CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Higher Order Derivatives

If the functi...

Question

If the function f(x) = 2 tan x + (2a + 1) log_e | sec x | + (a − 2) x is increasing on R, then
(a) a ∈ (1/2, ∞)
(b) a ∈ (−1/2, 1/2)
(c) a = 1/2
(d) a ∈ R

Open in App

Solution

$f (x) = 2 \tan x + (2 a + 1) \log_{e} |\sec x| + (a - 2) x When \sec x > 0 \Rightarrow |\sec x| = \sec x \frac{d}{d x} \{f (x)\} = 2 \sec^{2} x + (2 a + 1) \frac{1}{\sec x} \times \sec x \tan x + (a - 2) = 2 \sec^{2} x + (2 a + 1) \tan x + (a - 2) For f (x) to be increasing, 2 s e c^{2} x + (2 a + 1) \tan x + (a - 2) ⩾ 0 \Rightarrow 2 + 2 \tan^{2} x + (2 a + 1) \tan x + a - 2 ⩾ 0 \Rightarrow 2 \tan^{2} x + (2 a + 1) \tan x + a ⩾ 0 Its discriminant ⩽ 0 [∵ a x^{2} + b x + c ⩾ 0 \Rightarrow b^{2} - 4 a c ⩽ 0] \Rightarrow {(2 a + 1)}^{2} - 4.2 . a ⩽ 0 \Rightarrow 4 a^{2} - 4 a + 1 ⩽ 0 \Rightarrow {(2 a - 1)}^{2} ⩽ 0 {(2 a - 1)}^{2} < 0 cannot be possible . ∴ {(2 a - 1)}^{2} = 0 \Rightarrow a = \frac{1}{2} When \sec x < 0 \Rightarrow |\sec x| = - \sec x \frac{d}{d x} \{f (x)\} = 2 \sec^{2} x + (2 a + 1) \frac{1}{- \sec x} \times \sec x \tan x + (a - 2) = 2 \sec^{2} x - (2 a + 1) \tan x + (a - 2) For f (x) to be increasing, 2 s e c^{2} x - (2 a + 1) \tan x + (a - 2) ⩾ 0 \Rightarrow 2 + 2 \tan^{2} x - (2 a + 1) \tan x + a - 2 ⩾ 0 \Rightarrow 2 \tan^{2} x - (2 a + 1) \tan x + a ⩾ 0 Its discriminan t ⩽ 0 [∵ a x^{2} + b x + c ⩾ 0 \Rightarrow b^{2} - 4 a c ⩽ 0] \Rightarrow {\{- (2 a + 1)\}}^{2} - 4.2 . a ⩽ 0 \Rightarrow 4 a^{2} - 4 a + 1 ⩽ 0 \Rightarrow {(2 a - 1)}^{2} ⩽ 0 {(2 a - 1)}^{2} < 0 cannot be possible . ∴ {(2 a - 1)}^{2} = 0 \Rightarrow a = \frac{1}{2}$

Suggest Corrections

thumbs-up

0

Similar questions

f : R \to R^{+}

f (x + y) = f (x) f (y) \forall x, y \in R, f (0) = 1, f^{'} (0) = 2

Q. The range of the function

f (x) = \frac{x^{2} - x}{x^{2} + 2 x}

is
(a) R
(b) R − {1}
(c) R − {−1/2, 1}
(d) None of these

a, b \in R, (a \neq 0) .

If the function

f

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{2 x^{2}}{a}, 0 \leq x < 1 a, 1 \leq x < \sqrt{2} \frac{2 b^{2} - 4 b}{x^{3}}, \sqrt{2} \leq x < \infty \end{matrix}

is continuous in the interval

[0, \infty),

,then an ordered pair

(a, b)

f : R \to R

is a differentiable function such that

f^{'} (x) > 2 f (x)

x \in R

f (0) = 1,

Q. f a function

f : [- 2 a, 2 a] \to R

is an odd function such that

f (2 a - x) = f (x), \forall x ϵ [a, 2 a]

and left hand derivative at x = a is 0 then find left hand derivative at x = -a

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Higher Order Derivatives

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Higher Order Derivatives

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon