CBSE Important Questions Class 11 Maths Chapter 12 – Limits and Derivatives

Class 11 Maths Chapter 12 Limits and Derivatives is part of the CBSE Class 11 Maths syllabus. It is the first introduction to calculus. Students learn what a limit is, how to find left-hand and right-hand limits, the algebra of limits, and some standard limits involving trigonometric functions. The chapter then defines the derivative as a limit and gives the rules for differentiating sums, products and quotients.

These CBSE important questions for Class 11 Maths Chapter 12 cover evaluating limits by direct substitution and by factorisation, standard trigonometric limits, finding the derivative from first principles, the product rule and the quotient rule, and differentiating polynomial and trigonometric functions. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. Evaluate limx→3 (x + 3).

Answer: This is a polynomial, so we can find the limit by direct substitution.
limx→3 (x + 3) = 3 + 3 = 6
For every polynomial function, limx→a f(x) = f(a).

Q2. Evaluate limx→2 (x² − 4)/(x − 2).

Answer: Direct substitution gives 0/0, which is indeterminate, so we factorise first.
x² − 4 = (x − 2)(x + 2)
So the expression = (x − 2)(x + 2)/(x − 2) = x + 2, for x ≠ 2.
Now take the limit:
limx→2 (x + 2) = 4

Q3. Write the two standard trigonometric limits and use them to evaluate limx→0 (sin 4x)/x.

Answer: The two standard limits are:
• limx→0 (sin x)/x = 1
• limx→0 (1 − cos x)/x = 0
Evaluation: Multiply and divide by 4 to match the standard form:
limx→0 (sin 4x)/x
= limx→0 4 × (sin 4x)/(4x)
= 4 × 1
= 4

Q4. Find the derivative of f(x) = x² from first principles.

Answer: By definition,
f′(x) = limh→0 [f(x + h) − f(x)] / h
= limh→0 [(x + h)² − x²] / h
= limh→0 [x² + 2xh + h² − x²] / h
= limh→0 [2xh + h²] / h
= limh→0 (2x + h)
= 2x

Q5. Write the rules for differentiating a sum, a product and a quotient of two functions.

Answer: Let u and v be two functions of x.
• Sum rule: d/dx (u + v) = du/dx + dv/dx
• Product rule: d/dx (uv) = u (dv/dx) + v (du/dx)
• Quotient rule: d/dx (u/v) = [v (du/dx) − u (dv/dx)] / v²
Also useful is the power rule: d/dx (xⁿ) = n xn−1, and the derivative of a constant is 0.

Q6. Find the derivative of y = x³ − 5x² + 7x − 2.

Answer: Differentiate each term separately using the power rule:
dy/dx = 3x² − 5(2x) + 7(1) − 0
= 3x² − 10x + 7

Q7. Find the derivative of y = (x + 1)(x² + 2) using the product rule.

Answer: Let u = x + 1 and v = x² + 2.
du/dx = 1 and dv/dx = 2x
By the product rule:
dy/dx = u (dv/dx) + v (du/dx)
= (x + 1)(2x) + (x² + 2)(1)
= 2x² + 2x + x² + 2
= 3x² + 2x + 2

Q8. Find the derivative of y = (x² + 1)/(x − 2) using the quotient rule.

Answer: Let u = x² + 1 and v = x − 2.
du/dx = 2x and dv/dx = 1
By the quotient rule:
dy/dx = [v (du/dx) − u (dv/dx)] / v²
= [(x − 2)(2x) − (x² + 1)(1)] / (x − 2)²
= [2x² − 4x − x² − 1] / (x − 2)²
= (x² − 4x − 1) / (x − 2)²

Q9. Find the derivatives of sin x, cos x and tan x.

Answer: • d/dx (sin x) = cos x
• d/dx (cos x) = −sin x
• d/dx (tan x) = sec²x
Also: d/dx (cot x) = −cosec²x, d/dx (sec x) = sec x tan x, and d/dx (cosec x) = −cosec x cot x.

Q10. Evaluate limx→0 (1 − cos x)/x².

Answer: Use the identity 1 − cos x = 2sin²(x/2):
= limx→0 2sin²(x/2) / x²
Write x² as 4(x/2)²:
= limx→0 [2 / 4] × [sin(x/2)/(x/2)]²
= (1/2) × (1)²
= 1/2

Q11. Evaluate limx→1 (x¹⁵ − 1)/(x¹⁰ − 1).

Answer: Use the standard limit limx→a (xⁿ − aⁿ)/(x − a) = n an−1.
Divide both the numerator and the denominator by (x − 1):
= [lim (x¹⁵ − 1)/(x − 1)] ÷ [lim (x¹⁰ − 1)/(x − 1)]
= [15 × 1¹⁴] / [10 × 1⁹]
= 15/10
= 3/2

Q12. Find the derivative of y = x sin x.

Answer: This is a product of two functions, so we use the product rule.
Let u = x and v = sin x.
du/dx = 1 and dv/dx = cos x
dy/dx = u (dv/dx) + v (du/dx)
= x (cos x) + sin x (1)
= x cos x + sin x

Related Study Material on Extramarks

Q.1 If A(2, 2, 3) and B(13,3,13) are two points, the equation of set of points P such that 3PA = 2PB is

Marks:1

Ans

Let the co-ordinates of point P be (x, y, z).

Now, PA = {(x + 2)2+ (y – 2)2 + (z – 3)2}.

PA2 = (x + 2)2 + (y – 2)2 + (z – 3)2

PB = {(x – 13)2+ (y + 3)2 + (z – 13)2}

PB2 = (x – 13)2 + (y + 3)2 + (z – 13)2

Therefore, from the given condition,

3PA = 2PB 9PA2 = 4PB2

9[(x + 2)2 + (y – 2)2 +(z – 3)2] = 4[(x – 13)2 + (y + 3)2 +(z – 13)2]

On solving this equation we get,

5(x2 + y2 + z2) +140x – 60y – 50z – 1235 = 0.

Q.2 By using distance formula, prove that the three points ( – 4,6,10), (2,4,6) and (14, 0, – 2) are collinear.

Marks:4

Ans

Let points A(4,6,10), B(2,4,6) and C(14,0,2) are given. AB=(2+4)2+(46)2+(610)2=36+4+16=56=214 BC=(142)2+(04)2(26)2 =144+16+64=224=414 AC=(14+4)2+(06)2+(210)2=324+36+144=504=614 AB+BC=214+414=614= AC Thus, AtB and C are collinear.

Q.3 Find the ratio in which the line segment joining the points (4, 8, 10) and (6, 10, -8) is divided by the XZ-plane.

Marks:2

Ans

LetXZ-planedividesthesegmentjoiningA(4,8,10)andB(6,10,8) atP(x,y,z)in the ratiok:1.Then the coordinates of Pare1(4)+k(6)k+1,1(8)+k(10)k+1,1(10)k(8)k+14+6kk+1,8+10kk+1,108kk+1Since,

PliesonXZ-planeSo,itsY-coordinateiszero,i.e.,8+10kk+1=08+10k=010k=8k=810k=45Therefore, XZ-plane divides AB externallyin the ratio4:5.

Q.4 Name the octants in which the following points lie: (7,4,3) and (5,-3,2).

Points (7,4,3) and (5,-3,2) lie in V and VII Octants respectively.

Marks:1

Ans

Points (7,4,3) and (5,-3,2) lie in V and VII Octants respectively.

Q.5 Find the point in XY-plane which is equidistant from three points A(2,0,3), B(0,3,2) and C(0,0,1).

Marks:4

Ans

The z-coordinate = 0 on xy-plane.
Let P (x, y, 0) be a point on xy-plane such that PA = PB = PC
Now, PA = PB
PA2 = PB2 (x 2)2 + (y0)2 + (03)2 = (x0)2 + (y 3)2 + (02)2
4x6y = 0 or 2x 3y = 0 …(1)
PB = PC
PB2 = PC2
(x 0)2 + (y3)2 + (0 2)2 = (x 0)2 + (y0)2 + (01)2 6y + 12 = 0 or y = 2 …(2)
By (1), we get x = 3
Hence, the required point is (3, 2, 0).

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