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** Q.1. Let f (x) be a polynomial of degree 5 such that x =Â±1 are its critical points. Ifthen which one of the following is not true? (2020)****(1) f is an odd function.****(2) f(1) - 4f(-1) = 4.****(3) x = 1 is a point of maxima and x = âˆ’1 is a point of minimum of f.****(4) x = 1 is a point of minima and x = - 1 is a point of maxima of f.****Ans. **(4)

Given,

(1)

Let,

f(x) = 2x^{3} + ax^{4} + bx^{5} â‡’ f'(x) = 6x^{2} + 4ax^{3} + 5bx^{4}

Since x = Â±1 are the critical points of f (x), then

f'(1) = 6 + 4a + 5b = 0

â‡’ 4a + 5b = -6... (2)

Now, f'(-1) = 0 â‡’ 6 - 4a + 5b = 0

â‡’ -4a + 5b = - 6.... (3)

On solving Eqs. (2) and (3), we get

Therefore,

Hence, x = 1 is a point of maxima and x = -1 is a point of minima of f(x).**Q.2. Let f(x) = x cos ^{-1} then which of the following is true? (2020)**

(1) f' is increasing in and decreasing in

We have,

f(x) = x[Ï€ - cos

Now,

Hence, f'(x) is decreasing in and increasing in

The equation of the given curve is

x

Now, the slope of normal at (2,2) is

The equation of the normal is

y âˆ’2 = âˆ’1(xâˆ’2)â‡’ x+ y - 4 = 0

Hence, the length of perpendicular from origin to the normal is

Ans.

Given, f'(x) has a critical point x = 1, let

Now, f '(-1) = 0

Therefore,

Now, f(1) = - 6

â‡’ âˆ’11k + 6d = âˆ’36... (1)

Now, f(-1) = 10 â‡’ 5k + 6d = 60... (2)

From Eqs. (1) and (2), we get

k=5,d=6

f(x) = x

â‡’ f"(x) = 6x - 6 â‡’ f"(3) = 6 x 3 - 6 = 12

Hence, f(x) has a local minima at x =3.

Let the initial thickness of the ice is x cm, then

At x = 5 cm and = 50 cm

We have,

(1) a point of local minima.

(2) not a critical point.

(3) a point of local maxima.

(4) a point of inflection.

We have,

.... (since f (1) = 0)

(since g (1) = 0)

â‡’ F'(1) = 1.g (1) = 0.... (1)

â‡’ F" (x) = 2x g (x) + x

â‡’ F" (1) = 2g (1) + 1 . g ' (1) = f(1) = 3.... (2)

From Eqs. (1) and (2), it is clear that x = 1 is the point of minima for function F.

Volume of cone

...(2)

For maxima/minima,

Then, h = âˆš3 is point of maxima

Hence, the required maximum volume is,

Minimum value of det (A) is attained when sin

â‡’ d = -5 or 1

Also,

âˆ´ I is minimum when (a, b) = (-âˆš2, âˆš2)

(1) (2, 3e)

(2)

(3)

(4) (3, 6e)

Since (1, e) lies on the curvethen equation of tangent at (1, e) is

So, equation of tangent to the curve passes through the point

â‡’ f(x) is increasing function âˆ€ x > 0

(1) -122

(2) -222

(3) 122

(4) 222

f(x) = 3x(x - 3)

Now S = {x âˆˆ R : x

So x

â‡’ x âˆˆ [5, 6]

âˆ´ f(x) will have maximum value for x = 6

The maximum value of function is,

f(6) = 3 x 6 x 3 x 3 - 40 =

(2) f is a decreasing function of x

(3) f is not a continuous function of x

(4) f is neither increasing nor decreasing function of x

â‡’ f(x) is increasing function.

Hence, f(x) is increasing function.

(1) 36

(2) 20âˆš2

(3) 32

(4) 18âˆš3

Ans.

Solution.

Given, the equation of parabola is,

x

Area of the rectangle = (2t) (12 - t

At t = 2, area is maximum = 24(2) - 2(2)

= 48 - 16 = 32 sq. units

(1) 75/2

(2) 125/4

(3) 625/4

(4) 125/2

Parametric equations of the parabola y

For maximum area t = 1/2

(1) (3, 0)

(2) (âˆš3, 0)

(3) (-1, 2)

(4) (-âˆš2, 1)

Solution of equation

âˆ´ curve passes through point (1, -2)

â‡’ C = -9/4

Then, equation of curve

Since, above curve satisfies the point.

Hence, the curve passes through (âˆš3, o).

âˆµ Tangent to the given curve is parallel to line 2y = 4x + 1

âˆ´ Slope of tangent (m) = 2

Then, the equation of tangent will be of the form

y = 2x + c ...(1)

âˆµ Line (1) and curve y = x

âˆ´ 2x + c = x

x

âˆ´ D = 49 - 4(5 - c) = 0

â‡’

Hence, the equation of tangent:

(1) -7

(2) 5

(3) 7

(4) 6

Now,

â‡’

The shortest distance between line y = x and parabola = the distance LM between line y = x and tangent of parabola having slope 1.

Let equation of tangent of parabola having slope 1 is,

Distance between the line y - x = 0 and

Here at -2 & 1, f'(x) changes from negative value to positive value.

â‡’ -2 & 1 are local minimum points. At 0, f'(x) changes from positive value to negative value.

â‡’ 0 is the local maximum point.

Hence, S

(1) increasing on (0, 1) and decreasing on (1, 2).

(2) decreasing on (0, 2)

(3) decreasing on (0, 1) and increasing on (1,2).

(4) increasing on (0, 2)

But f" (x) > 0 â‡’ f'(x) is an increasing function

Hence, Ï† (x) is increasing on (1, 2) and decreasing on (0, 1).

(1) x log

(2) x log

(3) x

(4) x log

Equation (i) passes through the centre of the circle x

âˆ´ C = 2

Let radius of base and height of cylinder be r and h respectively.

...(1)

Now, volume of cylinder, V = Ï€r

Substitute the value of r

Differentiating w.r.t. h,

For maxima/minima,

â‡’ Volume is maximum when h =

(1) four irrational numbers.

(2) four rational numbers.

(3) two irrational and two rational numbers.

(4) two irrational and one rational number.

Since, function f(x) have local extreme points at x =

f(x) = K(x+ l)x(x - 1)

= K (x

(1) (-2,1)

(2) (-2,2)

(3) (2,-1)

(4) (2,-2)

y = x

Since, the point (1, -5) lies on the curve.

â‡’ 1 + a - b = - 5

â‡’ a - b = -6 ...(1)

Since, required line is perpendicular to y = x - 4, then slope of tangent at the point P (1, -5) = -1

âˆ´ 3 + a = - 1

â‡’ a = -4

â‡’ b = 2

the equation of the curve is y = x

â‡’ (2, -2) lies on the curve

Solution.

y = f(x) = x

f(1) = 1 - 1 -2 = -2, f(-1) = -1 -1+2 = 0

Since the tangent to the curve is parallel to the line segment joining the points (1, -2) and (-1, 0)

And their slopes are equal.

Hence, the required set

Given that water is poured into the tank at a constant rate of 5 m

Volume of the tank is,

...(i)

where r is radius and h is height at any time. By the diagram,

...(ii)

Differentiate eq. (i) w.r.t. â€˜tâ€™, we get

Putting h = 10, r = 5 and dV/dt = 5 in the above equation.

Given functions are, f(x) = e

Given f(g (x)) is increasing function.

are either both positive or negative

These values of Î± and Î² satisfies |6Î± + 2Î²| = 19

Given that ice melts at a rate of 50 cm

Substitute r = 5,

(1) 3/2

(2) 8/5

(3) 6/5

(4) 2/3

a

Here, a is first term of A.P and d is common difference

Given function

Differentiating w. r. t. x,

[âˆ´ f(x) is increasing in [0, 3]]

(1) x - y + 4 = 0

(2) x + y + 4 = 0

(3) x - 2y + 16 = 0

(4) 2x - y + 2 = 0

Given curves, y

Equation of tangent to the given parabola;

âˆµ This is common tangent.

âˆ´ equation of common tangent is

Solution.

Tangent to the curve y = (x - 2)

(1) 3

(2) -3

(3) -2âˆš2

(4) 2âˆš2

Ans.

Solution.

Hence, local minimum value is 2âˆš2

(1) 8âˆš2 Ï€

(2) 6âˆš2 Ï€

(3) 6âˆš3 Ï€

(4) 8âˆš3 Ï€

Ans.

Solution.

where r is radius and h is height of coin

For maxima or minima put f'(x) = 0

(1) 30

(2) 12.5

(3) 10

(4) 25

Ans.

2r + Î¸r =20 ... (i)

A to be maximum

Hence for r = 5, A is maximum

Now, 10 + Î¸Â·5 = 20 â‡’ Î¸ = 2 (radian)

(1) decreasing in R

(2) increasing in R

(3) increasing in (0, âˆž) and decreasing in (â€“âˆž, 0)

(4) decreasing in (0, âˆž) and increasing in (â€“âˆž, 0)

f(x) = x

f'(x) = 3x

â‡’x âˆˆ Ï•

Let y = f(x) be a curve

slope of tangent = f'(x)

Equation of tangent (Y-y) = f'(x) (X- x)

Put Y = 0

y = 1/x

y = 1/x

(1) 4/9

(2) 6/7

(3) 1/4

(4) 2/9

Ans.

(1) 7/4

(2) 15/4

(3) 9/4

(4) 1/4

Ans.

M = maximum value = 17/4

m = minimum value = 2

M-m = 17/4 - 2 = 9/4.

(1)

f(x) = sin

f'(x) = 4sin

= 4sinx cosx(sin

= - 2sin2x. cos2x

= - sin4x > 0

â‡’ sin4x < 0

â‡’ Ï€ < 4x < 2Ï€

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