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1. 5/4 2. 7 3. 4 4. 2 Solution: Given that F(x) = ∫ f(t) dt, where integral runs from 0 to x.
By Leibnitz rule, we have
F’(x) = f(x)
But F(x2) = x2 (1+x) = x2 + x3
Hence, this gives F(x) = x + x3/2
So, F’(x) = 1+ 3/2 x1/2
Hence, f(x) = F’(x) = 1+ 3/2 x1/2
Hence, f(4) = 1+ 3/2 . (4)1/2
This gives f(4) = 1+ 3/2 . 2 = 4
Hence, option (3) is correct.
1. ±1 2. ±1/ √2
3. ±1/ 2 4. 0 and 1
Solution: F(x) = ∫√2-t2 dt, where the integral runs from 1 to x
Hence, f’(x) = √2-x2
Another condition that is given in the question is that x2 – f’(x) = 0
Therefore, x2 = √2-x2
Hence, x4 = 2-x2
So, x4 + x2 – 2 = 0
This gives the value of x as ±1.
1. f(1/2) < 1/2 and f(1/3) > 1/3
2. f(1/2) > 1/2 and f(1/3) > 1/3
3. f(1/2) < 1/2 and f(1/3) < 1/3
4. f(1/2) > 1/2 and f(1/3) < 1/3
Solution: The given condition is that ∫ √1- (f’(t))2 dt = ∫ f(t) dt , where integral runs from 0 to x and 0 ≤ x ≤ 1.
Differentiating both sides with respect to x by using the Leibnitz rule, we get
√1- (f’(x))2 = f(x)
This gives f’(x) = ± √1- (f (x))2
∫ [f’(x) / √1- (f (x))2] dx = ± ∫ dx
This yields, sin-1 (f(x)) = ± x + c
Now, putting x = 0 we get,
sin-1 (f(0)) = c
Hence, since f(0) = 0, so we have c = sin-1 (0) = 0
Therefore, f(x) = ± sin x
But f(x) ≥ 0, ∀ x ∈ [0, 1]
Hence, f(x) = sin x
As we know that, sin x < x ∀ x > 0
Hence, sin (1/2) < 1/2 and sin (1/3) < 1/3
This gives, f (1/2) < 1/2 and f (1/3) < 1/3.
Hence, the correct option is (3). Click here to download
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