Important Questions: Limits & Derivatives | Mathematics (Maths) Class 11 - Commerce PDF Download

Q1: Find the positive integer “n” so that lim_{x → 3}[(xⁿ– 3ⁿ)/(x – 3)] = 108.
Ans: Given limit: limx → 3[(xn– 3n)/(x – 3)] = 108
Now, we have:
limx → 3[(xⁿ– 3ⁿ)/(x-3)] = n(3)^n-1
n(3)n-1 = 108
Now, this can be written as:
n(3)^n-1 = 4 (27) = 4(3)^4-1
Therefore, by comparing the exponents in the above equation, we get:
n = 4
Therefore, the value of positive integer “n” is 4.

Q2: Determine the derivative of cosx/(1+sin x).
Ans: Given function: cosx/(1+sin x)
Let y = cosx/(1+sin x)
Now, differentiate the function with respect to “x”, we get
dy/dx = (d/dx) (cos x/(1+sin x))
Now, use the u/v formula in the above form, we get
dy/dx = [(1+sin x)(-sin x) – (cos x)(cos x)]/(1+sin x)²
dy/dx = (-sin x – sin²x-cos²x)/(1+sin x)²
Now, take (-) outside from the numerator, we get:
dy/dx = -(sin x + sin².x + cos²x)/(1+sin x)²
We know that sin².x + cos²x = 1
By substituting this, we can get:
dy/dx = -(1+sin x)/(1+sin x)²
Cancel out (1+sin x) from both numerator and denominator, we get:
dy/dx = -1/(1+sin x)
Therefore, the derivative of cosx/(1+sin x) is -1/(1+sin x).

Q3: Evaluate the derivative of f(x) = sin²x using Leibnitz product rule.
Ans: Given function: f(x) = sin²x
Let y= sin²x
Now, by using Leibnitz product rule, we can write it as:
dy/dx = (d/dx) sin²x
Sin²x can be written as (sin x)(sin x)
Now, it becomes:
dy/dx = (d/dx) (sin x)(sin x)
dy/dx = (sin x)’(sin x) + (sin x)(sin x)’
dy/dx = cos x sin x + sin x cos x
dy/dx = 2 sin x cos x
dy/dx = sin 2x
Therefore, the derivative of the function sin²x is sin 2x.

Q4: Find the derivative of the function x²cos x.
Ans: Given function is x²cos x
Let y = x²cos x
Differentiate with respect to x on both sides.
Then, we get:
dy/dx = (d/dx)x²cos x
Now, using the formula, we can write the above form as:
dy/dx = x² (d/dx) cos x + cos x (d/dx)x²
Now, differentiate the function:
dy/dx = x² (-sin x) + cos x (2x)
Now, rearrange the terms, we will get:
dy/dx = 2x cos x – x² sin x

Q5: Find the derivative of f(x) = x³ using the first principle.
Ans: By definition,
f’(x) = lim_{h→ 0} [f(x+h)-f(x)]/h
Now, substitute f(x) = x³ in the above equation:
f’(x) = lim_{h→ 0} [(x+h)³-x³]/h
f’(x) = lim_{h→ 0} (x³+h³+3xh(x+h)-x³)/h
f’(x) = lim_{h→ 0} (h²+3x(x+h))
Substitute h = 0, we get:
f’(x) = 3x²
Therefore, the derivative of the function f’(x) = x³ is 3x².

Q6: limx→ 0 |x|/x is equal to:
Ans: The limit mentioned here is x→0
It has two possibilities:
Case 1: x→0⁺
Now, substitute the limit in the given function:
lim_{x→ 0}⁺ |x|/x = x/x = 1
Case 2: x→0^–
Now, substitute the limit in the given function:
lim_{x→ 0-} |x|/x = -x/x = -1
Hence, the result for both cases varies.