Find the intervals of monotonicity of the function: y = 10 4 x 3 - 9 x 2 + 6 x

Given, y = 10 4 x 3 - 9 x 2 + 6 x d y d x = - 10 12 x 2 - 18 x + 6 4 x 3 - 9 x 2 + 6 x 2 d y d x = - 60 x - 1 2 x - 1 4 x 3 - 9 x 2 + 6 x 2 d y d x = - 60 ( x - 1 ) ( 2 x - 1 ) x 2 4 x 2 - 9 x + 6 2 Now, discriminant D of 4 x 2 - 9 x + 6 is 81 - 4 ( 4 ) ( 6 ) = - 15 < 0 and a > 0 So, the quadratic equation 4 x 2 - 9 x + 6 always has positive value Now, function decreases if d y d x < 0 ⇒ - 60 x - 1 2 x - 1 x 2 < 0 ⇒ x - 1 2 x - 1 x 2 > 0 ; x ≠ 0 x ∈ - ∞ , 0 ∪ 0 , 1 2 ∪ 1 , + ∞ Hence,for x ∈ 1 2 , 1 it monotonically increases and for x ∈ - ∞ , 1 2 ∪ 1 , + ∞ - 0 it monotonically decreases.

Find the value of x > 1 for which the function F x = ∫ x x 2 1 t l n t - 1 32 d t is increasing and decreasing.

Given, F x = ∫ x x 2 1 t ln t - 1 32 d t By Newton's Leibnitz rule ⇒ F ' x = 1 x 2 ln x 2 - 1 32 × 2 x - 1 x ln x - 1 32 × 1 = 1 x ln x - 1 2 x + 1 2 32 2 × 32 x - 1 = 1 x ln x - 1 x + 1 2 32 To find critical points, we put F ' ( x ) = 0 ⇒ 1 x ln x - 1 x + 1 2 32 = 0 ⇒ x - 1 x + 1 2 = 32 ⇒ x 2 - 1 x + 1 = 32 ⇒ x 3 + x 2 - x - 33 = 0 ⇒ x - 3 x 2 + 4 x + 11 = 0 ⇒ x - 3 = 0 or x 2 + 4 x + 11 = 0 But x 2 + 4 x + 11 = 0 doesn't have real roots. And x - 3 = 0 ⇒ x = 3 Now, if 1 < x < 3 ⇒ F ' x < 0 And x > 3 ⇒ F ' x > 0 Hence, F x increasing in x ∈ 3 , ∞ & decreasing on x ∈ 1 , 3 .

If f x = a 2 - 1 3 x 3 + a - 1 x 2 + 2 x + 1 is monotonic increasing for every x ∈ R , then find the range of values of ' a ' .

Given, f x = a 2 - 1 3 x 3 + a - 1 x 2 + 2 x + 1 For a = 1 f x = 2 x + 1 ⇒ f is monotonic increasing If a ≠ 1 Differentiating f ( x ) w.r.t. x , we get f ' x = a 2 - 1 x 2 + 2 a - 1 x + 2 f ' x ≥ 0 ( ∴ ' f ' is monotonic increasing.) We know a quadratic equation is always positive if D ≤ 0 , a > 0 ⇒ D ≤ 0 & a 2 - 1 > 0 ⇒ 4 a - 1 2 - 8 a - 1 a + 1 ≤ 0 ⇒ a - 1 a - 1 - 2 a - 2 ≤ 0 ⇒ a - 1 a + 3 ≥ 0 ⇒ a ∈ ( - ∞ , - 3 ] ∪ [ 1 , ∞ )