JEE Mathematics Functions MCQs Set 14

Answer: (a) \( \{1,2,3\} \)

Explanation:
1. For \( {}^nP_r \) we need integers with \( n\ge r\ge0 \): \( x-3\ge0 \) and \( 7-x\ge x-3 \), i.e. \( 3\le x\le5 \).
2. \( x=3 \): \( {}^4P_0=1 \). \( x=4 \): \( {}^3P_1=3 \). \( x=5 \): \( {}^2P_2=2 \).
3. Range: \( \{1,2,3\} \).

Teacher's Note:
1. A permutation symbol needs integer arguments with n >= r >= 0.
2. Evaluate it at each allowed x.

Answer: (c) {0}

Explanation:
1. \( [x^2-x] \) is an integer k, and \( \tan(k\pi)=0 \).
2. The denominator \( 1+\sin(\cos x)\ge1+\sin(-1)>0 \), so the quotient is defined.
3. Hence \( f(x)=0 \) for all x and the range is \( \{0\} \).

Teacher's Note:
1. Spot that the argument of tan is an integer multiple of pi.
2. Check that the denominator never vanishes.

Answer: (a) \( \left[\sqrt{a-b},\sqrt{2(a-b)}\right] \)

Explanation:
1. The domain is \( [b,a] \).
2. \( f^2=(a-b)+2\sqrt{(a-x)(x-b)} \). The product \( (a-x)(x-b) \) is 0 at the end points and has maximum \( \left(\tfrac{a-b}2\right)^2 \) at the midpoint.
3. So \( f^2 \) ranges from \( a-b \) to \( 2(a-b) \), and f ranges over \( \left[\sqrt{a-b},\sqrt{2(a-b)}\right] \).

Teacher's Note:
1. Square the expression to expose the product term.
2. Use the fact that the product of two numbers with fixed sum is maximal when they are equal.

Answer: (c) \( |c|>|b|\sqrt2 \)

Explanation:
1. \( f(x)=(x+b)^2+2c^2-b^2 \) has minimum \( 2c^2-b^2 \).
2. \( g(x)=-(x+c)^2+b^2+c^2 \) has maximum \( b^2+c^2 \).
3. The condition \( 2c^2-b^2>b^2+c^2 \) gives \( c^2>2b^2 \), i.e. \( |c|>|b|\sqrt2 \).

Teacher's Note:
1. Complete the square in both quadratics.
2. Compare the minimum of the upward parabola with the maximum of the downward one.

Answer: (d) \( \left[\tfrac\pi4,\pi\right) \)

Explanation:
1. \( 2x-x^2=1-(x-1)^2\in(-\infty,1] \).
2. \( \cot^{-1}t \) is decreasing from \( \pi \) (as \( t\to-\infty \), not attained) to 0.
3. At \( t=1 \): \( \cot^{-1}1=\tfrac\pi4 \).
4. Range: \( \left[\tfrac\pi4,\pi\right) \).

Teacher's Note:
1. Find the range of the inner quadratic first.
2. Inverse cotangent is a decreasing function with values in (0, pi).

Answer: (d) \( \left(0,\tfrac{16}5\right] \)

Explanation:
1. By Cauchy-Schwarz, \( (a+b+c+d)^2\le4(a^2+b^2+c^2+d^2) \).
2. So \( (8-e)^2\le4(16-e^2) \), i.e. \( 5e^2-16e\le0 \), so \( 0\le e\le\tfrac{16}5 \).
3. e is positive, so \( e>0 \); the value \( e=\tfrac{16}5 \) occurs when \( a=b=c=d=\tfrac65 \).
4. Range: \( \left(0,\tfrac{16}5\right] \).

Teacher's Note:
1. Isolate e and apply Cauchy-Schwarz to the other four numbers.
2. Check that the end value is attained with positive numbers.

Answer: (b) \( [3,\infty),\ \{0\} \)

Explanation:
1. \( \dfrac{|x|}x=1 \) for \( x>0 \) and \( -1 \) for \( x < 0 \); a logarithm needs a positive argument, so \( x>0 \).
2. The base \( [x-1] \) must be positive and not 1, so \( [x-1]\ge2 \), i.e. \( x\ge3 \).
3. Then \( f(x)=\log_{[x-1]}1=0 \).
4. Domain \( [3,\infty) \), range \( \{0\} \).

Teacher's Note:
1. The fraction |x|/x is only 1 or -1.
2. A base of a logarithm must be positive and different from 1.

Answer: (d) \( (-2,1) \) and \( [-1,1] \)

Explanation:
1. We need \( 4-x^2>0 \) and \( 1-x>0 \) (the log argument is positive): \( -2 < x < 2 \) and \( x < 1 \), so the domain is \( (-2,1) \).
2. On \( (-2,1) \) the quotient \( \dfrac{\sqrt{4-x^2}}{1-x} \) is continuous, tends to 0 as \( x\to-2^+ \) and to \( +\infty \) as \( x\to1^- \), so it takes every positive value.
3. The logarithm therefore takes every real value and the sine takes every value in \( [-1,1] \).

Teacher's Note:
1. Use the limits at the ends of the domain to get the range of the quotient.
2. Every real number is a logarithm of some positive number.

Answer: (c) \( 2+x \) for \( 0\le x\le1 \); \( 2-x \) for \( 1 < x\le2 \); \( 4-x \) for \( 2 < x\le3 \)

Explanation:
1. For \( 0\le x\le1 \): \( f(x)=1+x\in[1,2] \), so \( f(f(x))=1+(1+x)=2+x \).
2. For \( 1 < x\le2 \): \( f(x)=1+x\in(2,3] \), so \( f(f(x))=3-(1+x)=2-x \).
3. For \( 2 < x\le3 \): \( f(x)=3-x\in[0,1) \), so \( f(f(x))=1+(3-x)=4-x \).

Teacher's Note:
1. Split the domain so that f(x) falls in a single branch.
2. The breakpoints are where f(x) crosses 2.

Answer: (a) 0

Explanation:
1. \( f(f(x))=\left(a-\left(a-x^n\right)\right)^{1/n}=x \).
2. \( g(x)-x=0 \) has no real roots and \( g(x)-x \) has positive leading coefficient, so \( g(x)>x \) for all real x.
3. Then \( g(g(x))>g(x)>x \), so \( g(g(x))-x>0 \) always.
4. Hence the equation has no real roots.

Teacher's Note:
1. Simplify f(f(x)) first: it is just x.
2. 'No real roots of g(x) = x' means the parabola stays above the line y = x.

Answer: (d) \( \left[-\tfrac\pi4,\tfrac\pi4\right] \)

Explanation:
1. \( g(f(x))=(\sin x+\cos x)^2-1=\sin2x \).
2. \( \sin2x \) is strictly increasing for \( 2x\in\left[-\tfrac\pi2,\tfrac\pi2\right] \), i.e. \( x\in\left[-\tfrac\pi4,\tfrac\pi4\right] \).
3. On the other intervals it repeats values, so it is not one-one there.

Teacher's Note:
1. Simplify the composite to a single trigonometric function.
2. A function is invertible on an interval where it is monotonic.

Answer: (d) odd and decreasing

Explanation:
1. Odd: \( f(-x)=-f(x) \). If \( y=f(x) \) then \( f(-x)=-y \), so \( f^{-1}(-y)=-x=-f^{-1}(y) \): \( f^{-1} \) is odd.
2. The graph of \( f^{-1} \) is the reflection of the graph of f in \( y=x \), and reflection preserves the property of decreasing.
3. So \( f^{-1} \) is odd and decreasing.

Teacher's Note:
1. Inverse functions inherit symmetry and monotonic direction.
2. Reflecting in y = x does not change the type of monotonicity.

Answer: (a) any rational number

Explanation:
1. \( \sin x \) has period \( 2\pi \) and \( \cos(ax) \) has period \( \tfrac{2\pi}{|a|} \) (for a not 0).
2. The sum is periodic exactly when the two periods have a common multiple, i.e. their ratio \( \tfrac1{|a|} \) is rational.
3. So a must be rational (a = 0 is included).

Teacher's Note:
1. A sum of periodic functions is periodic only if the periods are commensurable.
2. The ratio of the two periods must be a rational number.

Answer: (a) does not exist

Explanation:
1. \( \cos x \) has period \( 2\pi \) and \( \{x\} \) has period 1.
2. A common period T would need to be a multiple of both: \( T=2\pi m=n \) for positive integers m, n, so \( \pi=\tfrac n{2m} \) would be rational, which is false.
3. So f is not periodic.

Teacher's Note:
1. Check whether the two periods are commensurable.
2. Pi is irrational, so 2 pi and 1 have no common multiple.

Answer: (a) \( 2\pi \)

Explanation:
1. \( f(x+2\pi)=\sin(\cos x+x+2\pi)=f(x) \), so \( 2\pi \) is a period.
2. Test \( \pi \): \( f(x+\pi)=\sin(-\cos x+x+\pi)=-\sin(x-\cos x)\ne f(x) \) in general.
3. So the fundamental period is \( 2\pi \).

Teacher's Note:
1. Check the obvious candidate periods one by one.
2. Shifting by pi flips the sign of the cosine inside.

Answer: (b) \( 10\pi \)

Explanation:
1. \( \left|\sin^3\tfrac x2\right|=\left|\sin\tfrac x2\right|^3 \) has period \( 2\pi \).
2. \( \left|\cos^5\tfrac x5\right|=\left|\cos\tfrac x5\right|^5 \) has period \( 5\pi \).
3. The least common multiple of \( 2\pi \) and \( 5\pi \) is \( 10\pi \).

Teacher's Note:
1. Absolute values halve the period of an odd power of sine or cosine.
2. Then take the LCM of the two periods.

Answer: (b) 2p

Explanation:
1. \( 2-3f+3f^2-f^3=1-(f-1)^3 \).
2. Let \( u=f-1 \): \( u(x+p)=\left(1-u(x)^3\right)^{1/3} \).
3. Then \( u(x+2p)=\left(1-\left(1-u(x)^3\right)\right)^{1/3}=u(x) \).
4. So \( f(x+2p)=f(x) \) and the period is 2p.

Teacher's Note:
1. Recognise the cubic expression as 1 - (f - 1)^3.
2. Apply the relation twice.

Answer: (b) 13

Explanation:
1. \( f(2)=2f(0)-f(1)=4-3=1 \).
2. \( f(3)=2f(1)-f(2)=6-1=5 \).
3. \( f(4)=2f(2)-f(3)=2-5=-3 \).
4. \( f(5)=2f(3)-f(4)=10+3=13 \).

Teacher's Note:
1. A recurrence can be evaluated one step at a time.
2. Keep track of two consecutive values.

Answer: (c) {x}

Explanation:
1. \( \{x+r\}=\{x\} \) for every integer r.
2. The sum becomes \( 2000\cdot\dfrac{\{x\}}{2000}=\{x\} \).

Teacher's Note:
1. Adding an integer does not change the fractional part.
2. Then count the 2000 equal terms.

Answer: (a) 1

Explanation:
1. Divide by \( 5^x \): \( \left(\tfrac25\right)^x+\left(\tfrac35\right)^x+\left(\tfrac45\right)^x=1 \).
2. The left side is strictly decreasing in x, so it equals 1 at most once.
3. At \( x=2 \): \( 2^2+3^2+4^2=29>25=5^2 \); at \( x=3 \): \( 8+27+64=99 < 125 \). So the single root lies between 2 and 3.

Teacher's Note:
1. Divide by the largest base to get a monotonic function.
2. A strictly decreasing continuous function crosses 1 once if it takes values above and below 1.

Answer: (c) \( [1,4] \)

Explanation:
1. \( f'(x)=3x^2+2(a+2)x+3a \).
2. f is one-one exactly when \( f'(x)\ge0 \) for all x (a repeated root does not spoil strict monotonicity).
3. Discriminant \( \le0 \): \( 4(a+2)^2-36a\le0\Rightarrow a^2-5a+4\le0 \).
4. So \( 1\le a\le4 \).

Teacher's Note:
1. A cubic is one-one when its derivative never changes sign.
2. Use the discriminant of the derivative.

Answer: (b) [-1, 1]

Explanation:
1. For \( x\ge0 \): \( g(x)=\tfrac x3\ge0 \) and \( f(g(x))=3\cdot\tfrac x3=x \).
2. For \( x < 0 \): \( g(x)=x \) and \( f(x)=2x-x=x \).
3. So \( h(x)=x \) for all x, and the n-fold composite is x.
4. \( \sin^{-1}x \) has domain \( [-1,1] \).

Teacher's Note:
1. Simplify h first.
2. Once h is the identity, repeated composition changes nothing.

Answer: (b) a constant function

Explanation:
1. Divide by \( |x-y| \): \( \left|\dfrac{f(x)-f(y)}{x-y}\right|\le|x-y|^2 \).
2. Letting \( y\to x \) gives \( f'(x)=0 \) for all x.
3. So f is constant.

Teacher's Note:
1. Form the difference quotient and take the limit.
2. A function with zero derivative everywhere is constant.

Answer: (d) one-one and into

Explanation:
1. \( f'(x)=\dfrac1{(x+1)^2}>0 \), so f is one-one.
2. \( f(x)=1-\dfrac1{x+1} \) lies in \( [0,1) \), which is not all of R, so f is into.

Teacher's Note:
1. Rewrite the fraction as 1 minus a simpler fraction.
2. A bounded range means the function is into R.