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QUESTION: 1

If f(x) = , then

Solution:

QUESTION: 2

The number of points at which the function f(x) = max. {a – x, a + x, b} – ∞ < x < ∞, 0 < a < b cannot be differentiable is

Solution:

A function is not differentiable at the points where it has sharp-corner 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

QUESTION: 3

The function f(x) is defined by f(x) =

Solution:

lt x1^+ f(x) = log(4x -3) (x^{2}- 2x + 5)

= ln(x^{2} - 2x +5)/ln(4x-3)

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(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 multi[ly the denominator by (-4h)

ln(h → 0) ln(4 + h^{2})/[((1+4h)/-4h) * (-4h)

As we know that (1-4h)/(-4h) = 1

ln 1/(-4*0) = - ∞ (does not exist)

QUESTION: 4

Let f(x) = then

Solution:

QUESTION: 5

A point where function f(x) is not continuous where f(x) = [sin [x]] in (0, 2π) ; is ([ * ] denotes greatest integer ≤ x)

Solution:

QUESTION: 6

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)

Solution:

QUESTION: 7

If f is a real-valued differentiable function satisfying |f(x) – f(y)| ≤ (x – y)^{2}, x, y ∈ R and f(0) = 0, then f(1) equals

Solution:

QUESTION: 8

If f(x) = , then f([2x]) is (where [ * ] represent greatest integer function)

Solution:

QUESTION: 9

The value of f(0), so that the function, f(x)= becomes continuous for all x, is given by

Solution:

f(x)={(a^{2}−ax+x^{2})−(a^{2}+ax+x^{2})^1/2}/{(a+x -(a-x)^{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} + (a-x)^{1/2}]/[(a+x)-(a-x) * (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}

QUESTION: 10

If f(x) = then f(x) is (where { * } represents the fractional part function)

Solution:

QUESTION: 11

In order that function f (x) = (x + 1)^{cot x} is continuous at x = 0, f (0) must be defined as

Solution:

QUESTION: 12

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()

Solution:

QUESTION: 13

Given f(x)= then (where [*] represent the integral part function)

Solution:

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