Maths-
General
Easy
Question
Statement-I The equation has exactly one solution in [0, 2
].
Statement-II For equations of type to have real solutions in
should hold true.
- Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I.
- Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for Statement-I.
- Statement-I is true, Statement-II is false.
- Statement-I is false, Statement-II is true.
Hint:
In this question, given two statement. It is like assertion and reason. Statement1 is assertion and statement 2 is reason , Find the statement 1 is correct or not and the statement 2 correct or not if correct then is its correct explanation.
The correct answer is: Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for Statement-I.
Here , we have to find the which statement is correct and if its correct explanation or not.
Firstly ,
Statement-I: The equation √(3 ) cosx- sinx=2 has exactly one solution in [ 0 , 2 π ]
We have,
√(3 ) cosx- sinx=2
⇒√(3 )/2 cosx-1/2 sinx=1
⇒cos〖π/6〗 cosx-sin〖π/6〗 sinx=1 [ since we know that sinπ/3 = √(3 )/2 and cos〖π/3〗 = 1/2 ]
⇒cos〖(π/6+ x )=1〗 [ since ,cos a cosb -sin a sinb = cos ( a + b ) ]
⇒cos〖(x+ π/6)=1〗
⇒ x + π/6 = 2nπ
⇒ x = 2nπ - π/6
For n = 0 , we have x = - π/6 , and ∉ ( 0,2π)
For n = 1 , we have x = 11 π /6 , and ∈ ( 0, 2π)
For n = 2, we have x = 23 π /6 and ∉ ( 0, 2π)
it have only one solution in [ 0 , 2π]
x = 11 π /6
Therefore, the Statement-I is true.
Now,
Statement-II : For equations of type acos θ + bsin θ = c to have real solution in [ 0 , 2π], |c| ≤ √(a2+ b2 ) should hold true.
We have ,
acos θ + bsin θ = c
⇒a /c cos θ + b/c sin θ = 1
Let a/c = sinα and b/c = cosα
We have,
⇒ sinα cos θ + cosα sin θ = 1
⇒ sin(θ + α ) = 1 [ since, sin a cosb + cosa sinb = sin( a+ b)]
Therefore, (θ + α ) is definitely lies in [ 0 , 2π]. Hence statement – II is true. But it is not correct explanation of statement – I .
The correct answer is , Statement-I is true, Statement-II is true ;Statement-II is NOT a correct explanation for Statement-I
In this question, we have to find the statements are the correct or not and statement 2 is correct explanation or not , is same like assertion and reason. Here, Start solving first Statement and try to prove it . Then solve the Statement-II . Remember cos a cosb -sin a sinb = cos ( a + b ) and sin a cosb + cosa sinb = sin( a+ b) .
Related Questions to study
maths-
Assertion : If A is a skew symmetric matrix of order 3 then its determinant should be zero.
Reason : If A is square matrix then det A = det
= det (–
)
Assertion : If A is a skew symmetric matrix of order 3 then its determinant should be zero.
Reason : If A is square matrix then det A = det
= det (–
)
maths-General
maths-
Assertion : There are only finitely many 2 × 2 matrices which commute with the matrix
![open square brackets table attributes columnalign center center columnspacing 1em end attributes row 1 2 row cell negative 1 end cell cell negative 1 end cell end table close square brackets](data:image/png;base64,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)
Reason : If A is non-singular then it commutes with I, adj A and A–1
Assertion : There are only finitely many 2 × 2 matrices which commute with the matrix
![open square brackets table attributes columnalign center center columnspacing 1em end attributes row 1 2 row cell negative 1 end cell cell negative 1 end cell end table close square brackets](data:image/png;base64,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)
Reason : If A is non-singular then it commutes with I, adj A and A–1
maths-General
maths-
Consider the system of equationsx – 2y + 3z = –1–x + y – 2z = kx – 3y + 4z = 1
Assertion : The system of equations has no solution for k
3.and
Reason : The determinant ![open vertical bar table row 1 3 cell negative 1 end cell row cell negative 1 end cell cell negative 2 end cell k row 1 4 1 end table close vertical bar](data:image/png;base64,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)
0, for k
3.
Consider the system of equationsx – 2y + 3z = –1–x + y – 2z = kx – 3y + 4z = 1
Assertion : The system of equations has no solution for k
3.and
Reason : The determinant ![open vertical bar table row 1 3 cell negative 1 end cell row cell negative 1 end cell cell negative 2 end cell k row 1 4 1 end table close vertical bar](data:image/png;base64,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)
0, for k
3.
maths-General
maths-
Assertion : If a, b, c are distinct and x, y, z are not all zero given that ax + by + cz = 0, bx + cy + az = 0, cx + ay + bz = 0 then a + b + c
0
Reason : a2 + b2 + c2 > ab + bc + ca if a, b, c are distinct
Assertion : If a, b, c are distinct and x, y, z are not all zero given that ax + by + cz = 0, bx + cy + az = 0, cx + ay + bz = 0 then a + b + c
0
Reason : a2 + b2 + c2 > ab + bc + ca if a, b, c are distinct
maths-General
maths-
Assertion : If A is a skew symmetric of order 3 then its determinant should be zero.
Reason : If A is square matrix then det A = det
= det (–
)
Assertion : If A is a skew symmetric of order 3 then its determinant should be zero.
Reason : If A is square matrix then det A = det
= det (–
)
maths-General
maths-
Assertion : There are only finitely many 2 ×2 matrices which commute with the matrix ![open square brackets table row 1 2 row cell negative 1 end cell cell negative 1 end cell end table close square brackets](data:image/png;base64,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)
Reason : If A is non-singular then it commutes with I, Adj A and A–1.
Assertion : There are only finitely many 2 ×2 matrices which commute with the matrix ![open square brackets table row 1 2 row cell negative 1 end cell cell negative 1 end cell end table close square brackets](data:image/png;base64,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)
Reason : If A is non-singular then it commutes with I, Adj A and A–1.
maths-General
maths-
Let a matrix A =
& P =
Q = PAPT where PT is transpose of matrix P. Find PT Q2005 P is
Let a matrix A =
& P =
Q = PAPT where PT is transpose of matrix P. Find PT Q2005 P is
maths-General
maths-
Let A =
& I =
and A–1 =
[A2 + cA + dI], find ordered pair (c, d) ?]
Let A =
& I =
and A–1 =
[A2 + cA + dI], find ordered pair (c, d) ?]
maths-General
maths-
= A & | A3 | = 125, then
is -
= A & | A3 | = 125, then
is -
maths-General
maths-
If A =
, B =
and A2 = B, then
If A =
, B =
and A2 = B, then
maths-General
Maths-
Let A be a square matrix all of whose entries are integers. Then which one of the following is true ?
Let A be a square matrix all of whose entries are integers. Then which one of the following is true ?
Maths-General
maths-
Let A =
If |A2| = 25, then |
| equals
Let A =
If |A2| = 25, then |
| equals
maths-General
Maths-
Let A =
and B =
, a, b
N. Then
Let A =
and B =
, a, b
N. Then
Maths-General
Maths-
If A and B are square matrices of size n × n such that A2 – B2 = (A – B) (A + B), then which of the following will be always true –
If A and B are square matrices of size n × n such that A2 – B2 = (A – B) (A + B), then which of the following will be always true –
Maths-General
Maths-
If A =
and I =
, then which one of the following holds for all n
1, by the principle of mathematical induction -
If A =
and I =
, then which one of the following holds for all n
1, by the principle of mathematical induction -
Maths-General