Limits And Derivatives - 2 - Free MCQ Practice Test with solutions, JEE

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

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

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

after rationalizing

B. 1/12

Let u = ∛(8 + h) and v = 2.

Use the identity (u - v)(u² + u v + v²) = u³ - v³.

Multiply numerator and denominator by u² + u v + v². The numerator becomes u³ - v³ = (8 + h) - 8 = h.

Therefore the expression simplifies to 1 / (u² + u v + v²).

As h → 0, we have u → 2. Substituting gives the denominator 2² + 2·2 + 4 = 4 + 4 + 4 = 12.

Hence the limit equals 1/12, so option B is correct.

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

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).

Option D is correct; the general formula is (-1)^n-1 (n-1)! / xⁿ.

For n = 1, the value is 1/x, which equals (-1)⁰ 0! / x¹.

Assume for some integer k ≥ 1 the expression for the k-th derivative is (-1)^k-1 (k-1)! / x^k.

Differentiate this assumed form: differentiate x^-k using the power rule to get -k x^-(k+1). Multiplying by the constant factor gives (-1)^k-1 (k-1)! · ( -k ) x^-(k+1).

Simplifying the signs and factorials yields (-1)^k k! / x^k+1, which matches the formula with n = k+1.

By verification for n = 1 and the inductive step, the formula holds for all integers n ≥ 1. Thus the n-th derivative is (-1)^n-1 (n-1)! / xⁿ, corresponding to Option D.