Quiz 3

Solution:

f(x) = |x − 2| + |2.5 − x| + |3.6 − x| = g(x) + |2.5 – x|, where g(x) = |x − 2| + |3.6 − x|

When 2 ≤ x ≤ 3.6, g(x) attains a fixed value.

This happens as in this range |x – 2| = x – 2 and |3.6 – x| = 3.6 – x

|x − 2| + |3.6 − x| = x – 2 + 3.6 – x = 1.6

 

When x < 2, x – 2 < 0, |x – 2| > 0

Also, as –x > –2, 3.6 – x > 3.6 – 2

3.6 – x > 1.6

|x − 2| + |3.6 − x| > 1.6

 

Similarly, when x > 3.6,

|3.6 – x| > 0 and |x – 2| > 1.6

|x − 2| + |3.6 − x| > 1.6

Thus we can say that g(x) has the minimum value in the range 2 ≤ x ≤ 3.6

As f(x) = g(x) + |2.5 – x|,  f(x) attains the minimum value when 2 ≤ x ≤ 3.6 and |2.5 – x| is minimum.

This happens when x = 2.5  f(x) attains minimum when x = 2.5 Hence, option 2.

 

Alternatively, f(x) = |x − 2| + |2.5 − x| + |3.6 − x| Substituting the value of x from the given options in the function,  when x = 2.3 f(x) = 0.3 + 0.2 + 1.3 = 1.8

 

when x = 2.5 f(x) = 0.5 + 0 + 1.1 = 1.6

 

when x = 2.7 f(x) = 0.7 + 0.2 + 0.9 = 1.8

 

Substituting any arbitrary real values of x in f(x), when x = 2 f(x) = 0 + 0.5 + 1.6 = 2.1

 

when x = 3 f(x) = 1 + 0.5 + 0.6 = 2.1

 

For any other value of x, f(x) will be greater than 1.6 Hence, f(x) is minimum when x = 2.5

Hence, option 2 is the correct one.

By option, if r = $\frac{1}{2}$

2a-b-c = 0

2b- c- a= 0

2c-b -a = 0

2(a+b+c) - (a+b+c)- (a+b+c)= 0

similarly r= -1 is also satisfied.

 

Solution-

y = $\frac{1}{2+\frac{1}{3+\frac{1}{2+\frac{1}{3+....}}}}$

Solution-

$\frac{1}{2+\frac{1}{3+y}}$=> y = \frac{3+y}{2y+7}

$2y^{2}+7y+3+y\Rightarrow 2y^{2}+6y-3=0$

$y = \frac{-6\pm \sqrt{4-2-3}}{4}= \frac{-60\pm \sqrt{60}}{4}= \frac{\sqrt{15}-3}{2}$