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The number of points at which the function f(x) = max. {a – x, a + x, b} – ∞ < x < ∞, 0 < a < b cannot be differentiable is
A function is not differentiable at the points where it has sharpcorner point
Clearly above function has two sharp corner, so it is not differentiable at these two points,
Other than these points, function linear continuous, so it is diferentiable .
Hence above function is not differentiable at two points
The function f(x) is defined by f(x) =
lt x1^+ f(x) = log(4x 3) (x^{2} 2x + 5)
= ln(x^{2}  2x +5)/ln(4x3)
lt(h → 0) ln(1+h)^{2}  2(1+h) + 5)/ln(4(1+h)  3)
lt(h → 0) ln(1+h^{2} + 2h  2 2h + 5)/ln(4 + 4h  3)
ln(h → 0) ln(4 + h^{2})/(1+4h)
Divide and multiply the denominator by 4h
ln(h → 0) ln(4 + h^{2})/[((1+4h)/4h) * 4h]
As we know that (1+4h)/4h = 1
ln 4/(4*0) = + ∞ (does not exist)
lt x→ 1^ f(x) = log(4x 3) (x^{2} 2x + 5)
lt(h → 0) ln(1h)^{2}  2(1h) + 5)/ln(4(1h)  3)
lt(h → 0) ln(1+h^{2}  2h  2 + 2h + 5)/ln(4  4h  3)
ln(h → 0) ln(4 + h^{2})/(1  4h)
Divide and multi[ly the denominator by (4h)
ln(h → 0) ln(4 + h^{2})/[((1+4h)/4h) * (4h)
As we know that (14h)/(4h) = 1
ln 1/(4*0) =  ∞ (does not exist)
A point where function f(x) is not continuous where f(x) = [sin [x]] in (0, 2π) ; is ([ * ] denotes greatest integer ≤ x)
Let f(x) = [n + p sin x], x ∈ (0, π), n ∈ I and p is a prime number. Then number of points where f(x) is not differentiable is (where [ * ] denotes greatest integer function)
If f is a realvalued differentiable function satisfying f(x) – f(y) ≤ (x – y)^{2}, x, y ∈ R and f(0) = 0, then f(1) equals
If f(x) = , then f([2x]) is (where [ * ] represent greatest integer function)
The value of f(0), so that the function, f(x)= becomes continuous for all x, is given by
f(x)={(a^{2}−ax+x^{2})−(a^{2}+ax+x^{2})^1/2}/{(a+x (ax)^{1/2})^{1/2}}
on rationalising, we get
lom(x → 0) a^{2}  ax + x^{2} (a^{2} +ax + x^{2}) * (a+x)^{1/2} + (ax)^{1/2}]/[(a+x)(ax) * (a^{2}  ax + x^{2})^{1/2} + (a^{2} + ax + x^{2})]
lim(x → 0) [2ax * [(a)^{1/2} + (a)^{1/2}]]/[2x * (a^{2})^{1/2} * (a^{2})^{1/2}]
a/(a)^{1/2} = (a)^{1/2}
If f(x) = then f(x) is (where { * } represents the fractional part function)
In order that function f (x) = (x + 1)^{cot x} is continuous at x = 0, f (0) must be defined as
f is a continuous function on the real line. Given that x^{2} + (f(x) – 2) x – . f(x) + 2 – 3 = 0. then the value of f()
Given f(x)= then (where [*] represent the integral part function)
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209 videos218 docs139 tests
