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Answer 1

Final Answer : 4 : A) [-1,3]
5 : D) ±2

Understanding :4
1) Transformation of modulus graphs.

Method - 1 (see pic)
Method -2 (see below)
We know that,
if |f(x)| <= 1
f(x)=|x-1|-1
then,
-1<=f(x) <=1
=> -1<= |x-1|-1<=1
=> -1+1<=|x-1|<=1+1
=> 0<=|x-1|<=2
Since, MOD is always positive
=> |x-1|<=2
=> -2<=x-1<=2
=> -2+1<=x<= 2+1
=> -1<=x<=3
----------------------------
Understanding and steps :5
1) 4+√15 = 1/(4-√15)
Put (4+√15) ^x = t

For Calculation see pic

Answer 2

Given :

(2)

|x - 1| - 1| <= 1

= > |x - 1| <= 2

= > x - 1 <= 2, x - 1 >= -2

= > x <= 3, x >= -1


Therefore it lies in the range of [-1,3]




(3)

$= \ \textgreater \ (4 + \sqrt{15} )^x + (4 - \sqrt{15} )^x = 62$

$= \ \textgreater \ (4 + \sqrt{15} ) + (4 + \sqrt{15}) ^{-1} = 62$

Let a = $4 + \sqrt{15}$  ---- (1)

Now,

= > a + a^-1 = 62

= > a + 1/a = 62

= > a^2 + 1 = 62a

= > a^2 - 62a + 1 = 0

Now,

On solving with ax^2 + bx + c = 0, we get a = 1, b = -62, c = 1

x = -b + root b^2 - 4ac/(2a)  ----- (1)

x = -b - root b^2 - 4ac/2a   ----- (2)


On solving (1), we get

$= \ \textgreater \ \frac{-(-62) + \sqrt{(-62)^2 - 4(1)(1)} }{2(1)}$

$= \ \textgreater \ \frac{62 + \sqrt{3840} }{2}$

$= \ \textgreater \ \frac{2(8 \sqrt{15} + 31) }{2}$

$= \ \textgreater \ 8 \sqrt{15} + 31$


Therefore the 2nd equation will be : $31 - 8 \sqrt{15}$


Now,

Substitute in (1), we get

$= \ \textgreater \ (4 + \sqrt{15} )^x = (8 \sqrt{15} + 31 )$

$= \ \textgreater \ (4 + \sqrt{15} )^x = (4 + \sqrt{15})^2$

x = 2.


Substitute in (1), we get:

$= \ \textgreater \ (4 + \sqrt{15} )^x = 31 - 8 \sqrt{5}$

$= \ \textgreater \ (4 + \sqrt{15} )^x = (4 + \sqrt{15}) ^{-2}$

x = -2.



Therefore x = +2, - 2.



Hope this helps!