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General
Easy
Question
If then the equation whose roots are



 none of these
Hint:
The definition of a quadratic as a seconddegree polynomial equation demands that at least one squared term must be included. It also goes by the name quadratic equations. Here we have to find the equation whose roots are .
The correct answer is:
A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given
Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, the equation is .
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Let p, q {1, 2, 3, 4}. Then number of equation of the form px^{2} + qx + 1 = 0, having real roots, is
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The cartesian equation of is
A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as .
Now we know that:
.
So applying this, we get:
Now lets substitute x=$rcosθ andy=rsinθ, we get:$
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as .
Now we know that:
.
So applying this, we get:
Now lets substitute x=$rcosθ andy=rsinθ, we get:$
The cartesian equation of is
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A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as .
Now we know that:
.
So applying this, we get:
Now lets substitute x=$rcosθ andy=rsinθ, we get:$
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as .
Now we know that:
.
So applying this, we get:
Now lets substitute x=$rcosθ andy=rsinθ, we get:$
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A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as .
Now we know that:
.
So applying this, we get:
Now lets substitute x=$rcosθ andy=rsinθ, we get:$
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as .
Now we know that:
.
So applying this, we get:
Now lets substitute x=$rcosθ andy=rsinθ, we get:$
The castesian equation of is
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A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as .
Now we know that:
.
So applying this, we get:
Now lets substitute x=$rcosθ andy=rsinθ, we get:$
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as .
Now we know that:
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So applying this, we get:
Now lets substitute x=$rcosθ andy=rsinθ, we get:$
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Which of the following curves represents correctly the oscillation given by
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A is a set containing n elements. A subset P_{1} is chosen, and A is reconstructed by replacing the elements of P_{1}. The same process is repeated for subsets P_{1}, P_{2}, … , P_{m}, with m > 1. The Number of ways of choosing P_{1}, P_{2}, …, P_{m} so that P_{1} P_{2} … P_{m}= A is 
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For each a_{i} (1 i n), either a_{i} P_{j} or a_{i} P_{j} (1 j m) . Thus, there are 2^{m} choices in which a_{i} (1 j n) may belong to the P_{j} s.
Also there is exactly one choice, viz., a_{i} P_{j} for j = 1, 2, …, m, for which a_{i} P_{1} P_{2} ... P_{m}.
Therefore, a_{i} P_{1} P_{2} …. P_{m} in (2^{m} – 1) ways . Since there are n elements in the set A, the number of ways of constructing subsets
P_{1}, P_{2}, ….. , P_{m} is (2^{m} – 1)^{n}
For each a_{i} (1 i n), either a_{i} P_{j} or a_{i} P_{j} (1 j m) . Thus, there are 2^{m} choices in which a_{i} (1 j n) may belong to the P_{j} s.
Also there is exactly one choice, viz., a_{i} P_{j} for j = 1, 2, …, m, for which a_{i} P_{1} P_{2} ... P_{m}.
Therefore, a_{i} P_{1} P_{2} …. P_{m} in (2^{m} – 1) ways . Since there are n elements in the set A, the number of ways of constructing subsets
P_{1}, P_{2}, ….. , P_{m} is (2^{m} – 1)^{n}
A is a set containing n elements. A subset P_{1} is chosen, and A is reconstructed by replacing the elements of P_{1}. The same process is repeated for subsets P_{1}, P_{2}, … , P_{m}, with m > 1. The Number of ways of choosing P_{1}, P_{2}, …, P_{m} so that P_{1} P_{2} … P_{m}= A is 
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For each a_{i} (1 i n), either a_{i} P_{j} or a_{i} P_{j} (1 j m) . Thus, there are 2^{m} choices in which a_{i} (1 j n) may belong to the P_{j} s.
Also there is exactly one choice, viz., a_{i} P_{j} for j = 1, 2, …, m, for which a_{i} P_{1} P_{2} ... P_{m}.
Therefore, a_{i} P_{1} P_{2} …. P_{m} in (2^{m} – 1) ways . Since there are n elements in the set A, the number of ways of constructing subsets
P_{1}, P_{2}, ….. , P_{m} is (2^{m} – 1)^{n}
For each a_{i} (1 i n), either a_{i} P_{j} or a_{i} P_{j} (1 j m) . Thus, there are 2^{m} choices in which a_{i} (1 j n) may belong to the P_{j} s.
Also there is exactly one choice, viz., a_{i} P_{j} for j = 1, 2, …, m, for which a_{i} P_{1} P_{2} ... P_{m}.
Therefore, a_{i} P_{1} P_{2} …. P_{m} in (2^{m} – 1) ways . Since there are n elements in the set A, the number of ways of constructing subsets
P_{1}, P_{2}, ….. , P_{m} is (2^{m} – 1)^{n}
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x k –k x k ….(1)
& y k –k y k ….(2)
& x – y k y – x k –k y – x k x – k y x + k ….(3)
Number of points having integral coordinates
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& y k –k y k ….(2)
& x – y k y – x k –k y – x k x – k y x + k ….(3)
Number of points having integral coordinates
= (2k + 1)^{2} – 2
= (3k^{2} + 3k + 1).
The number of points in the Cartesian plane with integral coordinates satisfying the inequalities x k, y k, x – y k ; is
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& y k –k y k ….(2)
& x – y k y – x k –k y – x k x – k y x + k ….(3)
Number of points having integral coordinates
= (2k + 1)^{2} – 2
= (3k^{2} + 3k + 1).
& y k –k y k ….(2)
& x – y k y – x k –k y – x k x – k y x + k ….(3)
Number of points having integral coordinates
= (2k + 1)^{2} – 2
= (3k^{2} + 3k + 1).