JEE Mathematics Limits and Derivatives MCQs Set 02

Answer: (c) \( \pm1 \)

Explanation:
1. The left limit is \( \lim \) kx / \( \sin x = k \).
2. The right limit is \( \lim x / \sin \) kx = \( 1/k \).
3. So \( k = 1/k \), which gives \( k^2 = 1 \) and \( k = +1 \) or \( -1 \).

Teacher's Note:
1. Identify the type of limit (indeterminate form, standard limit, greatest integer) before choosing a method.
2. Standard expansions such as \( \sin x \sim x \) and \( (a^x - 1) \sim x \ln a \) for small \( x \) make these limits quick.

Answer: (c) 19

Explanation:
1. By the product rule, \( f'/f = \alpha'/\alpha + \beta'/\beta + \gamma'/\gamma \) (when none is zero).
2. At \( x = 2 \): \( 18 = 3 + (-4) + k \).
3. So \( k = 19 \).

Teacher's Note:
1. Logarithmic differentiation turns a product into a sum of ratios.
2. Check: \( 3 - 4 + 19 = 18 \).

Answer: (d) 0

Explanation:
1. The denominator contains \( 3^{2n+3} = 27 \cdot 9^n \), which grows fastest.
2. In the numerator the largest terms are \( 5^{n+1} \) and \( 2^{2n} = 4^n \), both much smaller than \( 9^n \).
3. Dividing top and bottom by \( 9^n \), every term of the numerator tends to 0 while the denominator tends to 27. The limit is 0.

Teacher's Note:
1. Compare bases: \( 9 \gt 5 \gt 4 \gt 3 \gt 2 \), so \( 9^n \) dominates.
2. Divide by the fastest growing term.

Answer: (c) 100

Explanation:
1. \( f'(x) = x^{99} + x^{98} + ... + x + 1 \).
2. \( f'(1) = 100 \) (one for each of the 100 terms) and \( f'(0) = 1 \).
3. So \( m = 100 \).

Teacher's Note:
1. Differentiating \( x^k/k \) gives \( x^{k-1} \), so the derivative is a simple geometric-type sum.
2. \( f'(1) \) counts the number of terms.

Answer: (b) -1

Explanation:
1. \( (1 + x^{1/4})(1 - x^{1/4}) = 1 - x^{1/2} \).
2. Then \( y = (1 - x^{1/2})(1 + x^{1/2}) = 1 - x \).
3. So \( dy/dx = -1 \).

Teacher's Note:
1. Use difference of squares twice before differentiating.
2. Simplify first; it avoids the product rule.

Answer: (b) f(x)

Explanation:
1. Divide numerator and denominator by \( e^{nx} \).
2. This gives \( (f(x) + g(x)e^{-nx}) / (1 + e^{-nx}) \).
3. As \( n \) tends to \( \infty \) with \( x \gt 0, e^{-nx} \) tends to 0, and \( g \) is bounded, so the limit is \( f(x) \).

Teacher's Note:
1. Dividing by the dominant exponential is the standard method.
2. Boundedness of \( g \) makes \( g(x) e^{-nx} \) vanish.

Answer: (c) 5

Explanation:
1. If the limit is \( L \), then \( L = (1/3)(2L + 125/L^2) \).
2. \( 3L = 2L + 125/L^2 \), so \( L = 125/L^2 \), hence \( L^3 = 125 \).
3. \( L = 5 \).

Teacher's Note:
1. For a recurrence with a limit, replace both \( b_n \) and \( b_{n+1} \) by \( L \).
2. This is Newton's method for the cube root of 125.

Answer: (b) equal to 0

Explanation:
1. Near x = 0 (x not 0), cos x is slightly less than 1, so \( [\cos x] = 0 \).
2. The expression becomes \( \sin 0 / (1 + 0) = 0 \).
3. Both one-sided limits are 0, so the limit is 0.

Teacher's Note:
1. For greatest integer functions, check the value of the inner expression just on each side of the point.
2. \( \cos x \lt 1 \) for small non-zero \( x \).

Answer: (c) \( (-1)^n(n-1)! \)

Explanation:
1. Write \( y = (x+1)(x+2)...(x+n) / x^n \).
2. At \( x = -1 \) the first factor \( (x+1) \) vanishes, so only its derivative contributes: \( dy/dx = (x+2)(x+3)...(x+n)/x^n \) at \( x = -1 \).
3. The numerator gives \( 1 \cdot 2 * ... * (n-1) = (n-1)! \) and the denominator gives \( (-1)^n \). So \( dy/dx = (-1)^n (n-1)! \).

Teacher's Note:
1. When one factor vanishes at the point, only the derivative of that factor matters.
2. Check \( n = 1 \): \( y = 1 + 1/x, dy/dx = -1/x^2 = -1 \) and \( (-1)^1 \cdot 0! = -1 \).

Answer: (b) \( 1-\frac1{x^2} \)

Explanation:
1. Expand: \( (sqrt x + 1/sqrt x)^2 = x + 2 + 1/x \).
2. Differentiate: \( 1 - 1/x^2 \).

Teacher's Note:
1. Expand before differentiating.
2. The constant 2 disappears.

Answer: (c) -3

Explanation:
1. \( f'(x) = 4x + 3 \).
2. \( f'(0) = 3 \) and \( f'(-1) = -1 \).
3. So \( 3 = m(-1) \), which gives \( m = -3 \).

Teacher's Note:
1. Differentiate once and substitute both values.
2. Check: \( -3 \cdot (-1) = 3 \).

Answer: (b) \( x^2-10x+21=0 \)

Explanation:
1. Left limit: \( 4 - 1 = 3 \).
2. Right limit: \( 2(2) + 3 = 7 \).
3. Sum of roots = 10 and product = 21, so \( x^2 - 10x + 21 = 0 \).

Teacher's Note:
1. For piecewise functions use the formula for the side from which \( x \) approaches.
2. Roots 3 and 7 give sum 10 and product 21.

Answer: (b) 2

Explanation:
1. The sequence decreases and stays above \( 2 (3, sqrt5, ...) \), so it converges.
2. Let the limit be \( L \): \( L = \sqrt{2 + L} \), so \( L^2 - L - 2 = 0 \).
3. \( (L - 2)(L + 1) = 0 \) gives \( L = 2 \), since \( L \) cannot be negative.

Teacher's Note:
1. Monotonic bounded sequences converge; then solve the fixed-point equation.
2. Reject the negative root since all terms are positive.

Answer: (c) 1

Explanation:
1. As \( x \) tends to \( \infty, x \sin(1/x) = \sin(1/x) / (1/x) \) tends to 1.
2. \( \sin(1/x^2) \) tends to \( \sin 0 = 0 \).
3. So the limit is \( 1 + 0 = 1 \).

Teacher's Note:
1. Put \( t = 1/x \) to use the standard limit \( \sin t / t = 1 \).
2. The definition at \( x = 0 \) is irrelevant for the limit at \( \infty \).

Answer: (c) -1

Explanation:
1. \( f(x) = x - [x] = {x} \), which lies in [0, 1,) and the fractional part of \( f(x) \) is \( f(x) \) itself.
2. So \( {f(x)}^{2n} \) tends to 0 for every \( x \), since \( 0 \le f(x) \lt 1 \).
3. \( g(x) = (0 - 1)/(0 + 1) = -1 \).

Teacher's Note:
1. Powers of a number in [0,1) tend to 0.
2. At integer \( x, f(x) = 0 \) and the expression is already \( -1 \).

Answer: (a) 1

Explanation:
1. For small non-zero \( x, 0 \lt \cos x \lt 1 \), so \( [\cos x] = 0 \).
2. The expression is \( |x|^0 = 1 \) for all small \( x \) not equal to 0.
3. So the limit is 1.

Teacher's Note:
1. Evaluate the exponent first when the base tends to 0.
2. Anything non-zero raised to power 0 is 1.

Answer: (b) \( a=4,b=-2 \)

Explanation:
1. Since \( (1 + x^2 + x^4)(1 - x^2 + x^4) = 1 + x^4 + x^8 \), the fraction equals \( 1 - x^2 + x^4 \).
2. Its derivative is \( -2x + 4x^3 \), so \( a = 4 \) and \( b = -2 \).

Teacher's Note:
1. Look for factorisations like \( x^8 + x^4 + 1 = (x^4 + x^2 + 1)(x^4 - x^2 + 1) \).
2. Simplify before differentiating.

Answer: (a) 0

Explanation:
1. Combine powers of \( \sin x \): the exponent is \( (m - n)(m + n) + (n - p)(n + p) + (p - m)(p + m) \).
2. This equals \( (m^2 - n^2) + (n^2 - p^2) + (p^2 - m^2) = 0 \).
3. So \( f(x) = 1 \) and \( f'(x) = 0 \).

Teacher's Note:
1. Collect exponents of a common base.
2. The sum of the exponents telescopes to zero.

Answer: (a) 0

Explanation:
1. Multiply the first term by \( x^{\alpha} / x^{\alpha} \): it becomes \( x^{\alpha} / (x^{\alpha} + x^{\beta} + x^{\gamma}) \).
2. Similarly the others become \( x^{\beta} / (...) \) and \( x^{\gamma} / (...) \), with the same denominator.
3. The sum is \( 1, a \) constant, so \( dy/dx = 0 \).

Teacher's Note:
1. Multiply each term by a suitable power to get a common denominator.
2. A constant function has zero derivative.

Answer: (a) 0

Explanation:
1. \( 0 \le \sin^2(n!) \le 1 \), so the expression lies between 0 and \( n^p/(n+1) \).
2. \( n^p/(n+1) \) tends to 0 because \( p \lt 1 \).
3. By the squeeze theorem, the limit is 0.

Teacher's Note:
1. Use the squeeze theorem when a bounded oscillating factor is multiplied by a vanishing one.
2. Power of \( n \) in the numerator \( (p \lt 1) \) is less than that of the denominator (1).

Answer: (c) 0

Explanation:
1. \( (-1)^[1/x] \) is either \( +1 \) or \( -1 \), so \( |f(x)| = |x| \).
2. Hence \( -|x| \le f(x) \le |x| \), and both bounds tend to 0.
3. By the squeeze theorem, the limit is 0.

Teacher's Note:
1. A bounded factor times a quantity tending to 0 tends to 0.
2. The sign pattern of \( (-1)^[1/x] \) does not matter.

Answer: (d) 1

Explanation:
1. Put \( x - 2 = h \). The numerator is \( \sin(e^h - 1) \), which is about \( e^h - 1 \), about \( h \).
2. The denominator is \( \ln(1 + h) \), about \( h \).
3. So the ratio tends to 1.

Teacher's Note:
1. Use \( \sin t \sim t, e^h - 1 \sim h \) and \( \ln(1 + h) \sim h \) as \( h \) tends to 0.
2. All three standard limits are of the form 0/0 giving 1.

Answer: (b) \( 8\sqrt2(\log3)^2 \)

Explanation:
1. Numerator: \( 27^x - 9^x - 3^x + 1 = (9^x - 1)(3^x - 1) \), which is about \( (2x \ln 3)(x \ln 3) = 2x^2 (\ln 3)^2 \).
2. Denominator: \( \sqrt{1 + \cos x} \) is about \( \sqrt{2 - x^2/2} = sqrt2 (1 - x^2/8) \), so the denominator is about \( sqrt2 x^2/8 \).
3. Ratio = \( 2 (\ln 3)^2 \cdot 8 / sqrt2 = 8 sqrt2 (\ln 3)^2 \).

Teacher's Note:
1. Factorise exponential expressions to use \( (a^x - 1)/x -\gt \ln a \).
2. Expand the root in the denominator with the binomial approximation.