JEE Mathematics Functions MCQs Set 17

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

Explanation:
1. We need \( -1\le x+[x]\le1 \). Write \( x=n+t \) with n an integer and \( 0\le t < 1 \); then \( x+[x]=2n+t \).
2. For \( n=0 \): \( x+[x]=t\in[0,1) \), which is allowed. For \( n\ge1 \): \( 2n+t\ge2 \), not allowed. For \( n\le-1 \): \( 2n+t < -1 \), not allowed.
3. So the domain is \( [0,1) \).

Teacher's Note:
1. Split x into integer part and fractional part.
2. Only the interval with n = 0 survives.

Answer: (a) \( [-1,\cos2] \)

Explanation:
1. \( x\in[-1,1] \) and \( \cos^{-1}x\in[0,\pi] \), so \( [\cos^{-1}x]\in\{0,1,2,3\} \).
2. \( \ln[\cos^{-1}x] \) needs \( [\cos^{-1}x]>0 \) and the denominator needs \( \ln[\cos^{-1}x]\ne0 \), i.e. \( [\cos^{-1}x]\ne1 \).
3. So \( [\cos^{-1}x]=2 \) or 3, i.e. \( 2\le\cos^{-1}x\le\pi \). Since cosine decreases, \( -1\le x\le\cos2 \).

Teacher's Note:
1. List the possible integer values of the inner function first.
2. Remember that cos is a decreasing function on [0, pi].

Answer: (b) \( \left[\sqrt2,\dfrac{1+\sqrt5}2\right] \)

Explanation:
1. \( -1\le x-x^2\le1 \): the right inequality always holds, the left gives \( x^2-x-1\le0 \), i.e. \( \dfrac{1-\sqrt5}2\le x\le\dfrac{1+\sqrt5}2 \).
2. \( 1-\dfrac1{|x|}\ge0 \) gives \( |x|\ge1 \).
3. \( [x^2-1]\ne0 \) means \( x^2-1\notin[0,1) \), so \( x^2\notin[1,2) \). With \( |x|\ge1 \) this gives \( |x|\ge\sqrt2 \).
4. Combine: \( \sqrt2\le x\le\dfrac{1+\sqrt5}2 \) (the negative part \( x\ge\frac{1-\sqrt5}2\approx-0.62 \) contradicts \( |x|\ge\sqrt2 \)).

Teacher's Note:
1. Find the domain of each of the three terms and intersect.
2. Numerical check: root 2 is about 1.414 and (1 + root 5)/2 is about 1.618.

Answer: (d) \( a=b=c \)

Explanation:
1. \( f'(x)=2x\left(e^{x^2}-e^{-x^2}\right)\ge0 \) on \( [0,1] \), so the maximum is \( a=f(1)=e+\tfrac1e \).
2. \( g'(x)=e^{x^2}\left(1+2x^2\right)-2xe^{-x^2}>0 \) because \( e^{x^2}\ge1 \) and \( 1+2x^2-2x>0 \). So g increases and \( b=g(1)=e+\tfrac1e \).
3. \( h'(x)=2xe^{x^2}\left(1+x^2\right)-2xe^{-x^2}=2x\left[e^{x^2}(1+x^2)-e^{-x^2}\right]\ge0 \), so \( c=h(1)=e+\tfrac1e \).
4. Hence \( a=b=c \).

Teacher's Note:
1. Check the sign of the derivative on [0, 1].
2. All three functions are increasing and have the same value at x = 1.

Answer: (b) \( \left[0,\dfrac\pi2\right) \)

Explanation:
1. Let \( t=x^2\ge0 \) and \( u^2=\dfrac{2t+1}{(t+1)^2} \).
2. \( \dfrac{d(u^2)}{dt}=\dfrac{-2t}{(t+1)^3}\le0 \), so \( u^2 \) decreases from 1 (at \( t=0 \)) towards 0 (as \( t\to\infty \)).
3. So \( u\in(0,1] \) and \( \cos^{-1}u\in\left[0,\dfrac\pi2\right) \).

Teacher's Note:
1. Substitute t = x^2 and study the monotonicity.
2. The value u = 0 is never reached.

Answer: (c) \( -1-x \) for \( -1\le x\le0 \); \( x-1 \) for \( 0 < x\le2 \)

Explanation:
1. For \( f(g(x)) \) we need \( g(x)\in[0,4] \), i.e. \( 2-|x|\ge0 \), so \( x\in[-1,2] \).
2. \( f(g(x))=-1+|2-g(x)|=-1+|x| \).
3. For \( -1\le x\le0 \): \( -1-x \). For \( 0 < x\le2 \): \( x-1 \).

Teacher's Note:
1. First restrict x so that g(x) lies in the domain of f.
2. Then simplify the modulus.

Answer: (c) \( \dfrac{x+\sqrt{x^2-4}}2 \)

Explanation:
1. Put \( y=x+\tfrac1x \): \( x^2-yx+1=0 \), so \( x=\dfrac{y\pm\sqrt{y^2-4}}2 \).
2. The domain of f is \( x\ge1 \), and \( \dfrac{y-\sqrt{y^2-4}}2\le1 \), so we must take the plus sign.
3. Hence \( f^{-1}(x)=\dfrac{x+\sqrt{x^2-4}}2 \).

Teacher's Note:
1. Solve the quadratic and select the root that lies in the domain.
2. The two roots are reciprocals of each other, so one is at least 1 and the other at most 1.

Answer: (a) 7

Explanation:
1. By the Leibniz rule \( g'(x)=f(x+n) \).
2. Since n is a multiple of the period 3, \( f(x+n)=f(x) \). So \( g'\left(\tfrac73\right)=f\left(\tfrac73\right) \).
3. \( f\left(\tfrac73\right)=f\left(\tfrac73-3\right)=f\left(-\tfrac23\right)=7 \).

Teacher's Note:
1. Differentiate the integral with respect to the upper limit.
2. Reduce the argument by one full period.

Answer: (b) q

Explanation:
1. Put \( y=1 \) and let \( f(1)=c \): \( f(cx)=x^p \), so \( f(x)=\left(\tfrac xc\right)^p \).
2. Then \( f\left(xf(y)\right)=\left(\dfrac{x}{c}\cdot\dfrac{y^p}{c^p}\right)^p=x^py^{p^2}c^{-p^2-p} \).
3. Comparing with \( x^py^q \): \( q=p^2 \) (and \( c=1 \)). So \( p^2=q \).

Teacher's Note:
1. Find the form of f from y = 1.
2. Compare the powers of y on both sides.

Answer: (a) an even function

Explanation:
1. Put \( y=0 \): \( 2f(x)=2f(x)f(0) \), so \( f(0)=1 \) (taking x with \( f(x)\ne0 \)).
2. Put \( x=0 \): \( f(y)+f(-y)=2f(0)f(y)=2f(y) \), so \( f(-y)=f(y) \).
3. Hence f is even.

Teacher's Note:
1. Substitute y = 0 and then x = 0.
2. The cosine function satisfies this relation.

Answer: (b) 4

Explanation:
1. Let \( g(x)=\dfrac{x+59}{x-1} \). \( g(7)=\dfrac{66}6=11 \) and \( g(11)=\dfrac{70}{10}=7 \).
2. Put \( x=7 \): \( 3f(7)+2f(11)=100 \). Put \( x=11 \): \( 3f(11)+2f(7)=140 \).
3. Multiply the first by 3 and the second by 2 and subtract: \( 9f(7)-4f(7)=300-280 \), so \( 5f(7)=20 \) and \( f(7)=4 \).

Teacher's Note:
1. Note that g swaps 7 and 11.
2. Solve the two simultaneous equations.

Answer: (c) \( 2a \)

Explanation:
1. Let \( u(x)=f(x)-\tfrac12 \). Then \( u(x+a)=\sqrt{\tfrac14-u(x)^2}\ge0 \) for every x.
2. Then \( u(x+2a)=\sqrt{\tfrac14-u(x+a)^2}=\sqrt{u(x)^2}=|u(x)| \).
3. Since \( u(x)=u((x-a)+a)\ge0 \), we get \( u(x+2a)=u(x) \). So f has period 2a.

Teacher's Note:
1. Apply the relation twice.
2. The square root makes every shifted value non-negative.

Answer: (c) 1

Explanation:
1. \( x^2+4x+5=(x+2)^2+1>0 \), so \( \operatorname{sgn}(\cdot)=1 \).
2. At \( x=2 \): \( \min(4,2,1)=1 \).

Teacher's Note:
1. Evaluate the signum function first.
2. Then take the smallest of the three numbers.

Answer: (a) \( (n+1)f_1(n)-n \)

Explanation:
1. The term \( \tfrac1k \) appears in \( f_1(k),f_1(k+1),\dots,f_1(n) \), i.e. \( n-k+1 \) times.
2. The sum equals \( \sum_{k=1}^n\dfrac{n-k+1}k=(n+1)\sum_{k=1}^n\dfrac1k-n=(n+1)f_1(n)-n \).

Teacher's Note:
1. Count how many times each fraction occurs.
2. Check with n = 2: 1 + 3/2 = 5/2 and 3(3/2) - 2 = 5/2.

Answer: (d) \( \left\{2m\pi+\dfrac\pi2\right\}\cup\left[2m\pi+\dfrac{7\pi}6,\ 2m\pi+\dfrac{11\pi}6\right] \)

Explanation:
1. We need \( 3|\sin x|-\sin x-2\ge0 \). Let \( s=\sin x \).
2. If \( s\ge0 \): \( 2s-2\ge0 \) gives \( s=1 \), i.e. \( x=2m\pi+\tfrac\pi2 \).
3. If \( s < 0 \): \( -4s-2\ge0 \) gives \( s\le-\tfrac12 \), i.e. \( x\in\left[2m\pi+\tfrac{7\pi}6,2m\pi+\tfrac{11\pi}6\right] \).
4. The domain is the union of both sets.

Teacher's Note:
1. Treat the cases sin x >= 0 and sin x < 0 separately.
2. The domain is the union, not just one of the two pieces.

Answer: (c) \( (-\infty,-7]\cup[7,\infty) \)

Explanation:
1. We need \( \left|[\,|x|-1\,]\right|>5 \), i.e. \( [\,|x|-1\,]\ge6 \) or \( [\,|x|-1\,]\le-6 \).
2. \( [\,|x|-1\,]\ge6 \) means \( |x|-1\ge6 \), i.e. \( |x|\ge7 \).
3. \( [\,|x|-1\,]\le-6 \) means \( |x|-1 < -5 \), i.e. \( |x| < -4 \), which is impossible.
4. So \( |x|\ge7 \): \( x\le-7 \) or \( x\ge7 \).

Teacher's Note:
1. The greatest integer is at least 6 exactly when the number itself is at least 6.
2. The expression must be strictly greater than 5 since it sits under a root in a denominator.

Answer: (a) \( x \)

Explanation:
1. Take \( q(x)=c \), a constant. Then \( p(c)=c \) for every real c.
2. A polynomial equal to c at every real c is \( p(x)=x \). It also satisfies the condition: \( p(q)=q=q(p) \).

Teacher's Note:
1. A constant polynomial is a convenient test case.
2. A polynomial with infinitely many fixed points must be x.

Answer: (d) \( 0,\ 2,\ -2 \)

Explanation:
1. Integral a with \( 4-a^2\ge0 \) means \( a\in\{0,\pm1,\pm2\} \).
2. \( a=0 \): \( \cos2x \) has period \( \pi \) and \( \sin x \) has period \( 2\pi \), so f is periodic. \( a=\pm2 \): \( f=\sin x+1 \) is periodic.
3. \( a=\pm1 \): the periods \( 2\pi \) and \( \tfrac{2\pi}{\sqrt3} \) have an irrational ratio, so f is not periodic.

Teacher's Note:
1. The sum of two periodic functions is periodic only if the periods have a rational ratio.
2. Check each integer separately.

Answer: (c) an onto function

Explanation:
1. Comparing degrees: \( n=(n-1)+(n-2) \), so \( n=3 \).
2. Take \( f(x)=\dfrac{x^3}{18} \): \( f'=\tfrac{x^2}6 \), \( f''=\tfrac x3 \), and \( f'f''=\tfrac{x^3}{18} \). So such a function exists.
3. Every cubic is continuous and tends to \( \pm\infty \) at the two ends, so it takes every real value: f is onto.

Teacher's Note:
1. Find the degree by comparing the degrees on both sides.
2. An odd-degree polynomial is onto R.

Answer: (a) \( R-\left\{(2n+1)\dfrac\pi4\right\} \) and \( \{1\} \)

Explanation:
1. \( f(x)=1-2\sin^2x=\cos2x=g(x) \). So \( H(x)=1 \) wherever \( g(x)\ne0 \).
2. \( g(x)=0 \) when \( 2x=(2n+1)\tfrac\pi2 \), i.e. \( x=(2n+1)\tfrac\pi4 \). These are removed from the domain, and the range is \( \{1\} \).

Teacher's Note:
1. Notice that f and g are the same function.
2. The domain excludes the zeros of g.

Answer: (b) onto

Explanation:
1. \( F(x)=2\cos2x \), whose range is exactly \( [-2,2] \), so F is onto.
2. It repeats values (period \( \pi \)), so it is not one-one.

Teacher's Note:
1. Find the range of F from the formula.
2. A periodic function is many-one.

Answer: (c) The union of the domains of G(x) and H(x) is the set of all real numbers

Explanation:
1. \( G(x)=0 \) for all real x (domain R, range \( \{0\} \)); H has domain \( R-\{(2n+1)\pi/4\} \) and range \( \{1\} \).
2. The domains differ and the ranges differ, but the union of the domains is R.

Teacher's Note:
1. Write down G and H explicitly.
2. G is the zero function.

Answer: (a) the number of solutions is 10 and their sum is \( 25\pi \)

Explanation:
1. \( F-G=2\cos2x=0 \) gives \( x=\dfrac\pi4+\dfrac{k\pi}2 \). In \( [0,5\pi] \): \( k=0,1,\dots,9 \), so there are 10 solutions in an A.P. with common difference \( \tfrac\pi2 \) (not \( \tfrac\pi4 \)).
2. Sum \( =10\cdot\dfrac\pi4+\dfrac\pi2\cdot45=\dfrac{5\pi}2+\dfrac{45\pi}2=25\pi \).

Teacher's Note:
1. The solutions are spaced half a period apart.
2. The last solution is 19 pi/4, just below 5 pi.