If f(x) =then (1/5) f(0) is equal toa)1b)0c)2d)6Correct answer is option 'B'. Can you explain this answer?

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JEE Exam > JEE Questions > If f(x) =then (1/5) f(0) is equal toa)1b)0c)2...

If f(x) =

then (1/5) f'(0) is equal to

a)
1
b)
0
c)
2
d)
6

Correct answer is option 'B'. Can you explain this answer?

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If f(x) =then (1/5) f(0) is equal toa)1b)0c)2d)6Correct answer is opti...

R₂ → R₂ - R₁, R₃ → R₃ - R₁

On applying determinant formula, we get

2cos⁴x * (0-(-9)) - 2sin⁴x * (-9-0) + (3 + sin²2x) * (9-0)

2cos⁴x * 9 - 2sin⁴x * (-9) + (3 + sin²2x) * 9

18cos⁴x +18 sin⁴x + (3 + sin²2x) * 9

18cos⁴x + 18sin⁴x + 27 + 9sin²2x

We are to find:

(1/5)·f'(0)

Step 1: Differentiate f(x)

We differentiate each term with respect to x:

Term 1: 18cos⁴x

Use the chain rule:
d/dx [cos⁴x] = 4cos³x * (−sinx) = −4cos³x sinx
So:
d/dx [18cos⁴x] = 18 × (−4cos³x sinx) = −72cos³x sinx

Term 2: 18sin⁴x

d/dx [sin⁴x] = 4sin³x * cosx
So:
d/dx [18sin⁴x] = 18 × 4sin³x cosx = 72sin³x cosx

Term 3: 27 → derivative is 0

Term 4: 9sin²(2x)

Use chain rule:
d/dx [sin²(2x)] = 2sin(2x) * cos(2x) * 2 = 4sin(2x)cos(2x)
So:
d/dx [9sin²(2x)] = 9 × 4sin(2x)cos(2x) = 36sin(2x)cos(2x)

Step 2: Combine derivatives

So,

f′(x) = −72cos³x sinx + 72sin³x cosx + 36sin(2x)cos(2x)

Step 3: Evaluate f′(0)

Plug in x = 0:

sin(0) = 0
cos(0) = 1
sin(2×0) = 0
cos(2×0) = 1

Now evaluate each term:

−72cos³(0) sin(0) = −72(1)(0) = 0
72sin³(0) cos(0) = 72(0)(1) = 0
36sin(0)cos(0) = 36(0)(1) = 0

So,

f′(0) = 0

Step 4: Final Answer

(1/5) × f′(0) = (1/5) × 0 = 0

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