JEE Mathematics Relations and Functions MCQs Set 06

Answer: (d) \( [2,\infty) \)

Explanation:
1. The denominator \( x^2+2x+8=(x+1)^2+7 \) is always positive.
2. We need x - 1 > 0 and \( -\log_{0.3}(x-1)\ge0 \), that is \( \log_{0.3}(x-1)\le0 \).
3. Since the base 0.3 is less than 1, this means x - 1 >= 1, so x >= 2.

Teacher's Note:
1. The logarithm base is below 1, so the inequality reverses.
2. Check the sign of the quadratic under the root.

Answer: (d) \( \left[\ln\frac{11}{3},\infty\right) \)

Explanation:
1. The quadratic has minimum value at x = 2/3: \( 3\cdot\frac49-\frac83+5=\frac{11}{3} \).
2. It takes every value from 11/3 to infinity.
3. So the range of the logarithm is \( [\ln\frac{11}3,\infty) \).

Teacher's Note:
1. Complete the square: 3(x - 2/3)^2 + 11/3.
2. ln is increasing.

Answer: (d) \( (1,\infty) \)

Explanation:
1. Put t = 2^x > 0, so f = t^2 + t + 1.
2. For t > 0 this is increasing and tends to 1 as t tends to 0 (never reaching it).
3. So the range is (1, infinity).

Teacher's Note:
1. t = 2^x takes all positive values.
2. The value at t = 0 is excluded.

Answer: (d) \( [0,2] \)

Explanation:
1. sqrt(2)(sin x - cos x) = 2 sin(x - pi/4), which takes all values in [-2, 2].
2. So the argument of the logarithm lies in [1, 5].
3. log to base sqrt(5): log(1) = 0 and log(5) = 2, so the range is [0, 2].

Teacher's Note:
1. Write sin x - cos x as sqrt(2) sin(x - pi/4).
2. log base sqrt(5) of 5 is 2.

Answer: (d) Neither one-one nor onto

Explanation:
1. The function is a ratio of two quadratics, so values are repeated: it is many-one. For example the equation f(x) = k reduces to a quadratic in x with two roots for suitable k.
2. The function is bounded (it tends to 2/7 at infinity), so it cannot cover every real number: it is into.
3. Hence f is neither one-one nor onto.

Teacher's Note:
1. A bounded function cannot be onto R.
2. A quadratic equation f(x) = k generally has two roots.

Answer: (d) One-one and onto

Explanation:
1. f'(x) = 3x^2 + 2x + 3 + cos x >= 3x^2 + 2x + 2 > 0 since the discriminant 4 - 24 < 0.
2. So f is strictly increasing, hence one-one.
3. f tends to minus infinity and plus infinity at the ends and is continuous, so it is onto.

Teacher's Note:
1. Use cos x >= -1.
2. Strictly increasing continuous functions with unbounded ends are bijections of R.

Answer: (d) \( (-1,0)\cup(1,2)\cup(2,\infty) \)

Explanation:
1. The logarithm needs x^3 - x = x(x - 1)(x + 1) > 0, which holds for x in (-1, 0) and (1, infinity).
2. The fraction needs x not equal to 2 or -2.
3. Removing x = 2 gives (-1,0) U (1,2) U (2,infinity).

Teacher's Note:
1. Sign chart of x(x - 1)(x + 1).
2. Exclude the zeros of 4 - x^2 that lie in the set.

Answer: (d) One-one but not onto

Explanation:
1. f'(x) = 1/(1 + x)^2 > 0, so f is increasing and one-one.
2. f(0) = 0 and f(x) approaches 1 but never reaches it, so the range is [0, 1).
3. The range is smaller than the co-domain [0, infinity), so f is not onto.

Teacher's Note:
1. Find the range and compare with the co-domain.
2. x/(1 + x) < 1 for all x >= 0.

Answer: (d) \( R-(0,1] \)

Explanation:
1. Let t = (x - 2)^2 >= 0. Then (x - 1)(x - 3) = t - 1, with t not equal to 1.
2. f = t/(t - 1) = 1 + 1/(t - 1).
3. For 0 <= t < 1, f takes values in (-infinity, 0]; for t > 1, f takes values in (1, infinity). So the range is R - (0, 1].

Teacher's Note:
1. (x - 1)(x - 3) = (x - 2)^2 - 1.
2. Treat the two cases t < 1 and t > 1 separately.

Answer: (d) R

Explanation:
1. Set y = f(x): y x^2 - (4y + 1)x + (3y + 2) = 0.
2. Its discriminant is (4y + 1)^2 - 4y(3y + 2) = 4y^2 + 1 > 0 for all y, so real x exists for every y (for y = 0, x = 2 works).
3. (x = 1 and x = 3 are never obtained as solutions for any finite y.) Hence the range is R.

Teacher's Note:
1. Treat y as a parameter and require a real root x.
2. The discriminant simplifies to 4y^2 + 1.

Answer: (d) One-one and onto

Explanation:
1. For rational x: (f - g)(x) = 0 - x = -x, which is rational.
2. For irrational x: (f - g)(x) = x - 0 = x, which is irrational.
3. Rationals map one-one onto rationals, and irrationals map one-one onto irrationals, so the function is a bijection of R.

Teacher's Note:
1. Compute f - g separately on rationals and irrationals.
2. The images of the two parts do not overlap.

Answer: (d) \( f(x)=\sin^2x+\cos^2x,\ g(x)=1 \)

Explanation:
1. Two functions are identical if they have the same domain and the same rule.
2. sin^2 x + cos^2 x = 1 for all real x, so both functions have domain R.
3. In the other pairs the domains differ: sec^2 - tan^2 excludes odd multiples of pi/2, cosec^2 - cot^2 excludes multiples of pi, and ln x^2 has domain x not equal to 0 while 2 ln x needs x > 0.

Teacher's Note:
1. Compare domains before comparing the rules.
2. Look for points where one function is undefined.

Answer: (d) \( [-6,1] \)

Explanation:
1. f(g(x)) is defined when g(x) lies in [-5, 7].
2. |2x + 5| is never negative, so we need |2x + 5| <= 7.
3. This gives -6 <= x <= 1.

Teacher's Note:
1. The values of g must lie in the domain of f.
2. Absolute value is at least 0.

Answer: (d) 1

Explanation:
1. g(x) = 1 + {x}, where {x} is the fractional part, so g(x) >= 1 > 0 for every x.
2. f of any positive number is 1.
3. Hence f(g(x)) = 1.

Teacher's Note:
1. x - [x] is the fractional part, between 0 and 1.
2. f is the signum function.

Answer: (d) Odd

Explanation:
1. Replace x by -x: sin(-x) = -sin x, so f(-x) = log((1 - sin x)/(1 + sin x)).
2. This is log of the reciprocal, hence equals -f(x).
3. So f is odd.

Teacher's Note:
1. Reciprocal inside a logarithm changes the sign.
2. The domain is symmetric about 0.

Answer: (d) Odd

Explanation:
1. For x not an integer, [-x] = -[x] - 1.
2. So f(-x) = -[x] - 1 + 1/2 = -[x] - 1/2 = -(f(x)).
3. Hence f is an odd function.

Teacher's Note:
1. Use [-x] = -[x] - 1 for non-integers.
2. The domain excludes integers and is symmetric.

Answer: (d) \( f(2+x)=f(2-x) \)

Explanation:
1. Points at equal distances x on either side of x = 2 have the same height.
2. So f(2 + x) = f(2 - x).

Teacher's Note:
1. Reflect x = 2 + h into x = 2 - h.
2. f(-x) = f(x) would be symmetry about the y-axis.

Answer: (d) \( \pi \)

Explanation:
1. sin(x + pi) = -sin x, and sec is an even function.
2. So f(x + pi) = sec(-sin x) = sec(sin x) = f(x), so pi is a period.
3. No smaller period exists: f depends on |sin x|, which has fundamental period pi, and sec is one-one on [0, 1]. Hence the fundamental period is pi.

Teacher's Note:
1. Check whether shifting x by pi changes the value.
2. sec is even.

Answer: (d) \( a\in[4,5) \)

Explanation:
1. The fundamental period of sin(kx) is 2pi/k, with k = sqrt([a]).
2. Setting 2pi/sqrt([a]) = pi gives sqrt([a]) = 2, so [a] = 4.
3. [a] = 4 means a is in [4, 5).

Teacher's Note:
1. Period of sin(kx) is 2pi/k.
2. [a] = 4 is a condition on an interval of a.

Answer: (d) \( 6\pi \)

Explanation:
1. 2 + 2cos 2x = 4cos^2 x, so the curve is y = [2|cos x|].
2. [2|cos x|] equals 1 when 1/2 <= |cos x| < 1, equals 0 when |cos x| < 1/2 and equals 2 only at isolated points. In one period of length pi, |cos x| >= 1/2 on a total length 2pi/3.
3. The interval has length 9pi, which is 9 periods, so the area is 9 x 2pi/3 = 6pi.

Teacher's Note:
1. Simplify the root using cos 2x = 2cos^2 x - 1.
2. Find on what part of one period the value is 1.

Answer: (d) \( \frac12\ln\frac{1+x}{1-x} \)

Explanation:
1. Let y = (e^x - e^-x)/(e^x + e^-x) = (e^2x - 1)/(e^2x + 1).
2. Then e^2x = (1 + y)/(1 - y), so x = (1/2) ln((1 + y)/(1 - y)).
3. Swapping the variables, f^-1(x) = (1/2) ln((1 + x)/(1 - x)).

Teacher's Note:
1. Multiply the numerator and denominator by e^x.
2. Solve for e^2x.

Answer: (d) \( \frac{x+\sqrt{x^2-4}}{2} \)

Explanation:
1. Let y = x + 1/x, so x^2 - yx + 1 = 0 and x = (y +- sqrt(y^2 - 4))/2.
2. Since the domain is x >= 1, we take the larger root (y + sqrt(y^2 - 4))/2.
3. Swapping the variables, f^-1(x) = (x + sqrt(x^2 - 4))/2.

Teacher's Note:
1. Solve the quadratic for x and choose the root >= 1.
2. The smaller root is at most 1.

Answer: (d) f(x) is odd

Explanation:
1. The graph of f^-1 is always the mirror image of f about y = x.
2. Reflecting f in y = x and then in y = -x amounts to a reflection of f in the origin, so the graph of f must be unchanged by reflection in the origin.
3. That is the definition of an odd function: f is odd.

Teacher's Note:
1. Combine the two reflections: y = x then y = -x gives a half turn about the origin.
2. (a, b) maps to (-a, -b).

Answer: (d) \( [-1,\sqrt2] \)

Explanation:
1. The domain of f is [-1, 2]. f(f(x)) is defined only when f(x) lies in [-1, 2].
2. For x in [-1, 1], f(x) = x is in [-1, 1], fine. For 1 < x <= 2, f(x) = x^2 must be <= 2, so x <= sqrt(2).
3. So the domain of h is [-1, sqrt(2)], and the range of the inverse function equals this domain.

Teacher's Note:
1. The inverse function's range is the original function's domain.
2. Find where f(x) stays inside the domain of f.

Answer: (d) False, True, True

Explanation:
1. Statement 1 is false: for a decreasing function the graphs of f and f^-1 can meet away from y = x (for example f(x) = -x^3 meets f^-1 at the points where x = 0 and also at x = 1 and -1 where y = -x).
2. Statement 2 is true: swapping the coordinates gives a point of the inverse.
3. Statement 3 is true: the inverse is unique and its range is the domain of f.

Teacher's Note:
1. Try f(x) = 1 - x as a decreasing example: f = f^-1 everywhere.
2. Test the statements one at a time.