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QUESTION: 1

The critical point of f (x) = |x- 1| is

Solution:

f (x) = |x- 1| is not diff at x = 1 critical point is at x = 1

QUESTION: 2

IF f (x) = a log x + bx^{2} + x has extreme values at x = -1, x = 2 then a = ...., b = .....

Solution:

QUESTION: 3

If x is real then the minimum value of

Solution:

Min. value = f(1)

Max. Value = f(-1)

QUESTION: 4

The value of “a” for which the sum of the squares of the roots of the equation x^{2} - (a - 2) x - a - 1 = 0 assume the least value is

Solution:

QUESTION: 5

Greatest value of (1/x)^{x} is

Solution:

Let y = (1/x)^{x}

Taking ln both sides, we get ln y =x ln 1/x = x ln (x)^{-1 }

= -x ln x {Since,ln x^{a} = a ln x }

Now, differentiating both sides with respect to x we get,

1/y(dy/dx) = -1-ln x

⇒ (dy/dx) =(1/x)^{x} [-1-ln x} = 0 {Condition for y to be maximum}

⇒ln x = -1 {Since, (1/x)^{x} = 0}

⇒ x = e^{-1 } {Converting logarithmic equation to exponential equation}

⇒ So, y(max) = (e)^{1/e}

QUESTION: 6

A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x unit and circle of radius = r units. If the sum of the area of the square and the circle so found is minimum then

Solution:

Given 4x + 2πr = 2

For minimum value of area A

x = 2r

QUESTION: 7

The function f (x) = 4x^{5} - 25x^{4} + 40x^{3} - 10 has

Solution:

f '(x) = 20x^{4} -100x^{3} + 120x^{2} f '(x) = 0

x = 2, 3 one maximum and one minimum

QUESTION: 8

The function x^{4} - 62x^{2} + ax + 9 attains its maximum value at x = 1, on the interval [0, 2] then value of ‘a’ is

Solution:

f (x) = x^{4} - 62x^{2} + ax + 9

f '(x) = 4x^{3} -124x + a

f '(1) = 0 at x =1 4 -124 + a = 0

x = 120

QUESTION: 9

The longest distance of the point (a, 0) from the curve 2x^{2} + y^{2}- 2x is

Solution:

f (x) = (PA)^{2} = (x - a)^{2} +2x - 2x^{2}

f '(x) = 0x = (1-a)

QUESTION: 10

The point on the curve y = x^{2} which is nearest to (3, 0) is

Solution:

Let Z=(x-3)^{2}+y^{2 }= x^{2}-6x+9+x^{4}

Z is minimum dz/dz = 2x-6+4x^{3} = 0

∴ x = 1 ∴ y = 1, (1,1)

QUESTION: 11

The difference between the greatest and least value of the function f (x) = sin 2x - x on

Solution:

QUESTION: 12

The sides of a rectangle of the greatest area which can be inscribed into an ellipse

Solution:

QUESTION: 13

A window is in the shape of a rectangle surmounted by a semi circle. If the perimeter of the window is of fixed length „l‟ then the maximum area of the window is

Solution:

QUESTION: 14

The sum of the hypotenuse and a side of a right triangle is constant. If the area of the triangle is maximum then the angle between the hypotenuse and the given side is

Solution:

z+x=k z+zsinθ = k z = k/1+sinθ

QUESTION: 15

The number of linear functions which map from [-1,1] onto [0, 2] is

Solution:

Let f (x) = ax+ b be the required liner function.

If a > 0 then f is increa sin g, f(-1) = 0,f(1) = 2 ⇒ -a + b = 0,a +b = 2 ⇒ a = -1,b = 1

If a < 0 then f decrea sin g, f(-1) = 2,f (1) ⇒ = 0 -a + b = 2,a+b = 0 = ⇒ a = -1,b = 1

∴ The required linear function are f(x) = x+1,f(x) = -x+1.

QUESTION: 16

for x ∈ R Then f (2002) =

Solution:

(cos^{2}x + sin^{4}x) - (sin^{2}x + cos^{4}x) = (cos^{2}x - sin^{2}x) - (cos^{4}x - sin^{4}x)

= cos^{2}x - cos^{2}x = 0

QUESTION: 17

If f (0) = 0, f (1) = 1, f (2) = 2 and f (x) = f (x - 2) + f (x - 3) for x = 3, 4, 5,&.., then f(9) =

Solution:

Givenf (x) = f(x-2)+f(x-3)

∴ f(3) = f(1)+f(0) = 1 + 0 = 1,

f(4) = f(2)+f(1) = 2 + 1 = 3

f(5) = f(3)+f(2) = 1+ 2 = 3,f(6). f (4) + f(3) = 3+1 = 4,

f(7) = f(5)+f(4) = 3+3 = 6,f(8) = f(6)+f(5) = 4+3 =7,

f(9) = f(7)+f(6) = 6+4 = 10.

QUESTION: 18

Solution:

QUESTION: 19

If f (x) = cos [π^{2}] x + cos [- π^{2}] x where [x] is the step function, then

Solution:

cos 9x + cos (-10) x = cos 9x + cos10x

f (0) = cos0 + cos0 = 2,f (π/4) = cos9 π/4 + cos5π/2 = cos π/ 4 =

f (π/2) = cos9π/2 + cos5π = 0 -1 = -l,f (π) = cos9π + cos10π = - l +1 = 0.

QUESTION: 20

If f (x) is a polynomial fuction such that f (x) f (1/x) = f (x)+ f (1/x) and f (2) = 33 then f (x) =

Solution:

f (x) f (1/x) = f (x) + f (1/x) ⇒ f(x) = x^{n} +1.

f (2) = 33 ⇒ 2^{n} +1= 33 ⇒ 2^{n} = 32 = 2^{5 }⇒ n = 5 ⇒ f (x) = x^{5} +1.

QUESTION: 21

If f (x) is a function such that f (x + y) = f (x) f (y) and f (3) = 125 then f (x) =

Solution:

f(x+y) + f(x) + f(y) ⇒ f(x) = a^{x}.f(3)

= 125 ⇒ a^{3} = 125 ⇒ a = 5.

∴ f (x) = 5^{x}.

QUESTION: 22

If f : [2, ∞) → B defined by f (x) = x^{2} - 4x + 5is a bijection, then B =

Solution:

f (x) = x^{2} -4x +5 = (x -2)^{2} +1

x ∈ [2,∞) ⇒ x __>__ 2 ⇒ x - 2 __>__ 0 ⇒ (x - 2)^{2} +1 __>__ 1 ⇒ B = [1,∞).

QUESTION: 23

The function f : C → C defined by f (x) for x ∈ C where bd ≠ 0 reduces to a constant function if

Solution:

is a constant function ⇒ bc ad 0 ⇒ ad bc.

QUESTION: 24

A function f : N → Z defined by f(n) = (n - 1)/2 when n is odd and f(n) = -n/2 when n is even, is

Solution:

**Correct Answer :- c**

**Explanation : f(n-1)/2 ; n is odd -n/2 ; n is even**

**f(1) = 0 f(3) = 1**

**f(2) = -1 f(4) = -2**

**f(5) = 2 f(6) = -3**

**f(n) + f(n+1) = -1 n is odd**

**if = 0 n is even**

**Range of f(n) = Z**

**f(n) is one-one, therefore f(n1) = f(n2)**

**(n _{1}-1)/2 = -n_{2}/2**

**=> 1 < (n1 + n2)/2 is not equal to 1/2**

**Therefore f(n) is onto.**

QUESTION: 25

A = {x : |x| __<__ 1|} and f : A → A such that f (x) = x |x|, then

Solution:

Graph of y = f (x) is given in the figure.

∴ f is bijective.

QUESTION: 26

The domain of the function f = {(1, 3), (3, 5), (2, 6)} is

Solution:

Domain = {1, 3, 2}

QUESTION: 27

Solution:

∴ f is an even function.

QUESTION: 28

The range of f(x)

Solution:

If x > 0 then f x = 0. If x < 0 then f (x) = tanh x. If f (x) = -1then

ex - e^{|x|} = - e^{x} - e^{|x|} ⇒e^{x} = 0 ⇒ x ∉ R.

Range does not contain -1.

∴ Range = (-1, 0].

QUESTION: 29

The domain of log (x - 3) (5- x) is

Solution:

log (x - 3) (5 - x) exists ⇒ (x - 3) (5 - x) > 0

⇒ (x -3)(x -5) < 0 ⇒ 3< x < 5

∴ Domain = (3, 5).

QUESTION: 30

The domain of log_{10} (x^{3} - x) is

Solution:

log_{10} (x^{3} - x)is defmed ⇒ x^{3} - x > 0 ⇒ x(x^{2} - 1) > 0 ⇒ (x +1)x(x - 1) > 0

x < -1 ⇒ (x+1)x(x-1) < 0; -1 < x < 0 ⇒ (x+1)x(x-1) > 0

0 < x < 1 ⇒ (x+1)x(x-1) < 0; x > 1 ⇒ (x+1)x(x-1) > 0.

∴ Domain = (-1,0)∪(1, ∞)

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