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**Q.1. Let the function, f :[ 7,0] â†’ be continuous on [âˆ’7, 0] and differentiable on (âˆ’7, 0). If f (âˆ’7) = âˆ’3 and f'(x) < for all x âˆˆ(-7, 0), then for all such functions f, f(-1) + f(0) lies in the interval (2020)**

(2) [ -3,11]

(3) ( -âˆž ,11]

(4) [ -6, 20]

Using LMVT in [ -7, -1], we get

.....(1)

Using LMVT in [ -7, -0], we get

....(2)

From Eqs. (1) and (2), we get f(0) + f(-1)

Q.2. Let S be the set of points where the functionis not differentiable. Thenis equal to ________. (2020)

We have,

f (x) =

It is clear from the graph that the function is not differentiable at x =1,3 and 5.

Now,

= f(0) + f(2) + f (0) = 1 + 1 + 1 = 3

f(x) is a polynomial function, so it is continuous and differentiable in [0,1].

Now, f(x) = x

f(0) = 11 and f(1) = 1 âˆ’ 4 + 8 + 11 = 16

Using Lagrangeâ€™s mean value theorem, we get

Now,

We have,

The function is continuous, then

â‡’ k =3+2 =5

Given, Rolleâ€™s theorem holds for the function in the interval [3, 4], Then,

Now,

(1)

(2)

(3)

(4)

S is a set of all functions

If we consider a constant function, then option 1, 3 and 4 are incorrect.

For option 2:

This may not be true for all the function

If we apply LMVT in (c, 1) then

such that

None of the option are incorrect

Using LMVT forwe get

Also, using LMVT for x âˆˆ [c,b], we get

Since f"(x)< 0, then f'(x) is decreasing. Hence,

Given, the function f(x) is continuous at x = 0, then

f(0

Therefore,

Now,

Therefore, a + 3 = b = 1 â‡’ a = -2 and b = 1

â‡’ a+2b = âˆ’2+2= 0

We have,

On solving options (2),(3) and (4), we obtain the integer values and integral multiple of Ï€ in sin is always zero.

For option (1), we getHence, f (x) is discontinuous at x =

(1) 3/2

(2) 4/3

(3) 2/3

(4) 1/3

Given,

x

Now, using Eq. (1), we get

On comparing with Eq. (2), we get

Ans.

Given

Given

Let x = sin Î¸ and y = sin Ï•, then the equation becomes

.....(1)

Differentiating Eq. (1) w.r.t. x, we get

(1) 1/e

(2) 1/e

(3) e

(4) e

We have,

(1) 2Ï€/3

(2)

(3) 5Ï€/6

(4) Ï€/3

Ans.

Let tan

Now,

Now,

Given,

Therefore,

(1) 0

(2) 1/10

(3)

(4)

We have,

....

(1) exists and equals

(3) exists and equals

(4) does not exist

Ans.

** (2019)(1) - sin 1(2) 1(3) sin 1 (4) 0**

(2019)

(1) equals 1

(2) equals 0

(3) equals - 1

(4) does not exist

(1) continuous if a = 5 and b = 5

(2) continuous if a = - 5 and b = 10

(3) continuous if a = 0 and b = 5

(4) not continuous for any values of a and b

Let f(x) is continuous at x = 1, then

f(1

â‡’

Let f(x) is continuous at x = 3, then

f(3

â‡’

Let f(x) is continuous at x = 5, then

f(5

â‡’ b + 25 = 30

â‡’

From (1),

But

Hence, f(x) is not continuous for any values of a and b

Let S be the set of points in the interval (- 4, 4) at which f is not differentiable. Then S: (2019)

âˆµ f(x) is not differentiable at -2, -1,0, 1 and 2.

âˆ´ S = {-2,-1,0, 1,2}

(2) exists and equals 4.

(3) does not exist.

(4) exists and equals 0.

Solution of differential equation

(2) equals Ï€

(3) equals Ï€ + 1

(e) equals 0

Solution.

Since,

Hence, limit does not exist.

(1) not differentiable if f' (c) = 0

is continuous at x = 0, then the ordered pair (p, q) is equal to: (2019)

**Ans. **(3)**Solution.****Q.26. Let f : R â†’ R be a continuously differentiable function such that f(2) = 6 and f'(2) = 1/48. (2019)****(1) 18****(2) 24****(3) 12****(4) 36****Ans. **(1)**Solution.****Q.27. **** (2019)****(1) 4(2) 4âˆš2(3) 8âˆš2(4) 8**

Since, f(x) is continuous, then

(1) {5,10,15}

(2) {10,15}

(3) {5, 10,15,20}

(4) {10}

Since, f(x)= 15 - |(10-x)|

âˆ´ g(x) = f(f(x)) = 15 - |10 - [15 - |10-x|]|

= 15 â€” || 10 â€” x| â€” 5|

âˆ´ Then, the points where function g(x) is Non-differentiable are

10 - x = 0 and |10 â€” x| = 5

â‡’ x = 10 and x - 10 = Â± 5

â‡’ x = 10 and x = 15, 5

âˆµ Function is continuous at x = 5

LHL = RHL

(5 - Ï€) b + 3 = (5 - Ï€) a + 1

(1) 1

(2) e

(3) e

(4) e

Ans.

Solution.

(2) 2f'(2)

(4) 12f'(2)

Using L' Hospital rule and Leibnitz theorem, we get

Putting x = 2,

2f(2)f'(2) = 12f'(2) [âˆµ f(2) = 6]

(1) differentiable at all points

(2) not continuous

(3) not differentiable at two points

(4) not differentiable at one point

â‡’ g(x) is non-differentiable at x = 1

â‡’ g(x) is not differentiable at one point.

(1) Ï† (an empty set)

(2) {Ï€}

(3) {0}

(4) {0,Ï€}

Then, function f(x) is differentiable for all x < 0

Now check for x = 0

Then, function f(x) is differentiable for x = 0. So it is differentiable everywhere

Hence, k = Ï†**Q.39. ****Let S be the set of all points in (- Ï€ , Ï€) at which the function f(x) = min {sinx, cosx} is not differentiable. Then S is a subset of which of the following? (2019)****Ans. **(2)**Solution.**

âˆµ f(x) is not differentiable at **Q.40. Let f: [- 1, 3] â†’ R be defined as****where [t] denotes the greatest integer less than or equal to t. Then, f is discontinuous at: (2019)(1) only one point(2) only two points(3) only three points(4) four or more points**

f(x) is discontinuous at x = {0, 1, 3}

Hence, f(x) is discontinuous at only three points.

â‡’ 2 - a = 5 â‡’

Then a + b = -3-4 = -7

Given limit is,

By the graph f(x) is maximum at x = 2

âˆ´ Î± = 2

g (x) = |x + 1|

Graph of y = g (x)

By the graph g (x) is minimum at x = - 1

(1) five elements

(2) one element

(3) three elements

(4) two elements

Now, the graph of the function

From the graph, it is clear that f(x) is not differentiable at

Hence, K has exactly

(2) is equal to 50

(3) is equal to 120

(4) does not exist (in R)

Ans.

(1) Ï• (an empty set)

(2) {0}

(3) {Ï€}

(4) {0,Ï€}

Ans.

Solution.

(1) R Ã— [0, âˆž)

(2) R Ã— (-âˆž, 0)

(3) (-âˆž, 0) Ã— R

(4) [-âˆž, 0) Ã— R

Ans:

(1) Exists and is equal to -2

(2) Exists and is equal to 0

(3) Does not exist

(4) Exists and is equal to 2

Ans:

Solution:

= cos x âˆ’ tan x [x

= x

fâ€²(x) = 2x tan x + x

= 2 tan x + x sec

= 0 + 0 âˆ’ 2 + 0 = âˆ’2

(1) -1/6

(2) 1/6

(3) 1/3

(4) -1/3

Ans:

Here â€˜Lâ€™ is in the indeterminate i.e., 0/0

âˆ´ using the Lâ€™ Hospital rule we get:

(1) (2, 1)

(2) (3, 1)

(3) (3, 2)

(4)

Ans:

Solution:

If the function is continuous at x = 0, then

For the limit to exist, power of x in the numerator should be greater than or equal to the power of x in the denominator. Therefore, coefficient of x in numerator is equal to zero

â‡’ 3 - k = 0

â‡’ k = 3

So the limit reduces to

(1) 1/4

(2) 1/24

(3) 1/16

(4) 1/8

Ans.

Solution.

= 1/16.

(2) 1/2âˆš2

(3) âˆš3/2

(4) âˆš3

k + 2/5 = 1

k = 1 - 2/5 = 3/5

(1) 2

(2) 1

(3) 1/2

(4) 1/4

Ans.

Solution:

(1) g is not differentiable at x = 0

(2) g'(0) = cos (log 2)

(3) g'(0) = - cos(log 2)

(4) g is differentiable at x = 0 and g'(0) = -sin(log 2)

Ans.

Solution:

(1)

(2)

(3)

(4) 3 log 3 - 2

Ans.

Solution.

P = 27/e

**Q.58. If f(x) is a differentiable function in the interval (0, âˆž) such that f(1) = 1 and for each x > 0, then f(3/2) is equal to: (2016)****(1) 13/6****(2) 23/18****(3) 25/9****(4) 31/18****Ans. **(4)**Solution.**

Differentiate w.r.t. t**Q.59. (2016)****(1) 2/3****(2) 3/2 ****(3) 2****(4) 1/2****Ans. **(2)**Solution.**

must be of the from 1^{âˆž}

â‡’

= e^{2a} = e^{3}

a = 3/2**Q.60. Let a,b âˆˆ R, (a â‰ 0). If the function f defined as (2016)****Ans. **(1)**Solution.****Q.61. (2016)****(1) 2****(3) 1/2****(4) -2****Ans.** (4)**Solution.**

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