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Test: Limits And Derivatives - 2 - JEE MCQ


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25 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Limits And Derivatives - 2

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Test: Limits And Derivatives - 2 - Question 1

 is equal to 

Detailed Solution for Test: Limits And Derivatives - 2 - Question 1

lt h->0 [sin(x+h)½ - sin(x)½]/h
Differentiate it with ‘h’
lt h->0 {cos(x+h)½ * ½(x+h)1/2] - 0}/1
lt h->0 {cos(x+h)½ * ½(x+h)1/2]
lt h->0 cos(x)½ / 2(x)½

Test: Limits And Derivatives - 2 - Question 2

Detailed Solution for Test: Limits And Derivatives - 2 - Question 2

Divide the numerator and denominator by x, so that given function becomes
f(x)=1+sinx/x/(1+cosx/x)
Now as x→∞.sinx/x→0 , because sin x would oscillate between +1 and -1, which in either case divided by ∞ would be 0. Thus the limit of the numerator would be 1. Like wise the limit of the denominator would also be 1.
Thus limit as a whole would be 1

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Test: Limits And Derivatives - 2 - Question 3

Test: Limits And Derivatives - 2 - Question 4

is equal to

Detailed Solution for Test: Limits And Derivatives - 2 - Question 4

lim x→0 sin xn . (x)m . xn /(sin x)m.(x)m.xn
lim x→0 sin xn.(x)m.xn-m/xn.(sin x)m
Applying limits.
=0n-m = 0

Test: Limits And Derivatives - 2 - Question 5

Test: Limits And Derivatives - 2 - Question 6

If G(x) =  then  has the value

Detailed Solution for Test: Limits And Derivatives - 2 - Question 6



Test: Limits And Derivatives - 2 - Question 7

then dy/dx is equal to

Test: Limits And Derivatives - 2 - Question 8

If f be a function such that f (9) = 9 and f ‘ (9) = 3, then   is equal to

Detailed Solution for Test: Limits And Derivatives - 2 - Question 8


after rationalizing

Test: Limits And Derivatives - 2 - Question 9

  is equal to 

Test: Limits And Derivatives - 2 - Question 10

If f(x) = , x ∈ (0,1), then f'(x) is equal to 

Detailed Solution for Test: Limits And Derivatives - 2 - Question 10

f(x) = 
xn differentiation is nxn-1
 differentiation would be 1/2
-x2  differentiation would be -2x
 f'(x) = 

Test: Limits And Derivatives - 2 - Question 11

If y = sin-1x  and z = cos-1 then dy/dz = 

Test: Limits And Derivatives - 2 - Question 12

 is equal to 

Detailed Solution for Test: Limits And Derivatives - 2 - Question 12

Method 1: Using the Chain Rule and Inverse Trigonometric Derivatives

  1. Identify the inner function: u = (3x-x^3)/(1-3x^2)

  2. Differentiate the outer function: d/dx [tan^(-1)(u)] = 1/(1+u^2)

  3. Differentiate the inner function: du/dx = (3-3x^2)/(1-3x^2)^2

  4. Apply the chain rule: d/dx [tan^(-1)((3x-x^3)/(1-3x^2))] = 1/(1+u^2) * du/dx

    Substitute u and du/dx:

    = 1/(1+((3x-x^3)/(1-3x^2))^2) * (3-3x^2)/(1-3x^2)^2

    Simplify:

    = (3-3x^2)/((1-3x^2)^2 + (3x-x^3)^2)

    Further simplification:

    = 3/(1+x^2)

Method 2: Using the Substitution Method

  1. Substitute x = tan(theta):

    y = tan^(-1)((3tan(theta)-tan^3(theta))/(1-3tan^2(theta)))

  2. Simplify:

    y = tan^(-1)(tan(3theta))

    Using the property tan^(-1)(tan(x)) = x (with appropriate domain restrictions):

    y = 3theta

  3. Substitute back theta = tan^(-1)(x):

    y = 3tan^(-1)(x)

  4. Differentiate:

    dy/dx = 3/(1+x^2)

Method 3: Using the Formula for the Derivative of tan^(-1)(u/v)

  1. Directly apply the formula:

    d/dx [tan^(-1)(u/v)] = (vdu - udv)/(v^2 + u^2)

    where u = 3x - x^3 and v = 1 - 3x^2.

  2. Calculate du and dv:

    du = (3 - 3x^2)dx dv = -6xdx

  3. Substitute into the formula and simplify to get the same result as in Method 1.

Conclusion:

Regardless of the method used, the derivative of the given expression is:

3/(1+x^2)

Important Note:

The validity of the result depends on the domain of x. The expression (3x-x^3)/(1-3x^2) is undefined for x = ±1/sqrt(3). Therefore, the result is valid only for |x| < 1/sqrt(3).

Test: Limits And Derivatives - 2 - Question 13

The function, f(x) =  and f(a) = 0, is 

Test: Limits And Derivatives - 2 - Question 14

 is equal to 

Test: Limits And Derivatives - 2 - Question 15

 is equal to 

Test: Limits And Derivatives - 2 - Question 16

If sin x = then dx/dy is equal to 

Test: Limits And Derivatives - 2 - Question 17

 holds true for

Test: Limits And Derivatives - 2 - Question 18

Dervative  of tan  w.r.t  is

Test: Limits And Derivatives - 2 - Question 19

If y = log  then 

Test: Limits And Derivatives - 2 - Question 20

If y = log x , then yn = 

Test: Limits And Derivatives - 2 - Question 21

The derivative of sec-1  with respect to  at x = 1/x is 

Test: Limits And Derivatives - 2 - Question 22

 is equal to

Test: Limits And Derivatives - 2 - Question 23

 is eqaual to  

Test: Limits And Derivatives - 2 - Question 24

If y =  then dy/dx = 

Test: Limits And Derivatives - 2 - Question 25

then at x = 1, f(x) is

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