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The number of values of x in (0, 2pi) wh...

The number of values of `x` in `(0, 2pi)` where the function `f(x) = (tan x + cotx)/2-|(tan x-cotx)/2|` continuous but non-derivable :

A

3

B

4

C

0

D

1

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The correct Answer is:
To find the number of values of \( x \) in the interval \( (0, 2\pi) \) where the function \[ f(x) = \frac{\tan x + \cot x}{2} - \left| \frac{\tan x - \cot x}{2} \right| \] is continuous but not differentiable, we will analyze the function step by step. ### Step 1: Understanding the function The function \( f(x) \) consists of two parts: \( \frac{\tan x + \cot x}{2} \) and \( \left| \frac{\tan x - \cot x}{2} \right| \). We need to determine where the modulus function affects the differentiability of \( f(x) \). ### Step 2: Identify conditions for \( \tan x \) and \( \cot x \) The two cases we need to consider are: 1. When \( \tan x \geq \cot x \) 2. When \( \tan x < \cot x \) ### Step 3: Case 1 - \( \tan x \geq \cot x \) If \( \tan x \geq \cot x \), then: \[ \left| \frac{\tan x - \cot x}{2} \right| = \frac{\tan x - \cot x}{2} \] Thus, the function simplifies to: \[ f(x) = \frac{\tan x + \cot x}{2} - \frac{\tan x - \cot x}{2} = \cot x \] ### Step 4: Case 2 - \( \tan x < \cot x \) If \( \tan x < \cot x \), then: \[ \left| \frac{\tan x - \cot x}{2} \right| = -\frac{\tan x - \cot x}{2} \] Thus, the function simplifies to: \[ f(x) = \frac{\tan x + \cot x}{2} + \frac{\tan x - \cot x}{2} = \tan x \] ### Step 5: Identify points of non-differentiability The function \( f(x) \) will be continuous everywhere except at points where \( \tan x \) and \( \cot x \) are undefined, which occurs at \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \). The points where \( \tan x = \cot x \) occur at: \[ \tan x = 1 \Rightarrow x = \frac{\pi}{4}, \quad \tan x = -1 \Rightarrow x = \frac{5\pi}{4} \] ### Step 6: Summary of non-differentiable points The function \( f(x) \) is not differentiable at the following points in the interval \( (0, 2\pi) \): 1. \( x = \frac{\pi}{4} \) 2. \( x = \frac{3\pi}{4} \) 3. \( x = \frac{5\pi}{4} \) 4. \( x = \frac{7\pi}{4} \) ### Conclusion Thus, there are **4 values of \( x \)** in the interval \( (0, 2\pi) \) where the function \( f(x) \) is continuous but not differentiable.
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VK JAISWAL ENGLISH-CONTINUITY, DIFFERENTIABILITY AND DIFFERENTIATION-EXERCISE (SUBJECTIVE TYPE PROBLEMS)
  1. The number of values of x in (0, 2pi) where the function f(x) = (tan x...

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  2. Let f (x)= {{:(ax (x-1)+b,,, x lt 1),( x+2,,, 1 le x le 3),(px ^(2) +q...

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  3. If y= sin (8 sin ^(-1) x ) then (1-x ^(2)) (d^(2)y)/(dx ^(2))-x (dy)/...

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  4. If y ^(2) =4ax, then (d^(2) y)/(dx ^(2))=(ka ^(2))/( y ^(2)), where k ...

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  5. The number of values of x , x ∈ [-2,3] where f (x) =[x ^(2)] sin (pix)...

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  6. If f (x) is continous and differentiable in [-3,9] and f'(x) in [-2,8]...

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  7. In f (x)= [{:(cos x ^(2),, x lt 0), ( sin x ^(3) -|x ^(3)-1|,, x ge 0)...

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  8. Consider f(x) =x^(2)+ax+3 and g(x) =x+band F(x) = lim( n to oo) (f...

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  9. Let f (x)= {{:(2-x"," , -3 le x le 0),( x-2"," , 0 lt x lt 4):} Then f...

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  10. If f (x) +2 f (1-x) =x ^(2) +2 AA x in R and f (x) is a differentiable...

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  11. Let f (x)= signum (x) and g (x) =x (x ^(2) -10x+21), then the number o...

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  12. If (d^(2))/(d x ^(2))((sin ^(4)x+ sin ^(2)x+1)/(sin ^(2)x + si n x+1))...

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  13. f (x) =a cos (pix)+b, f'((1)/(2))=pi and int (1//2)^(3//2) f (x) dx =2...

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  14. Let alpha (x) = f(x) -f (2x) and beta (x) =f (x) -f (4x) and alpha '(1...

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  15. Let f (x) =-4.e ^((1-x)/(2))+ (x ^(3))/(3 ) + (x ^(2))/(2)+ x+1 and g ...

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  16. If y=3^(2 sin ^(-1)) then |((x ^(2) -1) y^('') +xy')/(y)| is equal to

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  17. Let f (x)=x+ (x ^(2))/(2 )+ (x ^(3))/(3 )+ (x ^(4))/(4 ) +(x ^(5))/(5)...

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  18. In f (x)= [{:(cos x ^(2),, x lt 0), ( sin x ^(3) -|x ^(3)-1|,, x ge 0)...

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  19. For the curve sinx+siny=1 lying in first quadrant. If underset(xrarr0...

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  20. Let f (x) = x tan ^(-1) (x^(2)) + x^(4) Let f ^(k) (x) denotes k ^(th)...

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  21. If x = cos theta and y = sin^(3) theta, then |(yd ^(2)y)/(dx ^(2))+((d...

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