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y=sin x-a sin2x-1/3 sin 3x + 2ax, then y...

`y=sin x-a sin2x-1/3 sin 3x + 2ax,` then y increases for all values of x if

A

`a=1`

B

` a gt 1`

C

` a lt 0`

D

` 0 lt a lt 1`

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To solve the problem of determining the condition on \( a \) for which the function \[ y = \sin x - a \sin 2x - \frac{1}{3} \sin 3x + 2ax \] is increasing for all values of \( x \), we need to follow these steps: ### Step 1: Find the derivative of \( y \) To determine when \( y \) is increasing, we first need to find the derivative \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \cos x - 2a \cos 2x - \sin 3x + 2a \] ### Step 2: Set the condition for \( y \) to be increasing For \( y \) to be increasing for all \( x \), we need \( \frac{dy}{dx} \) to be greater than or equal to 0 for all \( x \): \[ \cos x - 2a \cos 2x - \frac{1}{3} \sin 3x + 2a \geq 0 \] ### Step 3: Analyze the terms The terms \( \cos x \) and \( \sin 3x \) oscillate between -1 and 1. The term \( 2a \) is constant. Therefore, we need to analyze the worst-case scenarios for \( \cos 2x \) and \( \sin 3x \). 1. **Maximum value of \( \cos 2x \)** is 1. 2. **Minimum value of \( \sin 3x \)** is -1. Substituting these values into the inequality gives us: \[ \cos x - 2a(1) - \frac{1}{3}(-1) + 2a \geq 0 \] This simplifies to: \[ \cos x - 2a + \frac{1}{3} + 2a \geq 0 \] Which further simplifies to: \[ \cos x + \frac{1}{3} \geq 0 \] ### Step 4: Find the maximum value of \( \cos x \) The maximum value of \( \cos x \) is 1, thus: \[ 1 + \frac{1}{3} \geq 0 \] This is always true, but we need to ensure that the term involving \( a \) does not violate the condition for all \( x \). ### Step 5: Analyze the minimum value of \( \cos x \) The minimum value of \( \cos x \) is -1. Therefore, we set: \[ -1 + \frac{1}{3} + 2a \geq 0 \] This simplifies to: \[ - \frac{2}{3} + 2a \geq 0 \] ### Step 6: Solve for \( a \) Rearranging gives: \[ 2a \geq \frac{2}{3} \] Dividing both sides by 2: \[ a \geq \frac{1}{3} \] ### Conclusion Thus, the condition for \( y \) to be increasing for all values of \( x \) is: \[ a \geq \frac{1}{3} \]
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ML KHANNA-FUNCTIONS-PROBLEM SET (4)
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  2. The value of a for which the function f (x) = sin x-Cos x-ax+b decreas...

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  3. y=sin x-a sin2x-1/3 sin 3x + 2ax, then y increases for all values of x...

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  4. The set of values of a 'for which the function f (x) = x^2 + ax +1 is ...

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  5. The length of a longest interval in which the function 3 sin x - 4 sin...

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  6. If f(x) = 2x + cot^(-1) x + log ( sqrt(1+x^2 )-x) then f(x)

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  7. Let h(x) = f (x) -(f (x))^2+ (f (x))^3 for every real number x: Then

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  8. If f(x) = x^(3) + bx^(2) + cx+d and 0 lt b^2 lt c, then in (-oo,...

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  9. The function f (x)=2 log (x - 2) - x^2 + 4x +1 increases on the interv...

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  10. On which of the following intervals is the function f(x)=2x^2 - log |...

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  11. y=[ x(x - 3)^2 ] increases for all values of x lying in the ...

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  12. Let f(x)=inte^x (x - 1)(x-2) dx. Then f decreases in the interval

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  13. Let f (x) = x^3 + ax^2 + bx + 5 sin^2 x be an increasing function on t...

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  14. If f(x) =( lamda^2-1)/( lamda^2 +1) x^3 - 3x +5 is a decreasing f...

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  15. Let f (x) = x^3 +6x^2 + px + 2. If the largest possible interval in w...

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  16. If x^(2)/(f(4a))+y^(2)/(f(a^(2)-5)) represents an ellipse with major a...

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  17. The function f (x) = loge (x^3 + sqrt(x^6 +1)) is of the following t...

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  18. The function f(x) = cos (pi / x) is decreasing in the interval

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  19. The function f (x) = sin^4 x+cos^4 x increases if

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  20. Let f (x) = {{:( x^(3) - x^(2) + 10 x- 5 , "," , x le 1), ( -2x + log ...

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