Let f and g be twice differentiable even functions on (-2, 2) such that
and
Then, the minimum number of solutions of
is equal to……….
Let f and g be twice differentiable even functions on (-2, 2) such that and
Then, the minimum number of solutions of is equal to……….
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1 Answer
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f (x) is an even function
So, f (x) has at least four roots in (-2, 2)
So, g (x) has at least two roots in (-2, 2)
now number of roots of f (x)
It is same as number of roots of will have atleast 4 roots in (-2, 2)
Similar Questions for you
Let
So, f(x) = x
Now,
f(x) = αx – b
option (D) satisfies
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f' (0) = f' (6) = f' (2) = f' (4) = f' (5) = f' (1) = 0
and from equation (1) we get f' (3) = -f' (3)
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h (x) = f' (x) . f' (x)
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h (x) = 0 has 13 roots in x
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1 + x? - x? = a? (1+x)? + a? (1+x) + a? (1+x)² . + a? (1+x)?
Differentiate
4x³ - 5x? = a? + 2a? (1+x) + 3a? (1+x)².
12x² - 20x³ = 2a? + 6a? (1+x).
Put x = -1
12 + 20 = 2a? ⇒ a? = 16
Kindly go through the solution
The functional equation f (x+y) = f (x)f (y) implies f (x) = a? for some constant a.
Then f' (x) = a? ln (a).
Given f' (0) = 3, we have a? ln (a) = 3 ⇒ ln (a) = 3 ⇒ a = e³.
So, f (x) = e³?
We need to evaluate the limit: lim (x→0) (f (x)-1)/x = lim (x→0) (e³? -1)/x.
Using the standard limit lim (u→0) (e? -1)/u = 1, we can write:
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