Sol : Option C
Let α,β are the roots of the equation, then
α+β = -w/6 ,αβ = 1/6 , α-β = 1/6 and w>0
Solving α= (-w+1)/12 , β = (-w-1)/12,
Therefore, (1-w)/12 × (-1-w)/12 = 1/6
Solving w= ± 5 as w>0, w= 5
Thus, option C is the answer
Q2. If the roots x1 and x2 of the quadratic equation satisfy the condition 7x2 – 4x1 = 47. The quadratic expression is x2 – 2x + c = 0, find the value of c.
A. – 15
B. 15
C. – 6
D. None of these
Sol : Option A x1 + x2= + 2/1 = + 2 ............ (1),
Also – 4x1 + 7x2= 47 ......... (2)
Solving (1) & (2) we get x 1 = -3 and x2= 5
∴ x1x2 = -15 = c
Q.3. If (x2 – y2) = 16 and xy = -15. Which of the following is a possible value of (x + y), if (x + y) is a positive number.
A. 3
B. 2
C. 5
D. None of these
Sol : Option B
x2 – y 2 = 16 and xy = -15, (x + y) >0 =?
x= -15/y, (-15/y)2 – y 2 = 16 , 225/y2- y2 = 16
y4 + 16y2 – 225 = 0
y2= -25, 9 or y = ± 3(avoiding complex roots)
Putting values x = ±5 for y = 3 and x = ±5 for y = -3
Therefore (x+y) = 5+3 = 8, -5 +3 =-2
Or (x+y) = 5-3 =2, -5 -3 = -8
Therefore, x+y = 2 or 8 (because x+y > 0).
Q.4. From any two numbers x and y, we define x * y = x + 0.5 y – xy. Suppose that both x and y are greater than 0.5, then x * x < y ×y if :
A. 1 > x > y
B. x > 1 > y
C. y > 1 > x
D. 1 > y > x
Sol : Option D x * x = x + 0.5 x – x2 = 1.5 x – x2 Y * y = y + 0.5 y – y2 = 1.5 y – y2
So if x * x < y * y, then 1.5 x – x2 < 1.5 y – y2
< 1.5 (y - x) or (y – x) x (y + x) < 1.5 (y – x). Now this is valid only if (y – x) is not equal to 0. Also, if(y – x) is a negative number, the equality sign will change.
To maintain the inequality, y – x has to be > 0, i.e. y > x.
And if (y – x) > 0, we can cancel out this factor without changing the equality sign.
Therefore, we have (x + y) < 1.5
Since x and y are both greater than 0.5, (x + y) < 1.5 only if both x and y are less than 1.
Hence we have 1 > y > x.
Q.5. The sum of the squares of two consecutive natural numbers is 85. Find those numbers.
A. 6, 7
B. 5, 8
C. 6, 8
D. -8, 6
Sol : Option A
Two consecutive natural nos. are x and x + 1.
⇒ x2 + (x + 1)2= 85 ⇒ x = 6. ∴nos. are 6 and 7.
Q.6. The sum of the squares of two consecutive odd natural numbers is 130. Find those numbers.
A. 3, 5
B. 7, 9
C. 5, 7
D. 7, -9
Sol : Option B
Two consecutive odd natural nos. are x and x + 2
⇒ x2 + (x + 2)2 = 130 ⇒ x = 7 ∴ nos. are 7 & 9.
Q.7. The sum of a natural number and its reciprocal is 50/7. What is the number?
A. 7
B. 10
C. 1/5
D. 5
Sol : Option A
x+ 1/x = 50/7 ⇒ x = 7
Q8. The difference between a natural number and twice its reciprocal is 47/7. What is that number?
A. 8
B. 9
C. 7
D. 1/7
Sol : Option C
x- 2/x = 47/7 ⇒ x = 7
Q9. The length of a rectangular field is less than twice its breadth by 5 metres. Its area is 700 square metres. Find the length and breadth of the field.
_10.jpg?null)
_3.jpg?null)
-icon.jpg?null&itok=2FjKP8-O "Chain Rule Practice Problems"){kind=icon}
