Hard Type
Q1. Prove that a2 + b2 + c2 + 2abc < 2 where a, b, c, are sides of  ABC such that a + b + c = 2
Ans: a + b + c = 2
 1 - a + 1 - b + 1 - c = 1
x + y + z = 1 where x = 1 - a, y = 1 - b, z = 1 - c
Sine in
Sum of two sides > third side
a + b > c 0 < c < 1
Similarly 0 < a, b < 1
hence 0 < x, y, z < 1
Now,
a2 + b2 + c2 + 2abc = (1 - x)2 + (1 - y)2 + (1 - z)2 + 2(1 - x)(1 - y)(1 - z)
= 3 - 2(x + y + z) + x2 + y2 + z2 + 2{1 - (x + y + z) + (xy + yz + zx) - xyz}
= 1 + x2 + y2 + z2 - 2xyz + 2(xy + yz + zx)
= 1 + (x + y + z)2 - 2xyz
= 2 - 2xyz < 2 as 0 < x, y, z < 1
 a2 + b2 + c2 + 2abc < 2
Dumb Question: How 0 < c < 1 ?
Ans: Since gn ,
Sum of two sides > third side
a + b > c ............................................. (i)
or a + b + c > 2c
2 > 2c c < 1 ..................................... (ii)
Since c is side of
c > 0 ............................................... (iii)
From (ii) & (iii)
0 < c < 1
Q.2. A function R  R is f(x) =  . Find integral value of  for which f is onto.
Ans: Since f : R R is onto mapping
Range = codomain = R
assumes all real values of x
Let y =
x2( + 8y) + 6x(1 - y) - (8 - y) = 0 v y  R
D  0
36(1 - y)2 + 4( + 8y)(8 + y) 0
[y2(8 + 9) + y(2 + 46) + (8 + 9) 0
ax2 + bx + c 0 v x R if a > 0 & D  0
(2 + 46)2 - 4(8 + 9)(8 + 9) 0 & (8 + 9) > 0
(2 + 46)2) - [2(8 + 9)]2 0 & > - 
(2 + 46 - 16 - 18)(2 + 46 + 16 + 18) 0 & > -
( - 14)( - 2)( + 8)2 0 and > -
[2, 14]  {-8} and > -
[2, 14]
Q.3. Solve the eq. [x]{x} = x
Ans: x = [x] + {x} .................................. (i)
[x]{x} = [x] + {x} {x} =  ............................. (ii)
[x]  1 in eq. (ii)
But if [x] = 1, then
{x} = x which is donje only when,
x [0, 1) ................................................. (iii)
& [x] = 1 [1, 2) ................................. (iv)
From (iii) & (iv) no value of x,
when [x] = 1
As we know
{x} [0, 1)
0 < 1
< 1 & 0
 < 0 & {[x] 0 or [x] > 1}
[x] < 1 & {[x] 0 or [x] > 1}
[x] 0
x = [x] + {x} = [x] +
x =  , where x takes values less than 1.
x = , where x < 1
x = , where [x] is any non positive integer.
Q4. Sketch y = (x - 1)(x - 2)
Ans: y = (x - 1)(x - 2)
(i) put y = 0 x = 1, 2
(ii) y = x2 - 3x + 2
 = 2x - 3 & 
as > 0
minimaq at x = 3/2.
(iii) Increases when x > 3/2 & decreases when x < 3/2.
y = x2 - 3x + 2
when x = 3/2
y = 
Q5. Sketch curve y = (x - 1)(x - 2)(x - 3)
Ans: y = (x - 1)(x - 2)(x - 3)
(i) Put y = 0 x = 1, 2, 3
(ii) y = x3 - 6x2 + 11x - 6
= 3x2 - 12x + 11 & = 6x - 12
when = 0 x = 
Maxima when x =  as = - 2
Minima when x =  as = 2
(iii) = 3x2 - 12x + 11
= 3(x - )(x - )
> 0 or Increases when
x < or x >
Decreases when < 0
or < x <
(iv) Concave upwards when x > 2 & concave down when x < 2.
x = 2 is point where concavity of curve changes.
Key Words:
* Function.
* Domain.
* Codomain.
* Range.
* Identity Function.
* Constant Function.
* Logarithmic Function.
* Modulus Function.
* Signum Function.
* Greatest Integer Function.
* Fractional Part Function.
* Odd & Even Function.
* Periodic Function.
* Composite Function.
* Mapping.
* Bijective.
* Surjective.
* Injective.
* Inverse of Function. |
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