CBSE Class 11 Mathematics Limits and Derivatives JEE Previous Year Questions Set 01

Answer: (d) does not exist

Explanation:
1. \( \sqrt{1 - \cos 2x} = \sqrt{2\sin^2 x} = \sqrt{2}\,|\sin x| \), so the expression = \( \frac{|\sin x|}{x} \)
2. RHL (x → 0⁺): \( \frac{\sin x}{x} \to 1 \); LHL (x → 0⁻): \( \frac{-\sin x}{x} \to -1 \)
3. LHL ≠ RHL, so the limit does not exist.

Teacher's Note:
1. √(sin²x) is |sin x|, not sin x; the modulus changes sign on the two sides of 0.
2. Whenever a modulus appears, check the left-hand and right-hand limits separately.

Answer: (c) −4

Explanation:
1. x f(2) − 2f(x) = (x − 2)f(2) − 2[f(x) − f(2)] (add and subtract 2f(2))
2. Expression = \( f(2) - 2 \cdot \frac{f(x) - f(2)}{x - 2} \to f(2) - 2f'(2) \)
3. = 4 − 2 × 4 = −4

Teacher's Note:
1. Adding and subtracting a suitable term creates the first-principle form (f(x) − f(a))/(x − a).
2. That ratio tends to f′(a), so given values of f and f′ are enough to finish.

Answer: (c) \( \frac{1}{32} \)

Explanation:
1. \( \frac{1 - \tan\frac{x}{2}}{1 + \tan\frac{x}{2}} = \tan\left( \frac{\pi}{4} - \frac{x}{2} \right) \). Put y = π/4 − x/2 → 0, so x = π/2 − 2y and π − 2x = 4y.
2. 1 − sin x = 1 − cos 2y = 2 sin² y
3. Expression = \( \frac{\tan y \cdot 2\sin^2 y}{64 y^3} = \frac{2}{64} \cdot \frac{\tan y}{y} \cdot \left( \frac{\sin y}{y} \right)^2 \to \frac{1}{32} \)

Teacher's Note:
1. (1 − tan A)/(1 + tan A) = tan(π/4 − A) is a handy identity for such problems.
2. Choosing the new variable so that every factor becomes a multiple of y makes the cubic denominator easy to handle.

Answer: (d) \( \frac{a^2}{2}(\alpha - \beta)^2 \)

Explanation:
1. ax² + bx + c = a(x − α)(x − β) = θ (say), and θ → 0 as x → α.
2. 1 − cos θ = 2 sin²(θ/2), so the expression = \( 2\left( \frac{\sin\frac{\theta}{2}}{\frac{\theta}{2}} \right)^2 \cdot \frac{a^2 (x - \alpha)^2 (x - \beta)^2}{4(x - \alpha)^2} \)
3. Limit = \( 2 \cdot \frac{a^2(\alpha - \beta)^2}{4} = \frac{a^2}{2}(\alpha - \beta)^2 \)

Teacher's Note:
1. Writing the quadratic in factor form a(x − α)(x − β) shows why the angle tends to 0.
2. The result (1 − cos θ)/θ² → 1/2 is the key standard limit here.

Answer: (d) does not exist

Explanation:
1. \( \sqrt{1 - \cos 2(x - 2)} = \sqrt{2}\,|\sin(x - 2)| \)
2. RHL: \( \sqrt{2} \cdot \frac{\sin(x - 2)}{x - 2} \to \sqrt{2} \); LHL: \( -\sqrt{2} \cdot \frac{\sin(x - 2)}{x - 2} \to -\sqrt{2} \)
3. LHL ≠ RHL, so the limit does not exist.

Teacher's Note:
1. The square root of a square always gives a modulus, which behaves differently on the two sides.
2. A limit exists only when the left-hand and right-hand limits are equal.

Answer: (a) 3

Explanation:
1. \( [f(x)]^2 - 9 = \frac{[f(x)]^2 - 9}{\sqrt{|x - 5|}} \times \sqrt{|x - 5|} \to 0 \times 0 = 0 \)
2. So \( \left[\lim_{x \to 5} f(x)\right]^2 = 9 \), i.e. the limit is 3 or −3.
3. Since f(x) ≥ 0, the limit is 3.

Teacher's Note:
1. Multiplying and dividing by the same quantity lets us use the product rule of limits.
2. The given range [0, ∞) of f is what rules out the value −3.

Answer: (d) 2a f(a) − a² f′(a)

Explanation:
1. x² f(a) − a² f(x) = (x² − a²) f(a) − a²[f(x) − f(a)] (add and subtract a² f(a))
2. Expression = \( (x + a) f(a) - a^2 \cdot \frac{f(x) - f(a)}{x - a} \)
3. Limit = 2a f(a) − a² f′(a)

Teacher's Note:
1. Adding and subtracting a² f(a) separates the expression into two easy limits.
2. (f(x) − f(a))/(x − a) → f′(a) is the definition of the derivative.

Answer: (d) 2

Explanation:
1. 1 − cos 2x = 2 sin² x
2. Expression = \( 2\left( \frac{\sin x}{x} \right)^2 \cdot \frac{4x}{\tan 4x} \cdot \frac{3 + \cos x}{4} \)
3. Limit = \( 2 \times 1 \times 1 \times \frac{4}{4} = 2 \)

Teacher's Note:
1. Pair x with tan 4x by writing x tan 4x = (1/4)(4x) tan 4x.
2. Factors such as (3 + cos x) that do not vanish can be found by direct substitution.

Answer: (a) π

Explanation:
1. π cos² x = π − π sin² x, and sin(π − θ) = sin θ, so sin(π cos² x) = sin(π sin² x).
2. Expression = \( \frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} \cdot \pi \cdot \left( \frac{\sin x}{x} \right)^2 \)
3. Limit = 1 × π × 1 = π

Teacher's Note:
1. The angle π cos² x tends to π, not 0, so first convert it into an angle that tends to 0.
2. Then the standard result sin θ/θ → 1 can be applied with θ = π sin² x.

Answer: (d) 3

Explanation:
1. x² + (k − 2)x − 2k = (x − 2)(x + k) and x² − 4x + 4 = (x − 2)²
2. Expression = \( \frac{\tan(x - 2)}{x - 2} \cdot (x + k) \to 1 \times (2 + k) \)
3. 2 + k = 5, so k = 3

Teacher's Note:
1. Factorising both quadratics shows the factor (x − 2) needed for tan(x − 2)/(x − 2).
2. Use tan(x − a)/(x − a) → 1 as x → a.

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

Explanation:
1. tan 2x − 2 tan x = \( \frac{2\tan x}{1 - \tan^2 x} - 2\tan x = \frac{2\tan^3 x}{1 - \tan^2 x} \)
2. (1 − cos 2x)² = 4 sin⁴ x, so the expression = \( \frac{1}{2} \cdot \left( \frac{\tan x}{x} \right)^3 \cdot \left( \frac{x}{\sin x} \right)^4 \cdot \frac{1}{1 - \tan^2 x} \)
3. Limit = \( \frac{1}{2} \)

Teacher's Note:
1. Using the double-angle formula for tan 2x reduces the numerator to a single power of tan x.
2. Count powers: x tan³x on top and sin⁴x below are both of order x⁴.

Answer: (a) −2

Explanation:
1. 2x tan x − x tan 2x = −x(tan 2x − 2 tan x) = \( -\frac{2x\tan^3 x}{1 - \tan^2 x} \)
2. Expression = \( -\frac{4\sin^4 x(1 - \tan^2 x)}{2x\tan^3 x} = -2 \left( \frac{\sin x}{x} \right)^4 \left( \frac{x}{\tan x} \right)^3 (1 - \tan^2 x) \)
3. Limit = −2

Teacher's Note:
1. This is the reciprocal of a similar expression with the sign changed, so the answer is −1 ÷ (1/2) = −2.
2. Keep track of the minus sign that appears when the order of terms in the denominator is reversed.

Answer: (b) \( \frac{1}{16} \)

Explanation:
1. cot x − cos x = \( \frac{\cos x(1 - \sin x)}{\sin x} \). Put x = π/2 − h: cos x = sin h, sin x = cos h, π − 2x = 2h.
2. Expression = \( \frac{\sin h(1 - \cos h)}{8h^3 \cos h} = \frac{1}{8} \cdot \frac{\sin h}{h} \cdot \frac{1 - \cos h}{h^2} \cdot \frac{1}{\cos h} \)
3. Limit = \( \frac{1}{8} \times 1 \times \frac{1}{2} \times 1 = \frac{1}{16} \)

Teacher's Note:
1. Shift the point π/2 to 0 with x = π/2 − h so standard limits can be used.
2. Remember (1 − cos h)/h² → 1/2.

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

Explanation:
1. Rationalise numerator: \( \sqrt{3x} - 3 = \frac{3(x - 3)}{\sqrt{3x} + 3} \)
2. Rationalise denominator: \( \sqrt{2x - 4} - \sqrt{2} = \frac{2(x - 3)}{\sqrt{2x - 4} + \sqrt{2}} \)
3. Expression = \( \frac{3(\sqrt{2x - 4} + \sqrt{2})}{2(\sqrt{3x} + 3)} \to \frac{3 \times 2\sqrt{2}}{2 \times 6} = \frac{1}{\sqrt{2}} \)

Teacher's Note:
1. Rationalise both numerator and denominator so the common factor (x − 3) appears in each.
2. Simplify √2/2 to 1/√2 to match the options.

Answer: (c) \( -\frac{1}{6} \)

Explanation:
1. Let t = (27 + x)^1/3; as x → 0, t → 3.
2. Expression = \( \frac{t - 3}{9 - t^2} = \frac{t - 3}{(3 - t)(3 + t)} = \frac{-1}{3 + t} \)
3. Limit = \( -\frac{1}{6} \)

Teacher's Note:
1. A substitution for the cube root turns the expression into a simple rational function.
2. Note that 9 − t² = (3 − t)(3 + t) = −(t − 3)(t + 3).

Answer: (c) 5

Explanation:
1. g(x)f(a) − g(a)f(x) = f(a)[g(x) − g(a)] − g(a)[f(x) − f(a)] (add and subtract f(a)g(a))
2. Limit = f(a) g′(a) − g(a) f′(a)
3. = 2 × 2 − (−1) × 1 = 5

Teacher's Note:
1. Adding and subtracting f(a)g(a) creates two first-principle derivative ratios.
2. Substitute the given values only at the end to avoid sign errors.

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

Explanation:
1. Sum = \( \frac{1 + 2 + \ldots + n}{1 - n^2} = \frac{n(n + 1)}{2(1 - n)(1 + n)} = \frac{n}{2(1 - n)} \)
2. Divide by n: \( \frac{1}{2\left(\frac{1}{n} - 1\right)} \to -\frac{1}{2} \)

Teacher's Note:
1. Take the common denominator out first; the numerator is then just 1 + 2 + … + n.
2. Do not take the limit term by term: each term tends to 0 but there are n of them.

Answer: (d) \( n + \frac{1}{n} \)

Explanation:
1. Expression = \( \left\{ (a - n)n - \frac{\tan x}{x} \right\} \cdot \frac{\sin nx}{x} \)
2. As x → 0 this tends to {(a − n)n − 1} · n.
3. Since n ≠ 0, (a − n)n = 1, so a = n + 1/n.

Teacher's Note:
1. Split x² as x · x and attach one x to each factor.
2. sin nx/x = n · (sin nx/nx) → n.

Answer: (b) a = 1, b = −4

Explanation:
1. Combine: \( \frac{(1 - a)x^2 + (1 - a - b)x + (1 - b)}{x + 1} \)
2. For a finite limit the x² term must vanish: a = 1.
3. Then \( \frac{-bx + (1 - b)}{x + 1} \to -b = 4 \), so b = −4.

Teacher's Note:
1. If the numerator's degree exceeds the denominator's, the limit at infinity is infinite, so extra coefficients must be zero.
2. With equal degrees the limit is the ratio of leading coefficients.

Answer: (b) \( -\frac{1}{2} \) and −1

Explanation:
1. Divide the equation by a (a ≠ 0) and put y = 1 + a → 1.
2. Using \( \lim_{y \to 1} \frac{y^n - 1}{y - 1} = n \): the coefficients tend to \( \frac{1}{3}, \frac{1}{2}, \frac{1}{6} \).
3. Limiting equation: \( \frac{x^2}{3} + \frac{x}{2} + \frac{1}{6} = 0 \Rightarrow 2x^2 + 3x + 1 = 0 \Rightarrow x = -\frac{1}{2}, -1 \)

Teacher's Note:
1. Dividing every coefficient by a does not change the roots of the equation.
2. The standard limit (xⁿ − aⁿ)/(x − a) → naⁿ⁻¹ works for fractional n as well.

Answer: (a) a = 2; (c) \( L = \frac{1}{64} \)

Explanation:
1. \( a - \sqrt{a^2 - x^2} = \frac{x^2}{a + \sqrt{a^2 - x^2}} \), so \( L = \lim_{x \to 0} \frac{\frac{1}{a + \sqrt{a^2 - x^2}} - \frac{1}{4}}{x^2} \)
2. For L finite the numerator must tend to 0: \( \frac{1}{2a} = \frac{1}{4} \), so a = 2.
3. Then \( L = \lim \frac{2 - \sqrt{4 - x^2}}{4x^2(2 + \sqrt{4 - x^2})} = \lim \frac{1}{4(2 + \sqrt{4 - x^2})^2} = \frac{1}{64} \)

Teacher's Note:
1. Rationalising a − √(a² − x²) brings out the factor x² and reduces the power in the denominator.
2. A finite limit with a denominator tending to 0 forces the numerator to tend to 0, which fixes a.

Answer: -1

Explanation:
1. For x < 0, |x|³ = −x³. Divide numerator and denominator by x³.
2. Expression = \( \frac{\frac{\sin(1/x)}{1/x} + \frac{1}{x}}{\frac{1}{x^3} - 1} \)
3. As x → −∞, 1/x → 0, so sin(1/x)/(1/x) → 1 and the limit = \( \frac{1 + 0}{0 - 1} = -1 \)

Teacher's Note:
1. Here the brackets are ordinary brackets, not the greatest integer function.
2. x⁴ sin(1/x)/x³ = x sin(1/x) = sin(1/x)/(1/x), which tends to 1 as x → ±∞.

Answer: 0

Explanation:
1. At x = 0 the numerator x − sin x = 0 and the denominator x + cos² x = 1 ≠ 0.
2. So the limit is found by direct substitution: \( \sqrt{\frac{0}{1}} = 0 \)

Teacher's Note:
1. Always try direct substitution first; it fails only when it gives an indeterminate form such as 0/0.
2. Near 0 the expression under the root is real only for x ≥ 0, so this is really a right-hand limit, and it equals 0.