JEE Mathematics Functions MCQs Set 03

Answer: (b) \( \lambda \) and \( \mu \) are integers

Explanation:
1. \( f(0) = \mu \) must be an integer.
2. \( f(1) = 1 + \lambda + \mu \) must be an integer, so \( \lambda \) is an integer.
3. Hence both are integers.

Teacher's Note:
1. Plug in the first few integers to force the coefficients.
2. An integral function can still have non-integer coefficients in general, but here the leading coefficient is 1.

Answer: (b) an integer

Explanation:
1. \( f(0) = c \) is an integer.
2. \( f(1) = a + b + c \) is an integer, so \( a + b = f(1) - c \) is an integer.
3. It need not be negative, so the answer is an integer.

Teacher's Note:
1. Use f(0) and f(1) first.
2. The difference of two integers is always an integer.

Answer: (c) \( \cos 4 \)

Explanation:
1. \( [\pi] = 3 \), so the first term is \( \cos\left(\frac{3\pi}{2}\right) = 0 \).
2. \( \pi x = \frac{\pi^2}{2} \approx 4.93 \), so \( [\pi x] = 4 \) and the second term is \( \cos 4 \).
3. Hence \( f(\pi/2) = \cos 4 \).

Teacher's Note:
1. [π] means the greatest integer not exceeding π, which is 3.
2. Angles here are in radians.

Answer: (c) -1

Explanation:
1. \( \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3} \).
2. \( f(-\sqrt{3}) = \sin(-\pi/3) = -\frac{\sqrt{3}}{2} \approx -0.866 \).
3. The greatest integer not exceeding \( -0.866 \) is \( -1 \).

Teacher's Note:
1. The greatest integer of a negative decimal goes down, not toward zero.
2. Find the inner value first.

Answer: (c) \( \frac{(a+1)f(x)+a-1}{(a-1)f(x)+a+1} \)

Explanation:
1. Put \( y = f(x) \); then \( x = \frac{1+y}{1-y} \).
2. \( f(ax) = \frac{ax-1}{ax+1} = \frac{a(1+y)-(1-y)}{a(1+y)+(1-y)} \).
3. This simplifies to \( \frac{(a+1)y + (a-1)}{(a-1)y + (a+1)} \).

Teacher's Note:
1. Solve for x in terms of f(x) first.
2. Then substitute into f(ax) and simplify.

Answer: (c) \( 3^{m-1} \)

Explanation:
1. \( f(2) = 2f(1) = 2 \), \( f(3) = 2(1+2) = 6 \), and in general \( f(n) = 2 \cdot 3^{n-2} \) for \( n \geq 2 \).
2. \( \sum_{n=2}^{m} 2 \cdot 3^{n-2} = 3^{m-1} - 1 \).
3. Adding \( f(1) = 1 \) gives \( 3^{m-1} \).

Teacher's Note:
1. Compute f(2), f(3) and spot the geometric pattern.
2. Add f(1) separately because the formula starts at n = 2.

Answer: (a) \( nf(1) \)

Explanation:
1. \( f(x+1) - f(x) = f(x) - f(x-1) \), so the values form an arithmetic progression.
2. With \( f(0) = 0 \) the common difference is \( f(1) \), so \( f(n) = nf(1) \).

Teacher's Note:
1. Equal consecutive differences mean an arithmetic progression.
2. The first term is f(0) = 0.

Answer: (a) \( \frac{2a+b}{2(a^2-b^2)} \)

Explanation:
1. Put \( u = x+1 \): \( af(u) + bf(1/u) = u - 1 \) and, replacing \( u \) by \( 1/u \), \( af(1/u) + bf(u) = \frac{1}{u} - 1 \).
2. Eliminating \( f(1/u) \): \( (a^2-b^2)f(u) = a(u-1) - b\left(\frac1u - 1\right) \).
3. At \( u = 2 \): \( f(2) = \frac{a + b/2}{a^2-b^2} = \frac{2a+b}{2(a^2-b^2)} \).

Teacher's Note:
1. Replace the variable by its reciprocal to get a second equation.
2. Solve the pair like simultaneous equations.

Answer: (b) nk

Explanation:
1. \( f(2) = 2f(1) \), \( f(3) = f(2)+f(1) = 3f(1) \), and so on.
2. Hence \( f(n) = nf(1) = nk \).

Teacher's Note:
1. Repeated addition gives multiples.
2. This is the Cauchy functional equation.

Answer: (a) \( \frac{3}{2}(3^n - 1) \)

Explanation:
1. \( f(r) = f(1)^r = 3^r \).
2. The sum is a geometric series: \( 3 + 3^2 + \dots + 3^n = \frac{3(3^n-1)}{3-1} = \frac{3}{2}(3^n-1) \).

Teacher's Note:
1. A product rule f(x+y) = f(x)f(y) means f is exponential.
2. Then use the GP sum formula.

Answer: (a) one

Explanation:
1. Try \( f(x) = -\frac{x^2}{2} + cx + 1 \); it satisfies the rule, and \( f(1) = 1 \) gives \( c = \frac12 \).
2. So \( f(n) = \frac{-n^2 + n + 2}{2} \). Setting \( f(n) = n \): \( n^2 + n - 2 = 0 \), so \( n = 1 \) or \( n = -2 \).
3. Only \( n = 1 \) is a natural number, so there is one solution.

Teacher's Note:
1. Guess a quadratic form from the xy term.
2. Reject roots that are not natural numbers.

Answer: (b) (2, 5]

Explanation:
1. The first root needs \( (x+2)(5-x) \geq 0 \), so \( -2 \leq x \leq 5 \).
2. The second needs \( x^2 - 4 > 0 \), so \( x > 2 \) or \( x \lt -2 \).
3. The intersection is \( (2, 5] \).

Teacher's Note:
1. Every square root and denominator gives a condition.
2. Take the intersection of all of them.

Answer: (d) \( (-\infty, +\infty) \)

Explanation:
1. The domain needs \( x^{12} - x^9 + x^4 - x + 1 > 0 \).
2. For \( x \leq 0 \) every term is non-negative and the constant is 1. For \( x \geq 1 \): \( x^9(x^3-1) + x(x^3-1) + 1 > 0 \). For \( 0 \lt x \lt 1 \): \( x^4 > x^9 \) and \( 1 - x > 0 \).
3. So the expression is always positive and the domain is all real numbers.

Teacher's Note:
1. Split the number line into ranges to prove positivity.
2. Negative powers like -1/2 need a strictly positive base.

Answer: (d) \( \left[\frac{1}{\sqrt2}, 1\right] \)

Explanation:
1. \( 1 - x^2 \geq 0 \) gives \( -1 \leq x \leq 1 \).
2. \( x \geq \sqrt{1-x^2} \) needs \( x \geq 0 \) and \( x^2 \geq 1 - x^2 \), i.e. \( x \geq \frac{1}{\sqrt2} \).
3. Domain: \( \left[\frac{1}{\sqrt2}, 1\right] \).

Teacher's Note:
1. A square root of a square root needs both conditions.
2. Square only when both sides are non-negative.

Answer: (d) [-1, 1]

Explanation:
1. The innermost root needs \( |x| \leq 1 \).
2. Then \( 0 \leq \sqrt{1-x^2} \leq 1 \), so \( 1 - \sqrt{1-x^2} \in [0, 1] \) and its root lies in \( [0, 1] \).
3. So \( 1 - \sqrt{\cdot} \geq 0 \) always holds, and the domain is \( [-1, 1] \).

Teacher's Note:
1. Work from the innermost root outward.
2. Each root of a number in [0,1] stays in [0,1].

Answer: (b) \( (0, +\infty) \)

Explanation:
1. The outer log needs \( \log_{10}(1+x^3) > 0 \).
2. That means \( 1 + x^3 > 1 \), so \( x^3 > 0 \) and \( x > 0 \).

Teacher's Note:
1. For a nested log, the inside of the outer log must be positive.
2. log10 of a number is positive only when the number exceeds 1.

Answer: (a) R

Explanation:
1. We need \( x^2 \geq [x] \).
2. If \( x \lt 0 \) then \( [x] \lt 0 \leq x^2 \). If \( 0 \leq x \lt 1 \) then \( [x] = 0 \). If \( x \geq 1 \) then \( x^2 \geq x \geq [x] \).
3. The condition holds for every real \( x \).

Teacher's Note:
1. Check negative, fractional and large x separately.
2. Remember [x] never exceeds x.

Answer: (a) \( (\sqrt2, +\infty) \)

Explanation:
1. The base needs \( x^2 - 1 > 0 \), \( x^2 - 1 \neq 1 \), and the argument needs \( x > 0 \): so \( x > 1 \), \( x \neq \sqrt2 \).
2. If \( x > \sqrt2 \) the base exceeds 1 and \( \log(x) \geq 0 \) because \( x > 1 \). If \( 1 \lt x \lt \sqrt2 \) the base is in \( (0,1) \) and \( \log \geq 0 \) would need \( x \leq 1 \), which fails.
3. Domain: \( (\sqrt2, +\infty) \).

Teacher's Note:
1. A base between 0 and 1 reverses the inequality.
2. Always exclude base 1.

Answer: (c) {2, 3}

Explanation:
1. \( ^nC_r \) needs \( r \leq n \): \( 2x - 1 \leq 16 - x \) gives \( x \leq 17/3 \).
2. \( ^nP_r \) needs \( r \leq n \): \( 4x - 5 \leq 20 - 3x \) gives \( x \leq 25/7 \approx 3.57 \). Also \( x \) must be a positive integer with \( 4x - 5 \geq 0 \), so \( x \geq 2 \).
3. Hence \( x \in \{2, 3\} \).

Teacher's Note:
1. n and r must be integers with r not more than n.
2. The tighter of the two upper bounds is used.

Answer: (c) {1, -1}

Explanation:
1. \( \left|\frac{1+x^2}{2x}\right| \leq 1 \) forces \( |1+x^2| \leq 2|x| \), i.e. \( (|x|-1)^2 \leq 0 \), so \( |x| = 1 \).
2. \( x = \pm1 \) also satisfies \( \sqrt{1-x^2} \). Domain: \( \{1, -1\} \).

Teacher's Note:
1. x + 1/x has magnitude at least 2 for real x.
2. Equality holds only at x = ±1.

Answer: (a) \( (-\infty, -3] \cup [3, +\infty) \)

Explanation:
1. \( \sec^{-1}(t) \) is defined only for \( |t| \geq 1 \). So \( \left|\frac{1-|x|}{2}\right| \geq 1 \).
2. \( 1 - |x| \geq 2 \) is impossible, and \( 1 - |x| \leq -2 \) gives \( |x| \geq 3 \).
3. The value of \( \sec^{-1} \) is always non-negative, so the root causes no further restriction.

Teacher's Note:
1. Inverse secant needs the argument outside (-1, 1).
2.
3. x
4. is never negative, which kills one case.

Answer: (b) \( x \in \left[-2, -\frac12\right] \cup \left[\frac12, 2\right] \)

Explanation:
1. \( \cos^{-1} \) needs \( -1 \leq \log_4 x^2 \leq 1 \), so \( \frac14 \leq x^2 \leq 4 \).
2. That gives \( \frac12 \leq |x| \leq 2 \), i.e. both intervals in option (b).

Teacher's Note:
1. x² allows negative x too.
2. Convert the log inequality to powers of 4.

Answer: (d) none of these

Explanation:
1. \( \log_e x \) needs \( x > 0 \).
2. The outer log needs \( |\log_e x| > 0 \), so \( x \neq 1 \).
3. Domain: \( (0, 1) \cup (1, \infty) \), which is none of the listed intervals.

Teacher's Note:
1. A modulus allows both signs, only zero is excluded.
2. Do not forget x > 0.

Answer: (b) \( R - Z \)

Explanation:
1. We need \( x - [x] > 0 \), i.e. the fractional part must be positive.
2. This fails exactly when \( x \) is an integer. Domain: \( R - Z \).

Teacher's Note:
1. x - [x] is the fractional part.
2. It is zero at every integer.

Answer: (a) \( [0, 1) \)

Explanation:
1. We need \( -1 \leq x + [x] \leq 1 \).
2. For \( 0 \leq x \lt 1 \): \( x + [x] = x \in [0, 1) \), valid. For \( -1 \leq x \lt 0 \): \( x + [x] = x - 1 \lt -1 \), invalid. \( x = 1 \) gives 2, invalid.
3. Domain: \( [0, 1) \).

Teacher's Note:
1. Test the range [x]=0 and [x]=-1 separately.
2. x + [x] is large in magnitude outside the interval.