ML AGGARWAL CLASS 12 Chapter 6 Applications of derivative Exercise 6.2 (CBSE)

applied math 6.1 ml aggarwal class 12 

 Exercise 6.1

Question 1

Solve for x

(i) x(x-2)(x-5)(x+3)>0

Sol :

Here, x=0,2,5,3

Let's check sign's of intervals 

-∞ and -3

Take any value between -∞ and -3

let say -10

and put it in x(x-2)(x-5)(x+3)

⇒-10(-10-2)(-10-5)(-10+3)

⇒-10(-12)(-15)(-7)

⇒12600

which means overall sign is (+ve) in interval -∞ and -3

In the same way, we going to find sign's for intervals

-3 and 0 (-ve)

0 and 2 (+ve)

2 and 5 (-ve)

5 and ∞ (+ve)

Since, the function is greater than 0, it include all positive intervals

So, x∈(-∞,-3)∪(0,2)∪(5,∞)

(ii) $x^{4}-5 x^{2}+4 \geqslant 0$

Sol :

Let $x^2=t$

$t^{2}-5 t+4 \geqslant 0$

$t^{2}-4 t-t+4 \geqslant 0$

(t-4)(t-1)≥0

$\left(x^{2}-4\right)\left(x^{2}-1\right) \geqslant 0$

(x-2)(x+2)(x-1)(x+1)≥0

So, x=2,-2,1-1

Since, the function is greater than 0, it include all positive intervals

x∈(-∞,-2]∪[-1,1]∪[2,∞)

Question 2

Find all real values of n :

(i) $x^{3}(x-1)(x-2)>0$

Sol :

Here , x=0,1,2

Since, the function is greater than 0, it include all positive intervals

Solution set : x∈(1,0)∪(2,∞)

(ii) $x^{2}(x-1)(x-2) \leq 0$

Sol :

Here, x=0,1,2

For (-∞,0) from inequality it is +ve

For (0,1) from inequality it is +ve

For (1,2) from inequality it is -ve

For (2,∞) from inequality it is +ve

Since, the function is less than 0, it include all negative intervals. Also, we include zero as well.

Solution set : x∈[1,2]∪{0}

Question 3

Solve for x

(i) $\frac{1}{x-2} \leq 1$

Sol :

$\frac{1}{x-2} \leq 1$ , x≠2

Multiplying both side by $(x-2)^{2}$

$(x-2)^{2} \times \frac{1}{(x-2)} \leq 1 \times(x-2)^{2}$

$(x-2) \leqslant x^{2}-4 x+4$

$0 \leq x^{2}-5 x+6$

$x^{2}-2x-3x+4 \geq 0$

(x-2)(x-3)≥0

Here, x=2,3

For (-∞,2) from inequality it is +ve

For (2,3) from inequality it is -ve

For (3,∞) from inequality it is +ve

Since, the function is greater than 0, it include all positive intervals

Solution set : x∈(-∞,2)∪[3,∞)

(ii) $\frac{(x+1)(x-3)}{(x+2)} \geqslant 0$

Sol :

$\frac{(x+1)(x-3)}{(x+2)} \geqslant 0$, x≠-2

Multiplying both side by (x+2)

(x+1)(x-3)(x+2)≥0

Here, x=-1,+3,-2

For (-∞,-2) from inequality it is -ve

For (-2,-1) from inequality it is +ve

For (-1,3) from inequality it is -ve

For (3,∞) from inequality it is +ve

Since, the function is greater than 0, it include all positive intervals

Solution set : x∈(-2,-1]∪[3,∞)

Question 1