Class 11 Mathematics Limits and Derivatives MCQs Set 20

Answer: (c) \( \frac{1}{2} \)

Explanation:
1. \( \operatorname{cosec} x - \cot x = \frac{1 - \cos x}{\sin x} = \frac{2\sin^2 \frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} = \tan\frac{x}{2} \)
2. \( \frac{\tan \frac{x}{2}}{x} = \frac{1}{2} \cdot \frac{\tan \frac{x}{2}}{\frac{x}{2}} \to \frac{1}{2} \)

Teacher's Note:
1. cosec x − cot x simplifies to tan(x/2), a useful identity to remember.
2. Adjust the denominator to x/2 to match the angle of tan.

Answer: (a) 2

Explanation:
1. \( 1 - \cos x = 2\sin^2\frac{x}{2} \)
2. \( \frac{x^2 \cos x}{2\sin^2 \frac{x}{2}} = \frac{\cos x}{2} \cdot \frac{4}{\left(\frac{\sin (x/2)}{x/2}\right)^2} \to \frac{1}{2} \times 4 = 2 \)

Teacher's Note:
1. Convert 1 − cos x into 2 sin²(x/2) so a standard sine limit appears.
2. Equivalently \( \frac{1 - \cos x}{x^2} \to \frac{1}{2} \), so the answer is 1 ÷ (1/2) = 2.

Answer: (a) 0

Explanation:
1. Substitute x = −1: 1 − 1 + 1 − 1 + … (10 terms).
2. The terms cancel in pairs, so the sum is 0.

Teacher's Note:
1. Count the terms: powers 0 to 9 give 10 terms, an even number.
2. Even powers of −1 give +1 and odd powers give −1.

Answer: (b) 100

Explanation:
1. Substitute x = 5: 25 × 4 = 100.

Teacher's Note:
1. Polynomial limits are found by direct substitution.
2. Keep the bracket (x − 1) = 4 separate before multiplying.

Answer: (d) 13

Explanation:
1. Substitute x = 2: 8 + 4 + 2 − 1 = 13.

Teacher's Note:
1. Substitute carefully term by term.
2. The constant −1 is unchanged by the limit.

Answer: (b) −28

Explanation:
1. Substitute x = −3: (−27) + (−3) + 2 = −28.

Teacher's Note:
1. An odd power of a negative number stays negative: (−3)³ = −27.
2. Bracket negative values while substituting.

Answer: (a) 192

Explanation:
1. Substitute x = 4: 256 − 64 = 192.

Teacher's Note:
1. x⁴ − x³ = x³(x − 1) = 64 × 3 = 192 is a quicker route.
2. Factorising first often reduces arithmetic.

Answer: (b) \( \frac{-1}{2} \)

Explanation:
1. h(−1) = −2 ≠ 0, so substitute: \( \frac{1 - 1 + 1}{-1 - 1} = -\frac{1}{2} \)

Teacher's Note:
1. (−1)¹⁰ = 1 and (−1)⁵ = −1.
2. Check the denominator is non-zero before substituting.

Answer: (a) \( \frac{7}{4} \)

Explanation:
1. Substitute x = −1: \( \frac{(-2)^2 + 3}{(1 + 1)^2} = \frac{7}{4} \)

Teacher's Note:
1. The denominator (x⁴ + 1)² is never zero, so substitution always works here.
2. (−1 − 1)² = 4, not 0.

Answer: (d) Does not exist

Explanation:
1. \( \frac{(x - 2)(x + 2)}{x(x - 2)^2} = \frac{x + 2}{x(x - 2)} \) for x ≠ 2
2. As x → 2 the numerator → 4 while the denominator → 0, so the limit does not exist.

Teacher's Note:
1. After cancelling the common factor, check again whether the denominator still vanishes.
2. A non-zero number divided by something tending to 0 has no finite limit.

Answer: (b) 1

Explanation:
1. At x = 1 both numerator and denominator are 0, so (x − 1) is a factor of each.
2. x⁷ − 2x⁵ + 1 = (x − 1)(x⁶ + x⁵ − x⁴ − x³ − x² − x − 1) and x³ − 3x² + 2 = (x − 1)(x² − 2x − 2).
3. At x = 1: (1 + 1 − 1 − 1 − 1 − 1 − 1)/(1 − 2 − 2) = (−3)/(−3) = 1.

Teacher's Note:
1. Use synthetic division by (x − 1) to find the other factor quickly.
2. The 0/0 form is a signal that x − a divides both numerator and denominator.

Answer: (b) 0

Explanation:
1. Rationalise: \( \frac{(1 + x^3) - (1 - x^3)}{x^2(\sqrt{1 + x^3} + \sqrt{1 - x^3})} = \frac{2x^3}{x^2(\sqrt{1 + x^3} + \sqrt{1 - x^3})} \)
2. \( = \frac{2x}{\sqrt{1 + x^3} + \sqrt{1 - x^3}} \to \frac{0}{2} = 0 \)

Teacher's Note:
1. Multiply by the conjugate to remove the difference of square roots.
2. After cancelling, one factor of x is left in the numerator, so the limit is 0.

Answer: (c) \( \frac{3}{5} \)

Explanation:
1. \( \frac{\sin 3x}{5x} = \frac{3}{5} \cdot \frac{\sin 3x}{3x} \to \frac{3}{5} \)

Teacher's Note:
1. Make the denominator equal to the angle 3x.
2. The ratio of coefficients gives the answer directly.

Answer: (b) 1

Explanation:
1. Put t = θ − b; as θ → b, t → 0.
2. \( \lim_{t \to 0} \frac{\tan t}{t} = 1 \)

Teacher's Note:
1. The standard limits work for any quantity tending to 0, not only x.
2. Shift the variable so that it tends to 0.

Answer: (a) 4

Explanation:
1. \( \tan 2x - \sin 2x = \tan 2x(1 - \cos 2x) = \tan 2x \cdot 2\sin^2 x \)
2. \( \frac{\tan 2x \cdot 2\sin^2 x}{x^3} = 2 \cdot \frac{\tan 2x}{2x} \cdot 2 \cdot \left(\frac{\sin x}{x}\right)^2 \to 4 \)

Teacher's Note:
1. Take tan 2x common; sin 2x = tan 2x cos 2x.
2. 1 − cos 2x = 2 sin² x supplies the remaining x².

Answer: (b) 1

Explanation:
1. \( 2\sin x - \sin 2x = 2\sin x(1 - \cos x) = 2\sin x \cdot 2\sin^2\frac{x}{2} \)
2. \( \frac{4 \sin x \sin^2 \frac{x}{2}}{x^3} = 4 \cdot \frac{\sin x}{x} \cdot \frac{1}{4}\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^2 \to 1 \)

Teacher's Note:
1. Use sin 2x = 2 sin x cos x to take 2 sin x common.
2. Then 1 − cos x = 2 sin²(x/2) gives the extra x² needed.

Answer: (a) \( \sqrt{2} \)

Explanation:
1. \( \sin x - \cos x = \sqrt{2}\sin\left(x - \frac{\pi}{4}\right) \)
2. Put \( t = x - \frac{\pi}{4} \to 0 \): \( \frac{\sqrt{2}\sin t}{t} \to \sqrt{2} \)

Teacher's Note:
1. sin x − cos x = √2 sin(x − π/4) follows from the compound angle formula.
2. After the shift t = x − π/4, the limit is the standard sin t / t.

Answer: (c) A is true, but R is false.

Explanation:
1. At x = 1 the expression is \( \frac{a + b + c}{c + b + a} = 1 \) since a + b + c ≠ 0; A is true.
2. \( \frac{\frac{1}{x} + \frac{1}{2}}{x + 2} = \frac{x + 2}{2x(x + 2)} = \frac{1}{2x} \to -\frac{1}{4} \), so R is false.

Teacher's Note:
1. Direct substitution works in A because the denominator is non-zero at x = 1.
2. Simplify the compound fraction in R before substituting; watch the sign of x = −2.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( \frac{\sin ax}{bx} = \frac{a}{b} \cdot \frac{\sin ax}{ax} \to \frac{a}{b} \times 1 \)
2. A is true, and it follows directly from the standard limit in R.

Teacher's Note:
1. Multiply and divide by a to create sin ax / ax.
2. R is the basic trigonometric limit used throughout this chapter.

Answer: (d) A is false, but R is true.

Explanation:
1. Divide by x: \( \frac{\frac{\sin ax}{x} + b}{a + \frac{\sin bx}{x}} \to \frac{a + b}{a + b} = 1 \), so A is false.
2. R: 5 + 5 + 1 = 11 by direct substitution, so R is true.

Teacher's Note:
1. Dividing each term by x reduces the expression to standard limits.
2. Polynomial limits are evaluated by substitution.

Answer: (d) A is false, but R is true.

Explanation:
1. Put h = π − x → 0: \( \frac{\sin h}{\pi h} = \frac{1}{\pi} \cdot \frac{\sin h}{h} \to \frac{1}{\pi} \). So A (value π) is false.
2. R: direct substitution gives \( \frac{\cos 0}{\pi} = \frac{1}{\pi} \), true.

Teacher's Note:
1. Substituting h = π − x converts the limit to the standard form sin h / h.
2. The factor π in the denominator stays as a constant 1/π.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( \frac{\sin ax}{\sin bx} = \frac{\sin ax}{ax} \cdot \frac{bx}{\sin bx} \cdot \frac{a}{b} \to \frac{a}{b} \)
2. A is true and is obtained using R, so R explains A.

Teacher's Note:
1. Create the standard form for both numerator and denominator.
2. Assume b ≠ 0 so that sin bx is not identically zero.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. \( \frac{\cos 2x - 1}{\cos x - 1} = \frac{2\sin^2 x}{2\sin^2 \frac{x}{2}} = \left(\frac{\sin x}{x}\right)^2 \cdot \left(\frac{x/2}{\sin (x/2)}\right)^2 \cdot 4 \to 4 \)
2. A is true; R is a true standard limit, but A uses sin limits, not tan x / x, so R does not explain A.

Teacher's Note:
1. 1 − cos θ = 2 sin²(θ/2) is the key identity for such ratios.
2. A true statement R explains A only if it is actually used to obtain A.

Answer: (c) A is true, but R is false.

Explanation:
1. \( \frac{x(a + \cos x)}{b \sin x} = \frac{a + \cos x}{b} \cdot \frac{x}{\sin x} \to \frac{a + 1}{b} \); A is true.
2. R: x sec x = x / cos x → 0/1 = 0, so R is false.

Teacher's Note:
1. Take x common in the numerator to expose x / sin x.
2. sec 0 = 1, so x sec x simply tends to 0.