Section - A

Q1) For what value of k will the following system of linear equations have an infinite number of solutions: 2x + 3y = 2; (k + 2)x + (2k + 1)y = 2(k - 1) ?  (Marks 2)
Ans1)  a₁/a₂ = b₁/b₂ = c₁/c₂
= 2/(k + 2) = 3/(2k + 1) = 2/(2(k - 1))
From (i) & (ii) and From (iii) & (iv)

.^.. k = 4

Q2) Find whether the number 6, 10, 14 and 22 are in proportional or not. If not what number be added to each number so that they become proportional ?  (Marks 2)
Ans2) Since 6 x 22 10 x 14
= 6, 10, 14 and 22 are not in proportional.
Let the number added be x
.^.. (6 + x)(22 + x) = (10 + x)(14 + x)
= 132 + 6x + 22x + x² = 140 + 10x + 14x + x²
= 132 + 28x = 140 + 24x
= 4x = 8
= x = 2

Q3) Reduce the following relation expression into lowest form:
(x⁴ - 10x² + 9)/(x³ + 4x² + 3x)  (Marks 2)
Ans3) x⁴ - 10x² + 9 = (x + 3)(x - 3)(x + 1)(x - 1)
x³ + 4x² + 3x = x(x + 3)(x + 1)
... (x⁴ - 10x² + 9)/(x³ + 4x² + 3x) = ((x + 3)(x - 3)(x + 1)(x - 1))/(x(x + 3)(x + 1))
= (x² - 4x + 3)/x

Q4) Solve for x and y
ax + by = a - b  - (i)
bx - ay = a + b  - (ii)  (Marks 2)
Ans4) Multiply (i) by a and (ii) by b and add
a²x + aby = a² - ab
b²x - aby = ab + b²
(a² + b²)x = a² + b²
= x = 1
subs x = 1 in (i)
a + by = a - b
y = -1
.^.. solution is x = 1, y = -1

Q5) Find the value of k such that sum of the roots of the quadratic equation
3x² + (2k + 1)x - (k + 5) = 0 is equal to the product of its roots.  (Marks 2)
Ans5)  Sum of the roots = -(2k + 1)/3
Product of the roots = -(k + 5)/3
.^.. -(2k + 1)/3 = -(k + 5)/3
= -2k - 1 = -k - 5
= 4 = k

Q6) Find two consecutive numbers, whose square have sum 85.  (Marks 2)
Ans6) Let the two consecutive number be x and x + 1
.^.. x² + (x + 1)² = 85
= x² + x² + 2x + 1 = 85
= 2x² + 2x - 84 = 0
x² + x - 42 = 0
(x + 7)(x - 6) = 0
x = -7 or x = 6
Two numbers are -7, -6 or 6, 7

Q7) Without using trignometric table, show that:
tan 7^o . tan 23^o . tan 60^o . tan 67^o . tan 83^o = 3  (Marks 2)
Ans7) LHS = tan 7^o . tan 23^o . tan 60^o . tan 67^o . tan 83^o 
= tan 7^o . tan 23^o . 3 . cot(90^o - 67^o) . cot (90^o - 83^o)  ^..^. tan 60^o = 3, tan = cot (90^o - ) 
= tan 7^o . tan 23^o . 3 . cot 23^o . cot 7^{o   .}.^. cot = 1/tan
= tan 7^o . tan 23^o . 3 . 1/tan 23^o . 1/tan 7^o 
= 3 = RHS

Q8) Show that (sin - 2sin³ )/(2cos³ - cos ) = tan .  (Marks 2)
Ans8) LHS = (sin - 2sin³ )/(2cos³ - cos )
= (sin (1 - 2sin² ))/(cos (2cos² - 1)
= tan = RHS   ^..^. cos² - sin² = 1 - 2sin² = 2cos² - 1 and sin² + cos² = 1

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