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## Algebra Level 3 Test 7

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Question 1 |

If pqr = 1, the value of the expression {1/( 1 + p + q

^{-1})} + {1/( 1 +q + r^{-1})} + {1/( 1 +r +p^{-1})} is equal top + q + r | |

1/ ( p +q + r) | |

1 | |

p ^{-1} q^{-1} r^{-1} |

Question 1 Explanation:

A shortcut to solve the question:

Assume p = q = r = 1, then the expression returns the value 1.

None of options other than option C return this value.

Hence, it is the only choice this satisfies the condition.

Assume p = q = r = 1, then the expression returns the value 1.

None of options other than option C return this value.

Hence, it is the only choice this satisfies the condition.

Question 2 |

Let p, q, r, s be four integers such that p+q+r+s = 4m+1 where m is a positive integer. Given m, which one of the following is necessarily true?

The minimum possible value of p ^{2} + q^{2} + r^{2 }+ s^{2} is 4m^{2}–2m+1 | |

The minimum possible value of p ^{2} + q^{2} + r^{2 }+ s^{2} is 4m^{2}+2m+1 | |

The maximum possible value of p ^{2} + q^{2} + r^{2 }+ s^{2} is 4m^{2}–2m+1 | |

The maximum possible value of p ^{2} + q^{2} + r^{2 }+ s^{2} is 4m^{2}+2m+1 |

Question 2 Explanation:

(p + q + r + s)

In that case 2(pq + pr + ps + qr + qs + rs) = 12(m + 0.25)

Thus, the minimum value of p

= (16m

= (16m

= 4m

Since it is an integer, the actual minimum value

= 4m

^{2}= (4m + 1)^{2}Thus, p^{2}+ q^{2}+ r^{2 }+ s^{2}+ 2(pq + pr + ps + qr + qs + rs) = 16m^{2}+ 8m + 1 p^{2}+ q^{2}+ r^{2 }+ s^{2}will have the minimum value if (pq + pr + ps + qr + qs + rs) is maximum. This is possible if p = q = r = s = (m + 0.25) … since p + q + r + s = 4m + 1In that case 2(pq + pr + ps + qr + qs + rs) = 12(m + 0.25)

^{2}= 12m^{2}+ 6m + 0.75Thus, the minimum value of p

^{2}+ q^{2}+ r^{2 }+ s^{2}= (16m

^{2}+ 8m + 1) – 2(ab + ac + ad + bc + bd + cd)= (16m

^{2}+ 8m + 1) – (12m^{2}+ 6m + 0.75)= 4m

^{2}+ 2m + 0.25Since it is an integer, the actual minimum value

= 4m

^{2}+ 2m + 1Question 3 |

Which one of the following conditions must p, q and r satisfy so that the following system of linear simultaneous equations has at least one solution, p +q +r is not equal to 0:

x+ 2y – 3z = p

2x + 6y – 11z = q

x – 2y + 7z = r

x+ 2y – 3z = p

2x + 6y – 11z = q

x – 2y + 7z = r

5p –2q – r = 0 | |

5p + 2q + r = 0 | |

5p + 2q – r = 0 | |

5p – 2q + r = 0 |

Question 3 Explanation:

Considering the expressions in the answer options, we need the value of 5p, 2q and r.

We know the from the conditions given above:

5p= 5x + 10y – 15z

2q= 4x + 12y – 22z

r= x – 2y + 7z

substituting these values in option a, we see that it is satisfied.

None of the other options are satisfied for these values of 5p, 2q and r.

Hence, the condition that is true is option A.

We know the from the conditions given above:

5p= 5x + 10y – 15z

2q= 4x + 12y – 22z

r= x – 2y + 7z

substituting these values in option a, we see that it is satisfied.

None of the other options are satisfied for these values of 5p, 2q and r.

Hence, the condition that is true is option A.

Question 4 |

Let p and q be the roots of the quadratic equation x

^{2}− (α − 2) x − α −1= 0 . What is the minimum possible value of p^{2}+ q^{2}?0 | |

3 | |

4 | |

5 |

Question 4 Explanation:

p + q = α –2 and pq = –α – 1

(p + q)

Thus (α –2)

p

p

p

Thus, minimum value of p

(p + q)

^{2}= p^{2}+ q^{2}+ 2pq,Thus (α –2)

^{2}= p^{2}+ q^{2}+ 2(–α – 1)p

^{2}+ q^{2}= α^{2}– 4α + 4 + 2α + 2p

^{2}+ q^{2}= α^{2}– 2α + 6p

^{2 }+ q^{2}= (α – 1)^{2}+ 5Thus, minimum value of p

^{2}+ q^{2}is 5.Question 5 |

The number of non-negative real roots of 2

^{x }– x – 1 = 0 equals 0 | |

1 | |

2 | |

3 |

Question 5 Explanation:

2

2

If we put x = 0, then this is satisfied and if we put x = 1, then also this is satisfied.

Now we put x = 2, then this equation is not satisfied.

The number of real roots are 2

^{x}– x – 1 = 02

^{x}– 1 = xIf we put x = 0, then this is satisfied and if we put x = 1, then also this is satisfied.

Now we put x = 2, then this equation is not satisfied.

The number of real roots are 2

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