer is option 'B'. Can you explain this answer?
In which quadrant are sin, cos and tan positive?
a)
IInd quadrant
b)
IVth quadrant
c)
IIIrd quadrant
d)
Ist quadrant
|
Nandini Patel answered |
For an angle in the fourth quadrant the point P has positive x coordinate and negative y coordinate. Therefore: In Quadrant IV, cos(θ) > 0, sin(θ) < 0 and tan(θ) < 0 (Cosine positive). The quadrants in which cosine, sine and tangent are positive are often remembered using a favorite mnemonic.
Sin(n+1)A sin(n+2)A + cos(n+1)A cos(n+2)A=- a)sinA
- b)sin2A
- c)cosA
- d)cos2A
Correct answer is option 'C'. Can you explain this answer?
Sin(n+1)A sin(n+2)A + cos(n+1)A cos(n+2)A=
a)
sinA
b)
sin2A
c)
cosA
d)
cos2A
|
|
Gaurav Kumar answered |
sin(n+1)Asin(n+2)A + cos(n+1)Acos(n+2)A = cos (n+1)Acos(n+2)A + sin(n+1)Asin(n+2)A = cos{A(n+2-n-1)} = cos (A.1) = cos A
What is the sign of the sec θ and cosec θ in second quadrant respectively?- a)positive and negative
- b)positive and positive
- c)negative and negative
- d)negative and positive
Correct answer is option 'D'. Can you explain this answer?
What is the sign of the sec θ and cosec θ in second quadrant respectively?
a)
positive and negative
b)
positive and positive
c)
negative and negative
d)
negative and positive
|
Preeti Iyer answered |
In quadrant sin, cos tan, cot, sec, cosec all +ve .In second quadrant sin and cosec are +ve. in 3rd quadrant tan and cot are positive.And in 4th cos and sec are +ve. 
cos(π/4 -x) cos ( π/4 -y)-sin(π/4-x) sin( π/4 -y)=- a)cos(x-y)
- b)sin(x-y)
- c)cos(x+y)
- d)sin(x+y)
Correct answer is option 'D'. Can you explain this answer?
cos(π/4 -x) cos ( π/4 -y)-sin(π/4-x) sin( π/4 -y)=
a)
cos(x-y)
b)
sin(x-y)
c)
cos(x+y)
d)
sin(x+y)
|
Infinity Academy answered |
Cos(π/4-x)cos (π/4-y) - sin (π/4-x) sin(π/4-y)
= CosA*Cos B - Sin A*Sin B
= Cos (A+B)
= cos(π/4-x+π/4-y)
= cos(π/2-x-y)
= cos{π/2 - (x+y)}
= sin(x+y)
= Cos (A+B)
= cos(π/4-x+π/4-y)
= cos(π/2-x-y)
= cos{π/2 - (x+y)}
= sin(x+y)
What is the range of cos function?
- a)[-1,0]
- b)[0,1]
- c)[-1,1]
- d)[-2,2]
Correct answer is option 'C'. Can you explain this answer?
What is the range of cos function?
a)
[-1,0]
b)
[0,1]
c)
[-1,1]
d)
[-2,2]
|
|
Om Desai answered |
Just look at the graph of cosine.
We know , Range of a function is the set of all possible outputs for that function. If you look at any 2π interval, the cosine function is periodic after every 2π. as you can see it value range between -1 to 1 along the y-axis . So the range for cos function is [-1,1]
Which of the following cannot be the value of cos θ .- a)1
- b)-1
- c)√2
- d)0
Correct answer is option 'C'. Can you explain this answer?
Which of the following cannot be the value of cos θ .
a)
1
b)
-1
c)
√2
d)
0
|
|
Naina Sharma answered |
√2 cannot be the value for Cosθ.
The values of Cos θ at different angles are given below :
Cos0°=1
Cos30°=√3/2
Cos45°=1/√2
Cos60°=1/2
Cos90°=0
Can you explain the answer of this question below:sin(60° + A) cos(30° – B) + cos(60° + A) sin(30° – B) is equal to- A:sin(A + B)
- B:sin(A – B)
- C:cos(A – B)
- D:cos(A + B)
The answer is c.
sin(60° + A) cos(30° – B) + cos(60° + A) sin(30° – B) is equal to
A:
sin(A + B)
B:
sin(A – B)
C:
cos(A – B)
D:
cos(A + B)
|
|
Lavanya Menon answered |
L.H.S. = sin(60+A)cos(30−B)+cos(60+A)sin(30−B)
= sin[(60+A)+(30−B)] (Using, sin(A+B)sinAcosB+cosAsinB)
= sin(90+A−B)
= sin(90+(A−B))
= cos(A−B) (Using, sin(90+θ)=cosθ) = R.H.S.Hence Proved.
= sin[(60+A)+(30−B)] (Using, sin(A+B)sinAcosB+cosAsinB)
= sin(90+A−B)
= sin(90+(A−B))
= cos(A−B) (Using, sin(90+θ)=cosθ) = R.H.S.Hence Proved.
Chose which of the following expressions equals sinA + cosA.- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Chose which of the following expressions equals sinA + cosA.
a)
b)
c)
d)
|
|
Lavanya Menon answered |
sinA + cosA = √2(sinA/√2+cosA/√2)
=√2(sinAcos(π/4)+cosAsin(π/4))
=√2sin(A+π/4)
by using sin(A+B) formula
=√2(sinAcos(π/4)+cosAsin(π/4))
=√2sin(A+π/4)
by using sin(A+B) formula
What is the sign of the sinA and tanA in third quadrant respectively- a)negative and positive
- b)positive and positive
- c)negative and negative.
- d)positive and negative
Correct answer is option 'A'. Can you explain this answer?
What is the sign of the sinA and tanA in third quadrant respectively
a)
negative and positive
b)
positive and positive
c)
negative and negative.
d)
positive and negative
|
Suresh Iyer answered |
The sign of the sinA and tanA in third quadrant is negative and positive.
If a, b, c, d are in H.P., then ab + bc + cd is- a)3 b d
- b)(a + b) (c + d)
- c)3 a d
- d)3 a c
Correct answer is option 'C'. Can you explain this answer?
If a, b, c, d are in H.P., then ab + bc + cd is
a)
3 b d
b)
(a + b) (c + d)
c)
3 a d
d)
3 a c
|
|
Rajesh Gupta answered |
Since a,b,c are in H.P, so b = 2ac/(a+c).
Also, b,c,d are in H.P, so c = 2bd/(b+d).
Therefore, (a+c)(b+d) = 2ac/b × 2bd/c
⇒ab+cb+ad+cd = 4ad
⇒ab+bc+cd = 3ad
Also, b,c,d are in H.P, so c = 2bd/(b+d).
Therefore, (a+c)(b+d) = 2ac/b × 2bd/c
⇒ab+cb+ad+cd = 4ad
⇒ab+bc+cd = 3ad
The sum of all 2-digited numbers which leave remainder 1 when divided by 3 is- a)1605
- b)1616
- c)1604
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The sum of all 2-digited numbers which leave remainder 1 when divided by 3 is
a)
1605
b)
1616
c)
1604
d)
none of these
|
|
Maheshwar Unni answered |
Problem:
The sum of all 2-digit numbers which leave a remainder of 1 when divided by 3 is:
a) 1605
b) 1616
c) 1604
d) none of these
Solution:
To find the sum of all 2-digit numbers that leave a remainder of 1 when divided by 3, we can follow these steps:
Step 1: Determine the range of 2-digit numbers:
The range of 2-digit numbers is from 10 to 99.
Step 2: Identify the numbers that satisfy the given condition:
To find the numbers that leave a remainder of 1 when divided by 3, we can use the modulo operation. If a number leaves a remainder of 1 when divided by 3, then it can be written as 3k + 1, where k is an integer.
We need to find all values of k such that 3k + 1 is a 2-digit number. This condition is satisfied when k = 3, 4, ..., 33.
Step 3: Calculate the sum:
To find the sum of these numbers, we can use the formula for the sum of an arithmetic series:
Sum = (n/2)(first term + last term)
In this case, the first term is 3 * 3 + 1 = 10 and the last term is 3 * 33 + 1 = 100. The number of terms, n, is 33 - 3 + 1 = 31.
Using the formula, the sum of these numbers is:
Sum = (31/2)(10 + 100) = 31 * 55 = 1705
Conclusion:
Therefore, the sum of all 2-digit numbers that leave a remainder of 1 when divided by 3 is 1705. However, none of the given answer choices match this result, so the correct answer is none of these.
The sum of all 2-digit numbers which leave a remainder of 1 when divided by 3 is:
a) 1605
b) 1616
c) 1604
d) none of these
Solution:
To find the sum of all 2-digit numbers that leave a remainder of 1 when divided by 3, we can follow these steps:
Step 1: Determine the range of 2-digit numbers:
The range of 2-digit numbers is from 10 to 99.
Step 2: Identify the numbers that satisfy the given condition:
To find the numbers that leave a remainder of 1 when divided by 3, we can use the modulo operation. If a number leaves a remainder of 1 when divided by 3, then it can be written as 3k + 1, where k is an integer.
We need to find all values of k such that 3k + 1 is a 2-digit number. This condition is satisfied when k = 3, 4, ..., 33.
Step 3: Calculate the sum:
To find the sum of these numbers, we can use the formula for the sum of an arithmetic series:
Sum = (n/2)(first term + last term)
In this case, the first term is 3 * 3 + 1 = 10 and the last term is 3 * 33 + 1 = 100. The number of terms, n, is 33 - 3 + 1 = 31.
Using the formula, the sum of these numbers is:
Sum = (31/2)(10 + 100) = 31 * 55 = 1705
Conclusion:
Therefore, the sum of all 2-digit numbers that leave a remainder of 1 when divided by 3 is 1705. However, none of the given answer choices match this result, so the correct answer is none of these.
In an A.P., sum of first p terms is equal to the sum of first q terms. Sum of its first p + q terms is- a)0
- b)- ( p + q)
- c)p + q
- d)none of these.
Correct answer is option 'A'. Can you explain this answer?
In an A.P., sum of first p terms is equal to the sum of first q terms. Sum of its first p + q terms is
a)
0
b)
- ( p + q)
c)
p + q
d)
none of these.
|
Trisha Vashisht answered |
Sp = Sq
⇒ p/2(2a+(p−1)d) = q/2(2a+(q−1)d)
⇒ p(2a+(p−1)d) = q(2a+(q−1)d)
⇒ 2ap + p2d − pd = 2aq + q2d − qd
⇒ 2a(p−q) + (p+q)(p−q)d − d(p−q) = 0
⇒ (p−q)[2a + (p+q)d − d] = 0
⇒ 2a + (p+q)d − d = 0
⇒ 2a + ((p+q) − 1)d = 0
⇒ Sp+q = 0
⇒ p/2(2a+(p−1)d) = q/2(2a+(q−1)d)
⇒ p(2a+(p−1)d) = q(2a+(q−1)d)
⇒ 2ap + p2d − pd = 2aq + q2d − qd
⇒ 2a(p−q) + (p+q)(p−q)d − d(p−q) = 0
⇒ (p−q)[2a + (p+q)d − d] = 0
⇒ 2a + (p+q)d − d = 0
⇒ 2a + ((p+q) − 1)d = 0
⇒ Sp+q = 0
The next term of the sequence 1, 5, 14, 30, 55, …… is
- a)95
- b)91
- c)90
- d)80
Correct answer is option 'B'. Can you explain this answer?
The next term of the sequence 1, 5, 14, 30, 55, …… is
a)
95
b)
91
c)
90
d)
80
|
Mihir Chaudhary answered |
The sequence is obtained by adding consecutive odd numbers, starting with 1.
1 + 0 = 1
1 + 3 = 4
4 + 5 = 9
9 + 7 = 16
16 + 9 = 25
So the next term in the sequence is 25.
Therefore, the complete sequence is:
1, 5, 14, 30, 55, 25
1 + 0 = 1
1 + 3 = 4
4 + 5 = 9
9 + 7 = 16
16 + 9 = 25
So the next term in the sequence is 25.
Therefore, the complete sequence is:
1, 5, 14, 30, 55, 25
If n is a +ve integer, then the binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are- a)additive inverse of each other
- b)multiplicative inverse of each other
- c)equal
- d)nothing can be said
Correct answer is option 'C'. Can you explain this answer?
If n is a +ve integer, then the binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are
a)
additive inverse of each other
b)
multiplicative inverse of each other
c)
equal
d)
nothing can be said
|
Hansa Sharma answered |
(x+a)n = nC0 xn + nC1 x(n-1) a1 + nC2 x(n-2) a2 + ..........+ nC(n-1) xa(n-1) + nCn an
Now, nC0 = nCn, nC1 = nCn-1, nC2 = nCn-2,........
therefore, nCr = nCn-r
The binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are equal.
Now, nC0 = nCn, nC1 = nCn-1, nC2 = nCn-2,........
therefore, nCr = nCn-r
The binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are equal.
The eleventh term of the sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, ….. is- a)89
- b)66
- c)72
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The eleventh term of the sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, ….. is
a)
89
b)
66
c)
72
d)
none of these
|
|
Nandini Iyer answered |
The sequence is the Fibonacci series
1+1 = 0
1+2 = 3
2+3 = 5
3+5 = 8
5+8 = 13
8+13 = 21
13+21 = 34
21+34 = 55
34+55 = 89
The 11th term will be 89.
1+1 = 0
1+2 = 3
2+3 = 5
3+5 = 8
5+8 = 13
8+13 = 21
13+21 = 34
21+34 = 55
34+55 = 89
The 11th term will be 89.
Which term of the sequence 8 – 6i, 7 – 4i, 6 – 2i, ….is a real number ?- a)7th
- b)6th
- c)5th
- d)4th
Correct answer is option 'D'. Can you explain this answer?
Which term of the sequence 8 – 6i, 7 – 4i, 6 – 2i, ….is a real number ?
a)
7th
b)
6th
c)
5th
d)
4th
|
|
Preeti Iyer answered |
a = 8−6i
d = 7−4i−8+6i
= −1+2i
an = a+(n−1)d
a+ib = 8−6i+(n−1)(−1+2i)
a+ib = 8−6i+(−1)(n−1)+(n−1)2i
= − 6+2(n−1)=0
= 2(n−1) = 6
n = 4
an = 8−6i+(4−1)(−1+2i)
= 8−6i−3+6i = 5
4th term = 5
d = 7−4i−8+6i
= −1+2i
an = a+(n−1)d
a+ib = 8−6i+(n−1)(−1+2i)
a+ib = 8−6i+(−1)(n−1)+(n−1)2i
= − 6+2(n−1)=0
= 2(n−1) = 6
n = 4
an = 8−6i+(4−1)(−1+2i)
= 8−6i−3+6i = 5
4th term = 5
If cos a + 2cos b + cos c = 2 then a, b, c are in- a)2b = a+c
- b)b2 = a x c
- c)a = b = c
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
If cos a + 2cos b + cos c = 2 then a, b, c are in
a)
2b = a+c
b)
b2 = a x c
c)
a = b = c
d)
none of these
|
|
Hansa Sharma answered |
cos A + 2 cos B + cos C = 2
⇒ cos A + cos C = 2(1 – cos B)
⇒ cos A + cos C = 2(1 – cos B)
2 cos((A + C)/2) × cos((A-C)/2 = 4 sin²(B/2)
2 sin(B/2) cos((A-C)/2) = 4 sin² (B/2)
cos((A-C)/2) = 2 sin (B/2)
cos((A-C)/2) = 2 cos((A+C)/2)
cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2)
cos((A-C)/2) = 2 sin (B/2)
cos((A-C)/2) = 2 cos((A+C)/2)
cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2)
2 sin(A/2) sin(C/2) = sin(B/2)
⇒ 2 {√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac
⇒ 2 {√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac
2(s – b) = b
a + b + c – 2b = b
a + c – b = b
a + c = 2b
a + b + c – 2b = b
a + c – b = b
a + c = 2b
The value of 
is:- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The value of is:
is:a)
b)
c)
d)
|
Samyak Jain answered |
1/2(cos π/2 + 1 -[cos(π/2-π/4)+1])=1/2+π/4
In an A.P., sum of first p terms is q and sum of first q terms is p. Sum of its p + q terms is- a)0
- b)- ( p + q)
- c)p + q
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
In an A.P., sum of first p terms is q and sum of first q terms is p. Sum of its p + q terms is
a)
0
b)
- ( p + q)
c)
p + q
d)
none of these
|
|
Om Desai answered |
Let the first term of the given AP be ‘a' and the common difference be ‘d'. Then, the sum of first ‘n' terms of the AP is given by:
Sn = n/2 {2a+(n-1)d} …….(1)
Here, it is given that:
Sp = q and Sq = p
Using (1), we get:-
q = p/2 {2a+(p-1)d}
and p = q/2 {2a+(q-1)d}
i.e. 2a + (p-1)d = 2q/p …..(2)
and 2a + (q-1)d = 2p/q …..(3)
Subtracting (3) from (2), we get:
(p - 1 - q + 1)d = 2q/p - 2p/q
So, d = 2(q2 - p2)/pq(p-q)
i.e. d = -2(p+q)/pq
Now, substituting the value of ‘d' in eq.n (2), we get:
2a + (p-1){-2(p+q)/pq} = 2q/p
i.e. 2a= 2q/p + 2(p-1)(p+q)/pq
This gives:
a = (p2 + q2 - p - q + pq)/pq
So, we have
Sp+q = (p+q)/2 { 2(p2+q2-p-q+pq)/pq - (p+q-1) 2 (p+q)/pq}
i.e. Sp+q = (p+q)/pq { p2+q2-p-q+pq-p2-pq-qp-q2+p+q}
So, Sp+q = -(p+q)
Sn = n/2 {2a+(n-1)d} …….(1)
Here, it is given that:
Sp = q and Sq = p
Using (1), we get:-
q = p/2 {2a+(p-1)d}
and p = q/2 {2a+(q-1)d}
i.e. 2a + (p-1)d = 2q/p …..(2)
and 2a + (q-1)d = 2p/q …..(3)
Subtracting (3) from (2), we get:
(p - 1 - q + 1)d = 2q/p - 2p/q
So, d = 2(q2 - p2)/pq(p-q)
i.e. d = -2(p+q)/pq
Now, substituting the value of ‘d' in eq.n (2), we get:
2a + (p-1){-2(p+q)/pq} = 2q/p
i.e. 2a= 2q/p + 2(p-1)(p+q)/pq
This gives:
a = (p2 + q2 - p - q + pq)/pq
So, we have
Sp+q = (p+q)/2 { 2(p2+q2-p-q+pq)/pq - (p+q-1) 2 (p+q)/pq}
i.e. Sp+q = (p+q)/pq { p2+q2-p-q+pq-p2-pq-qp-q2+p+q}
So, Sp+q = -(p+q)
pth term of an H.P. is qr and qth term is pr, then rth term of the H.P. is- a)pq
- b)1
- c)pqr2
- d)pqr
Correct answer is option 'A'. Can you explain this answer?
pth term of an H.P. is qr and qth term is pr, then rth term of the H.P. is
a)
pq
b)
1
c)
pqr2
d)
pqr
|
|
Raghav Bansal answered |
Given pth term of HP = qr
So pth term of AP = 1/qr
a+(p−1)d = 1/qr....(1)
and qth term of HP = pr
so qth term of AP = 1/pr
a + (q−1)d = 1/pr.....(2)
subtracting equation 1 and 2 we get,
(p−q)d = (p−q)/pqr
d = 1/pqr
Now from equation 1,
a = 1/qr − (p−1)/pqr
= (p−p+1)/pqr = 1/pqr
So rth term of AP = a+(r−1)d = 1/pqr + (r − 1)/pqr = 1/pq
So, rth term of HP = pq
So pth term of AP = 1/qr
a+(p−1)d = 1/qr....(1)
and qth term of HP = pr
so qth term of AP = 1/pr
a + (q−1)d = 1/pr.....(2)
subtracting equation 1 and 2 we get,
(p−q)d = (p−q)/pqr
d = 1/pqr
Now from equation 1,
a = 1/qr − (p−1)/pqr
= (p−p+1)/pqr = 1/pqr
So rth term of AP = a+(r−1)d = 1/pqr + (r − 1)/pqr = 1/pq
So, rth term of HP = pq
The number of numbers between n and n2 which are divisible by n is- a)n – 2
- b)n
- c)n – 1
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The number of numbers between n and n2 which are divisible by n is
a)
n – 2
b)
n
c)
n – 1
d)
none of these
|
|
Hansa Sharma answered |
Between n & n2, numbers divisible by n are:
2n, 3n, 4n, ….... (n – 1)n
No. of numbers = (n – 1) – 2 + 1 = n – 2
2n, 3n, 4n, ….... (n – 1)n
No. of numbers = (n – 1) – 2 + 1 = n – 2
The sum of 40 A.M.’s between two number is 120. The sum of 50 A.M.’s between them is equal to- a)150
- b)130
- c)160
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The sum of 40 A.M.’s between two number is 120. The sum of 50 A.M.’s between them is equal to
a)
150
b)
130
c)
160
d)
none of these
|
Mira Sharma answered |
Let A1, A2, A3, , A40 be 40 A.M's between two numbers 'a' and 'b'.
Then,
a, A1, A2, A3, , A40, b is an A.P. with common difference d = (b - a)/(n + 1) = (b - a)/41
[ where n = 40]
now,
A1, A2, A3, , A40 = 40/2( A1 + A40)
A1, A2, A3, , A40 = 40/2(a + b)
[ a, A1, A2, A3, , A40, b is an Ap then ,a + b = A1 + A40]
sum of 40A.M = 120(given)
120= 20(a + b)
=> 6 = a + b --(1)
Again ,
consider B1, B2, B50 be 50 A.M.'s between two numbers a and b.
Then, a, B1, B2,, B50, b will be in A.P. with common difference = ( b - a)/51
now , similarly,
B1, B2, ... , B50 = 50/2(B1 + B2)
= 25(6) -from(1)
= 150
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
a)
b)
c)
d)
|
Vikas Kapoor answered |
Option d is correct, because it is the property of definite integral
∫02a f(x) dx = ∫0a f(x) dx + ∫0a f(2a – x) dx
∫02a f(x) dx = ∫0a f(x) dx + ∫0a f(2a – x) dx
If a, 4, b are in A.P.; a, 2, b are in G.P.; then a, 1, b are in- a)G.P.
- b)H.P.
- c)A.P.
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
If a, 4, b are in A.P.; a, 2, b are in G.P.; then a, 1, b are in
a)
G.P.
b)
H.P.
c)
A.P.
d)
none of these
|
Lekshmi Chopra answered |
Given: a, 4, b are in A.P. and a, 2, b are in G.P.
To find: a, 1, b are in which progression.
Solution:
Let d be the common difference in the A.P.
Then, 4 - a = b - 4 = d ...(1)
Let r be the common ratio in the G.P.
Then, 2/a = b/2 = r ...(2)
From equation (1), we get:
b - a = 4 - 4a
a - 4a + b = 4
(1 - 4 + 1)r = 4 - a
-3r = 4 - a
r = (a - 4)/3
From equation (2), we get:
b = 2r/a
b = 2(a - 4)/(3a)
Substituting the value of b in equation (1), we get:
(a - 4)/(3a) - a = d
(a - 4) - 3a^2 d = 0
3a^2 - a + 4d = 0
Using the formula for the sum of n terms of an H.P., we get:
a, 1, b are in H.P. with sum 3/(a + 1/b) = 3/(a + 2/(a - 4)/3) = (3a(a - 4))/(4a^2 - 4a - 24)
Simplifying, we get:
3(a - 4)/(4a - 6) = 3(1 - 4/a)/(4 - 6/a)
Therefore, a, 1, b are in H.P.
Hence, the correct option is (b).
To find: a, 1, b are in which progression.
Solution:
Let d be the common difference in the A.P.
Then, 4 - a = b - 4 = d ...(1)
Let r be the common ratio in the G.P.
Then, 2/a = b/2 = r ...(2)
From equation (1), we get:
b - a = 4 - 4a
a - 4a + b = 4
(1 - 4 + 1)r = 4 - a
-3r = 4 - a
r = (a - 4)/3
From equation (2), we get:
b = 2r/a
b = 2(a - 4)/(3a)
Substituting the value of b in equation (1), we get:
(a - 4)/(3a) - a = d
(a - 4) - 3a^2 d = 0
3a^2 - a + 4d = 0
Using the formula for the sum of n terms of an H.P., we get:
a, 1, b are in H.P. with sum 3/(a + 1/b) = 3/(a + 2/(a - 4)/3) = (3a(a - 4))/(4a^2 - 4a - 24)
Simplifying, we get:
3(a - 4)/(4a - 6) = 3(1 - 4/a)/(4 - 6/a)
Therefore, a, 1, b are in H.P.
Hence, the correct option is (b).
In a triangle ABC, tan A/2 = 5/6, tan B/2 = 20/37, then tan C/2 is equal to:
- a)2/5
- b)100/222
- c)34/37
- d)5/2
Correct answer is option 'A'. Can you explain this answer?
In a triangle ABC, tan A/2 = 5/6, tan B/2 = 20/37, then tan C/2 is equal to:
a)
2/5
b)
100/222
c)
34/37
d)
5/2
|
Anjali Sharma answered |
In triangle ABC,
► Sum of all three angles = 1800
► A + B + C = 180 = A + B = 180 - C
► Sum of all three angles = 1800
► A + B + C = 180 = A + B = 180 - C
devide by 2 both side
A/2 + B/2 = 180/2 - C/2
tan x = - 5/12, x lies in the second quadrant. So sinx=?- a)5/13
- b)-5/13
- c)-12/13
- d)12/13
Correct answer is option 'A'. Can you explain this answer?
tan x = - 5/12, x lies in the second quadrant. So sinx=?
a)
5/13
b)
-5/13
c)
-12/13
d)
12/13
|
|
Raghav Bansal answered |
tanx = -5/12
Therefore perpendicular = -5, base = 12
Applying pythagoras theorem,
(hyp)2 = (per)2 + (base)2
⇒ (-5)2 + (12)2
hyp = [25+144]1/2
hyp = (169)1/2
hyp = 13
sinx = perpendicular/hypotenous
= -5/13
In second quadrant, only sin x, cosec x are positive
So. sinx = 5/13
Therefore perpendicular = -5, base = 12
Applying pythagoras theorem,
(hyp)2 = (per)2 + (base)2
⇒ (-5)2 + (12)2
hyp = [25+144]1/2
hyp = (169)1/2
hyp = 13
sinx = perpendicular/hypotenous
= -5/13
In second quadrant, only sin x, cosec x are positive
So. sinx = 5/13
Identify the odd one out from the following- a)sin2θ + cos2θ=1
- b)sec2θ = 1 + tan2θ
- c)sec2θ + tan2θ = 1
- d)cosec2 θ = 1 + cot2 θ
Correct answer is option 'C'. Can you explain this answer?
Identify the odd one out from the following
a)
sin2θ + cos2θ=1
b)
sec2θ = 1 + tan2θ
c)
sec2θ + tan2θ = 1
d)
cosec2 θ = 1 + cot2 θ
|
|
Sonal Mestry answered |
Bcoz in trigonometry, we have only 3 identities which are A, B and D respectively. but we trigo don't have C type identity or formula.
The coefficient of y in the expansion of (y² + c/y)5 is - a)10c
- b)10c²
- c)10c³
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
The coefficient of y in the expansion of (y² + c/y)5 is
a)
10c
b)
10c²
c)
10c³
d)
None of these
|
|
Varun Kapoor answered |
Given, binomial expression is (y² + c / y)5
Now, Tr+1 = 5Cr × (y²)5-r × (c / y)r
= 5Cr × y10-3r × Cr
Now, 10 – 3r = 1
⇒ 3r = 9
⇒ r = 3
So, the coefficient of y = 5C3 × c³ = 10c³
Now, Tr+1 = 5Cr × (y²)5-r × (c / y)r
= 5Cr × y10-3r × Cr
Now, 10 – 3r = 1
⇒ 3r = 9
⇒ r = 3
So, the coefficient of y = 5C3 × c³ = 10c³
is:
The next term of the sequence, 2, 6, 12, 20, …..is- a)30
- b)28
- c)40
- d)24
Correct answer is option 'A'. Can you explain this answer?
The next term of the sequence, 2, 6, 12, 20, …..is
a)
30
b)
28
c)
40
d)
24
|
|
Lavanya Menon answered |
The given sequence is based on the following pattern:
∴ Required number = 30.
∴ Required number = 30.
If 2nd, 3rd and 4th terms in the expansion of (x+a)n are 240, 720 and 1080 respectively, then the value of n is- a)15
- b)20
- c)5
- d)10
Correct answer is option 'C'. Can you explain this answer?
If 2nd, 3rd and 4th terms in the expansion of (x+a)n are 240, 720 and 1080 respectively, then the value of n is
a)
15
b)
20
c)
5
d)
10
|
Shiksha Academy answered |
General term Tr+1 of (x+y)n is given by
Tr+1 = nCr xn-r yr
T2 = nC2 xn-2 y = 240
T3 = nC3 xn-3 y2 = 720
T4 = nC4 xn-4 y3 = 1080
T3/T2 = [(n-1)/2] * [y/x] = 3......(1)
T4/T2 = {[(n-1)(n-2)]/(3*2)} * x2/y2 = 9/2
T4/T3 = [(n-2)/3] * [y/x] = 3/2...(2)
Dividing 1 by 2
[(n-1)/2] * [3/(n-2)] = 2
⇒ 3n−3 = 4n−8
⇒ 5 = n
Tr+1 = nCr xn-r yr
T2 = nC2 xn-2 y = 240
T3 = nC3 xn-3 y2 = 720
T4 = nC4 xn-4 y3 = 1080
T3/T2 = [(n-1)/2] * [y/x] = 3......(1)
T4/T2 = {[(n-1)(n-2)]/(3*2)} * x2/y2 = 9/2
T4/T3 = [(n-2)/3] * [y/x] = 3/2...(2)
Dividing 1 by 2
[(n-1)/2] * [3/(n-2)] = 2
⇒ 3n−3 = 4n−8
⇒ 5 = n
The number of terms common to the Arithmetic progressions 3, 7, 11, …., 407 and 2, 9, 16, …., 709 is- a)14
- b)21
- c)51
- d)28
Correct answer is option 'A'. Can you explain this answer?
The number of terms common to the Arithmetic progressions 3, 7, 11, …., 407 and 2, 9, 16, …., 709 is
a)
14
b)
21
c)
51
d)
28
|
|
Geetika Shah answered |
First A.P′s sequence is 3,7,11,....,407
General term will be 4k+3 k≤101
Second A.P's sequence is 2,9,16,....,709
General term will be 7p+2 p≤101
The common terms will be 51,79,...,28m + 51
28m + 51 ≤ 407
⟹ 28k ≤ 356
⟹ k ≤ 12.71
And adding the 2 starting number count of the 2 A.P′s i.e. 3 & 2
Number of common terms will be 12 + 2 = 14
General term will be 4k+3 k≤101
Second A.P's sequence is 2,9,16,....,709
General term will be 7p+2 p≤101
The common terms will be 51,79,...,28m + 51
28m + 51 ≤ 407
⟹ 28k ≤ 356
⟹ k ≤ 12.71
And adding the 2 starting number count of the 2 A.P′s i.e. 3 & 2
Number of common terms will be 12 + 2 = 14
If acos x + bsin x = c, then the value of (asin x – bcos x)² is:
- a)a² + b² + c²
- b)a² - b² - c²
- c)a² - b² + c²
- d)a² + b² - c²
Correct answer is option 'D'. Can you explain this answer?
If acos x + bsin x = c, then the value of (asin x – bcos x)² is:
a)
a² + b² + c²
b)
a² - b² - c²
c)
a² - b² + c²
d)
a² + b² - c²
|
|
Preeti Iyer answered |
(acos x + bsin x)² + (asin x – bcos x)² = a² + b²
⇒ c² + (asin x – bcos x)² = a² + b²
⇒ (asin x – bcos x)² = a² + b² – c²
⇒ c² + (asin x – bcos x)² = a² + b²
⇒ (asin x – bcos x)² = a² + b² – c²
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