JEE Mathematics Functions MCQs Set 23

Answer: (c) \( g(x) \) is neither even nor odd

Explanation:
1. \( g(x)=|x-1|+|x-2|+|x-3| \).
2. \( g(-x)=|x+1|+|x+2|+|x+3| \), which is neither \( g(x) \) nor \( -g(x) \) (for example \( g(0)=6 \) but \( g(-1)=9 \), \( g(1)=3 \)).
3. A sum of absolute values grows without bound, so it is not periodic.

Teacher's Note:
1. Test evenness and oddness with a couple of specific values such as g(1) and g(-1).

Answer: (a) even

Explanation:
1. \( \log\left(x+\sqrt{1+x^2}\right) \) is an odd function, since \( \left(x+\sqrt{1+x^2}\right)\left(-x+\sqrt{1+x^2}\right)=1 \).
2. \( \sec \) is even, so \( \sec(\text{odd function}) \) is even.

Teacher's Note:
1. An even function of an odd function is even.

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

Explanation:
1. Write \( f(x)=(x^2+1)+\frac{1}{x^2+1}-1 \).
2. By AM-GM, \( (x^2+1)+\frac1{x^2+1}\ge2 \) with equality at \( x=0 \).
3. So \( f(x)\ge1 \), and f takes every value from 1 to infinity.

Teacher's Note:
1. Adding and subtracting 1 turns the expression into the form t + 1/t.

Answer: (b) 48

Explanation:
1. \( f(x)+f(1-x)=\frac{4^x}{4^x+2}+\frac{4}{4+2\cdot4^x}=\frac{4^x}{4^x+2}+\frac{2}{4^x+2}=1 \).
2. The 96 terms form 48 pairs \( \left(\frac{k}{97},\frac{97-k}{97}\right) \), each pair summing to 1.
3. Sum = 48.

Teacher's Note:
1. When arguments add to 1, first check whether f(x) + f(1-x) is a constant.

Answer: (c) I, II

Explanation:
1. Putting \( x=y=0 \): \( 2k=2k^2 \), so \( k=0 \) or \( 1 \).
2. Putting \( x=0 \): \( f(y)+f(-y)=2kf(y) \).
3. For \( k=1 \): \( f(-y)=f(y) \), even. For \( k=0 \): \( f(-y)=-f(y) \), odd.
4. So I and II are correct.

Teacher's Note:
1. Substituting x = 0 gives a relation between f(y) and f(-y), which settles parity.

Answer: (c) 8

Explanation:
1. Replace x by \( x+2 \): \( f(x+2)+f(x+6)=f(x+4)+f(x+8) \).
2. The left side equals \( f(x)+f(x+4) \) by the given relation, so \( f(x)+f(x+4)=f(x+4)+f(x+8) \), giving \( f(x)=f(x+8) \).
3. The period is 8.

Teacher's Note:
1. Shifting the variable by 2 and comparing the two relations eliminates the middle terms.

Answer: (c) \( \frac{1}{\frac1\alpha+\frac1\beta+\frac1\gamma+\frac1\delta} \)

Explanation:
1. \( \log_ax=\alpha \Rightarrow \log_xa=\frac1\alpha \), and similarly for b, c and d.
2. \( \log_{abcd}x=\frac1{\log_x(abcd)}=\frac1{\log_xa+\log_xb+\log_xc+\log_xd} \).
3. So \( \log_{abcd}x=\frac1{\frac1\alpha+\frac1\beta+\frac1\gamma+\frac1\delta} \).

Teacher's Note:
1. The reciprocal rule log_a x = 1 / log_x a converts the product into a sum.

Answer: (c) 2

Explanation:
1. Let \( a=\log_2x \). The equation becomes \( \frac{2a}{a-1}+\frac{4a}{a+2}=\frac{9a}{a+1} \).
2. \( a=0 \) (that is x = 1) satisfies it and all bases are valid.
3. For \( a\ne0 \): \( 2(a+2)(a+1)+4(a-1)(a+1)=9(a-1)(a+2) \Rightarrow a^2+a-6=0 \Rightarrow a=2,-3 \), i.e. \( x=4,\frac18 \).
4. The integral solutions are \( x=1 \) and \( x=4 \).

Teacher's Note:
1. Convert every logarithm to base 2 using a = log_2 x, but remember to check the value a = 0 separately.

Answer: (b) \( \operatorname{sgn}(-x)\sqrt{\frac{|x|}{1-|x|}} \)

Explanation:
1. For \( x\ge0 \): \( y=-\frac{x^2}{1+x^2}\in(-1,0] \), so \( x^2=\frac{-y}{1+y}=\frac{|y|}{1-|y|} \).
2. For \( x<0 \): \( y=\frac{x^2}{1+x^2}\in(0,1) \), so \( x=-\sqrt{\frac{y}{1-y}} \).
3. Combining, \( f^{-1}(y)=\operatorname{sgn}(-y)\sqrt{\frac{|y|}{1-|y|}} \).

Teacher's Note:
1. Handle the two cases x of either sign separately, then combine with the sign function.

Answer: (a) I, II, III

Explanation:
1. For \( 02. The inner expression is \( \{5+x\}=x \), so \( f(x)=\sin x \).
3. The inverse is \( \sin^{-1}x \). Statement II is the same function, since \( \frac\pi2-\cos^{-1}x=\sin^{-1}x \).
4. The range of f is \( (0,\sin\frac\pi4) \subset(0,1) \), so \( \{x\}=x \) there and III is also correct. IV is not equal to \( \sin^{-1}x \).

Teacher's Note:
1. Evaluate fractional parts of numbers lying in (0, 1) as the numbers themselves.

Answer: (b) \( G(x) \) is an even function

Explanation:
1. \( \frac1{a^x-1}+\frac12=\frac{a^x+1}{2(a^x-1)} \), which is an odd function (replacing x by -x gives the negative).
2. The product of two odd functions is even, so \( G(x) \) is even.

Teacher's Note:
1. odd x odd = even; odd x even = odd; even x even = even.

Answer: (c) a constant number

Explanation:
1. Put \( m=2x+\frac y8 \), \( n=2x-\frac y8 \). Then \( x=\frac{m+n}4 \) and \( y=4(m-n) \), so \( xy=(m+n)(m-n)=m^2-n^2 \).
2. Thus \( f(m,n)=m^2-n^2 \) and \( f(n,m)=n^2-m^2 \).
3. The sum is 0, a constant.

Teacher's Note:
1. Solve for x and y in terms of the new variables m and n.

Answer: (a) f is odd and periodic

Explanation:
1. From the first relation, \( f(30-x)=f(x) \).
2. The second relation then gives \( f(30+x)=-f(30-x)=-f(x) \).
3. Hence \( f(x+60)=-f(x+30)=f(x) \), so f is periodic with period 60.
4. Also \( f(-x)=f(30+x)=-f(x) \), so f is odd.

Teacher's Note:
1. Combine the two symmetries (about x = 15 and about the point (30, 0)) to get both parity and period.

Answer: (b) \( \frac53 \)

Explanation:
1. \( f_1(x)=\frac{x-1}{x} \), \( f_2(x)=x \), and \( f_3=f_0 \): the sequence repeats with period 3.
2. \( f_{100}=f_1 \), so \( f_{100}(3)=\frac23 \).
3. \( f_1\left(\frac23\right)=\frac{-1/3}{2/3}=-\frac12 \) and \( f_2\left(\frac32\right)=\frac32 \).
4. Sum \( =\frac23-\frac12+\frac32=\frac53 \).

Teacher's Note:
1. Compute the first few iterates and look for a cycle.

Answer: (d) neither one-one nor onto

Explanation:
1. \( f(2)=f(-2) \), so f is not one-one.
2. The range is \( [-1,1) \), which is smaller than R, so f is not onto.

Teacher's Note:
1. An even function can never be one-one over all of R.

Answer: (a) (i) and (ii) only

Explanation:
1. \( \frac{a^x-1}{a^x+1} \) is odd.
2. If n is even, \( x^n \) is even, so f is odd (odd divided by even).
3. If n is odd, \( x^n \) is odd, so f is even (odd divided by odd).
4. So (i) and (ii) are correct.

Teacher's Note:
1. Parity of a quotient follows the same rules as that of a product.

Answer: (b) 1

Explanation:
1. Replace x by \( \frac{2002}x \): \( f\left(\frac{2002}x\right)+2f(x)=\frac{6006}x \).
2. Eliminating \( f\left(\frac{2002}x\right) \) gives \( -3f(x)=3x-\frac{12012}x \), so \( f(x)=\frac{4004}x-x \).
3. \( f(2)=2002-2=2000 \), so the ratio is 1.

Teacher's Note:
1. Swap x with 2002/x to get a second equation and solve the pair as simultaneous equations.

Answer: (b) 2

Explanation:
1. Let \( t=\log_2(16\sin^2x+1)\in[0,\log_217] \); we need \( 2-t>0 \), so \( 0\le t<2 \) and \( 2-t\in(0,2] \).
2. The largest value of f occurs at \( 2-t=2 \): \( \log_{\sqrt2}2=2 \).
3. As \( 2-t\to0^+ \), \( f\to-\infty \), so the range is \( (-\infty,2] \), giving a = 2.

Teacher's Note:
1. Find the range of the inner expression first, then apply the decreasing base-greater-than-1 logarithm.

Answer: (b) 4

Explanation:
1. For \( x\in[0,1),[1,2),[2,3),[3,4) \), \( [x]=0,1,2,3 \), and \( \sin\left(\frac\pi2[x]\right)=0,1,0,-1 \).
2. The pattern repeats every 4 units.

Teacher's Note:
1. Tabulate one cycle of the integer part and look for the repeat.

Answer: (b) 6

Explanation:
1. A one-one polynomial must be monotonic, so \( f'(x)=(4a-7)x^2+2(a-3)x+1 \) must keep one sign, so \( 4a-7>0 \) and \( D\le0 \).
2. \( D=4(a-3)^2-4(4a-7)=4(a^2-10a+16)\le0\Rightarrow2\le a\le8 \).
3. Both conditions hold for \( a\in[2,8] \), so \( |\lambda-\mu|=6 \).

Teacher's Note:
1. A cubic is invertible exactly when its derivative never changes sign.

Answer: (c) 5

Explanation:
1. \( f(x)-10x \) vanishes at 1, 2, 3, so \( f(x)=10x+(x-1)(x-2)(x-3)(x-k) \).
2. \( f(12)+f(-8)=40+990(12-k)+990(8+k)=40+19800=19840 \).
3. \( \lambda=\frac{19840}{3968}=5 \).

Teacher's Note:
1. Subtract the linear function that matches the three given values to expose the factorisation.

Answer: (c) 4

Explanation:
1. \( f(x)=2+\cos2x+\sqrt3\sin2x=2+2\sin\left(2x+\frac\pi6\right) \).
2. For \( x\in\left[\frac\pi6,\frac{2\pi}3\right] \), \( 2x+\frac\pi6\in\left[\frac\pi2,\frac{3\pi}2\right] \), where sine decreases from 1 to -1, so f is one-one with range \( [0,4] \).
3. For the inverse to exist, B must be the range, so \( a+b=0+4=4 \).

Teacher's Note:
1. The codomain must equal the range for an inverse to exist.

Answer: (b) 1

Explanation:
1. Put \( y=1 \): \( f(n+1)=f(n)+f(1)-n-1=f(n)-n \).
2. So f decreases from \( f(1)=1 \): \( f(2)=0 \), \( f(3)=-2 \), and so on.
3. \( f(n)=n \) holds only for \( n=1 \).

Teacher's Note:
1. Build the sequence f(1), f(2), f(3), ... from the recurrence and compare with n.

Answer: (b) 2

Explanation:
1. \( nx+n-[nx+n]=\{nx\} \), which has period \( \frac1n \).
2. \( \tan\frac{\pi x}2 \) has period 2.
3. The common period of \( \frac1n \) and 2 is 2 (since \( 2=2n\cdot\frac1n \)).

Teacher's Note:
1. A common period exists when the ratio of the individual periods is rational; take the smallest common multiple.

Answer: (c) 18

Explanation:
1. p is a quadratic with a single (repeated) root at -1, so \( p(x)=k(x+1)^2 \).
2. \( p(-2)=k=2 \), so \( p(x)=2(x+1)^2 \).
3. \( p(2)=2\cdot9=18 \).

Teacher's Note:
1. A quadratic with exactly one zero is a constant multiple of a perfect square.

Answer: (c) 8

Explanation:
1. Let \( A=(a,0) \) and \( B=(0,b) \) with \( \frac12|ab|=16 \), so \( |ab|=32 \).
2. The graph of \( f^{-1} \) is the reflection in y = x, so A' = (0, a) and B' = (b, 0), and the rectangle has sides |a| and |b|.
3. \( OC=\sqrt{a^2+b^2}\ge\sqrt{2|ab|}=8 \).

Teacher's Note:
1. Reflection in y = x swaps the intercepts; use AM-GM for the minimum.

Answer: (b) 1

Explanation:
1. \( f(f(x))=\frac{\alpha\cdot\frac{\alpha x}{x+1}}{\frac{\alpha x}{x+1}+1}=\frac{\alpha^2x}{(\alpha+1)x+1} \).
2. Setting this equal to x: \( \alpha^2=(\alpha+1)x+1 \) for all x, so \( \alpha+1=0 \) and \( \alpha^2=1 \).
3. \( \alpha=-1 \), so \( |\alpha|=1 \).

Teacher's Note:
1. For an identity in x, equate coefficients of like powers of x.