﻿ IIT - JEE Main Important Questions of Continuity and Differentiability | Zigya

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# Continuity and Differentiability

#### Multiple Choice Questions

1.

let  . If f(x) is continuous in  then f (π/4) is

• 1

• -1/2

• -1

• 1/2

B.

-1/2

Also f(x) is continuous in [0, π/ 2] ,
so f(x) will be continuous at π / 4 .
∴ Value of function = Value of limit ⇒ f(π/ 4) = −1/ 2
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2.

The differential equation representing the family of curves  , where c > 0, is a parameter, is of order and degree as follows:

• order 1, degree 2

• order 1, degree 1

• order 1, degree 3

• order 2, degree 2

C.

order 1, degree 3

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3.

The differential equation whose solution is Ax2 + By2 = 1, where A and B are arbitrary constants is of

• second order and second degree

• first order and second degree

• first order and first degree

• second order and first degree

D.

second order and first degree

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4.

The differential equation for the family of curves,x2 +y2 -2ay = 0  where a is an arbitrary constant is

• 2(x2-y2)y' = xy

• (x2+y2)y' = xy

• 2(x2+y2)y' = xy

• (x2- y2)y' = xy

A.

2(x2-y2)y' = xy

2x + 2yy′ - 2ay′ = 0

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5.

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

• -1

• 0

• 2

• 1

B.

0

⇒ |f′(x)| ≤ 0
⇒ f′(x) = 0
⇒ f(x) = constant As
f(0) = 0 ⇒ f(1) = 0
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6.

The solution of the differential equation ydx + (x + x2y) dy = 0 is

• -1/ XY  =C

• -1/XY + log y = C

• 1/XY + log y = C

• log y =Cx

B.

-1/XY + log y = C

y dx + x dy + x2y dy = 0

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7.

A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness than melts at a rate of 50 cm3 /min. When the thickness of ice is 5 cm, then the rate at which the thickness of ice decreases, is

B.

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8.

If then the solution of the equation is

C.

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9.

A function y = f(x) has a second order derivative f″(x) = 6(x – 1). If its graph passes through the point (2, 1) and at that point the tangent to the graph is y = 3x – 5, then the function is

• (x-1)2

• (x-1)3

• (x+1)3

• (x+1)2

B.

(x-1)3

f''(x) = 6(x - 1) f'(x) = 3(x - 1)2 + c ........ (i)
At the point (2, 1) the tangent to graph is y = 3x - 5 Slope of tangent = 3
∴ f'(2) = 3(2 - 1)2 + c = 3
3 + c = 3
⇒ c = 0
∴ From equation (i) f'(x) = 3(x - 1)2
f'(x) = 3(x - 1)2 f(x) = (x - 1)3 + k ...... (ii)
Since graph passes through (2, 1)
∴ 1 = (2 - 1)2 + k k = 0
∴ Equation of function is f(x) = (x - 1)3

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10.

Let f be differentiable for all x. If f(1) = - 2 and f′(x) ≥ 2 for x ∈ [1, 6] , then

• f(6) ≥ 8

• f(6) < 8

• f(6) < 5

• f(6) = 5

A.

f(6) ≥ 8

As f(1) = - 2 & f′(x) ≥ 2 ∀ x ∈ [1, 6]
Applying Lagrange’s mean value theorem

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