JEE Mathematics Functions MCQs Set 13

Answer: (a) one-one and onto

Explanation:
1. \( f'(x)=3x^2-6x+6=3\left[(x-1)^2+1\right]>0 \), so f is strictly increasing and hence one-one.
2. A cubic polynomial takes every real value, so f is onto.
3. f is one-one and onto.

Teacher's Note:
1. A positive derivative everywhere means strictly increasing.
2. Odd-degree polynomials are onto R.

Answer: (c) many-one and into

Explanation:
1. The denominator \( (x-4)^2+2 \) is never zero.
2. \( y(x^2-8x+18)=x^2+4x+30 \) gives \( (y-1)x^2-(8y+4)x+(18y-30)=0 \), and the discriminant condition \( y^2-32y+13\le0 \) restricts y to a bounded interval, so f is into.
3. Values inside that interval are obtained for two values of x, so f is many-one.

Teacher's Note:
1. Treat y = f(x) as a quadratic in x and use the discriminant.
2. The interval for y is bounded, so f cannot be onto R.

Answer: (a) one-one and onto

Explanation:
1. \( f'(x)=2+\cos x\ge1>0 \), so f is strictly increasing and one-one.
2. \( f(x)\to\pm\infty \) as \( x\to\pm\infty \) and f is continuous, so f is onto.

Teacher's Note:
1. Check the sign of the derivative for injectivity.
2. Continuity plus unbounded limits gives surjectivity.

Answer: (c) continuous if f is continuous

Explanation:
1. The modulus of a continuous function is continuous, so the third statement is true.
2. g takes only non-negative values, so it is never onto R, even if f is onto.
3. If \( f(1)=1 \) and \( f(2)=-1 \) (f can be one-one) then \( g(1)=g(2) \), so g need not be one-one.

Teacher's Note:
1. Think of the absolute value as folding the graph above the axis.
2. Folding loses injectivity and the negative part of the range.

Answer: (a) an even function

Explanation:
1. A continuous function with only rational values is constant (it cannot pass through irrational values between two rationals).
2. So \( f(x)=3 \) for all x.
3. \( f(-x)=f(x) \), so f is even. (It is not odd and not increasing.)

Teacher's Note:
1. Use the intermediate value property.
2. A non-zero constant is an even function.

Answer: (b) an odd function

Explanation:
1. \( f(-x)=\ln\left(-x+\sqrt{1+x^2}\right)=\ln\dfrac{1}{x+\sqrt{1+x^2}} \), because \( \left(\sqrt{1+x^2}-x\right)\left(\sqrt{1+x^2}+x\right)=1 \).
2. So \( f(-x)=-f(x) \), and f is odd.

Teacher's Note:
1. Rationalise to relate f(-x) and f(x).
2. This function is the inverse hyperbolic sine, which is odd.

Answer: (c) 0

Explanation:
1. Odd: \( f(-x)=-f(x) \); even: \( f(-x)=f(x) \).
2. Together they give \( f(x)=-f(x) \), so \( f(x)=0 \) for every x in the domain.
3. So \( f(3)-f(2)=0-0=0 \).

Teacher's Note:
1. Equate the two symmetry conditions.
2. The zero function is the only function that is both even and odd.

Answer: (b) -5 < K < 7

Explanation:
1. We need \( x^2+(k-1)x+9>0 \) for all real x.
2. This holds iff the discriminant is negative: \( (k-1)^2-36 < 0 \).
3. So \( -6 < k-1 < 6 \), i.e. \( -5 < k < 7 \).

Teacher's Note:
1. A parabola opening upwards lies above the axis when it has no real roots.
2. Negative discriminant is the condition.

Answer: (a) \( (100,\infty) \)

Explanation:
1. \( \sin x \) is odd, so \( \left[\tfrac{x^2}a\right] \) must be odd. But it is an even function of x, so it must be identically 0.
2. That needs \( 0\le\tfrac{x^2}a < 1 \) for all x in \( [-10,10] \); the largest value of \( x^2 \) is 100, so \( \tfrac{100}a < 1 \).
3. Hence \( a>100 \).

Teacher's Note:
1. An even function that is also odd must be zero.
2. Use the largest value of x squared on the interval.

Answer: (a) \( \phi \)

Explanation:
1. \( \sin^{-1}x \) requires \( -1\le x\le1 \).
2. \( \dfrac1{\sqrt{x-2}} \) requires \( x>2 \).
3. No real number satisfies both, so the domain is empty.

Teacher's Note:
1. Collect the condition from each term and intersect them.
2. Contradictory conditions give the empty set.

Answer: (b) \( (-3,\infty)-\{-1,-2\} \)

Explanation:
1. The logarithm needs \( x+3>0 \), i.e. \( x>-3 \).
2. The denominator \( (x+1)(x+2) \) must not vanish, so \( x\ne-1,-2 \).
3. Domain: \( (-3,\infty)-\{-1,-2\} \).

Teacher's Note:
1. A fraction needs a non-zero denominator, and a logarithm needs a positive argument.
2. Remove the zeros of the denominator that lie in the interval.

Answer: (d) \( \left[-\tfrac32,-1\right] \)

Explanation:
1. We need \( 0\le2x+3\le1 \).
2. Subtract 3: \( -3\le2x\le-2 \).
3. Divide by 2: \( -\tfrac32\le x\le-1 \).

Teacher's Note:
1. The inside expression must land in the domain of f.
2. Solve the double inequality step by step.

Answer: (a) [1, 2]

Explanation:
1. The square root needs \( \sin^{-1}(\log_2x)\ge0 \), i.e. \( 0\le\log_2x\le1 \) (the upper limit comes from the domain of inverse sine).
2. So \( 1\le x\le2 \).

Teacher's Note:
1. Inverse sine is non-negative exactly on [0, 1].
2. Convert the logarithm condition into one on x.

Answer: (b) \( [1,2)\cup(2,3] \)

Explanation:
1. \( |x|>0 \), so the sign depends on \( \log_{0.3}|x-2|\ge0 \).
2. The base 0.3 is below 1, so this needs \( 0 < |x-2|\le1 \).
3. \( |x-2|\le1 \) gives \( 1\le x\le3 \) and \( |x-2|\ne0 \) removes \( x=2 \).
4. Domain: \( [1,2)\cup(2,3] \).

Teacher's Note:
1. A logarithm to a base below 1 is non-negative when the argument is at most 1.
2. The argument must also stay positive.

Answer: (d) \( \left(10^{10},\infty\right) \)

Explanation:
1. \( \log_{10}x \) needs \( x>0 \).
2. The second log needs \( \log_{10}x>0 \), i.e. \( x>1 \).
3. The third needs \( \log_{10}\log_{10}x>0 \), i.e. \( \log_{10}x>1 \), \( x>10 \).
4. The fourth needs \( \log_{10}\log_{10}\log_{10}x>0 \), i.e. \( \log_{10}\log_{10}x>1 \), \( \log_{10}x>10 \), \( x>10^{10} \).

Teacher's Note:
1. Work from the outside in: each logarithm must have a positive argument.
2. Each extra log raises the bound to a power of 10.

Answer: (b) \( \left[2n\pi-\tfrac\pi3,\ 2n\pi+\tfrac\pi3\right] \)

Explanation:
1. \( 3+2\cos x\ge1>0 \), so the condition is \( \dfrac4{3+2\cos x}\le1 \).
2. That means \( 3+2\cos x\ge4 \), i.e. \( \cos x\ge\tfrac12 \).
3. So \( x\in\left[2n\pi-\tfrac\pi3,\,2n\pi+\tfrac\pi3\right] \).

Teacher's Note:
1. The denominator is always positive here, so the fraction is positive.
2. Then only the upper bound 1 matters.

Answer: (c) \( \left[-\tfrac{\sqrt3}2,0\right)\cup\left(0,\tfrac{\sqrt3}2\right] \)

Explanation:
1. We need \( -1\le[2-4x^2]\le1 \).
2. \( [t]\ge-1 \) means \( t\ge-1 \): \( 2-4x^2\ge-1 \), so \( x^2\le\tfrac34 \).
3. \( [t]\le1 \) means \( t < 2 \): \( 2-4x^2 < 2 \), so \( x\ne0 \).
4. Domain: \( \left[-\tfrac{\sqrt3}2,0\right)\cup\left(0,\tfrac{\sqrt3}2\right] \).

Teacher's Note:
1. Translate bounds on [t] into bounds on t: [t] < = 1 means t < 2.
2. Do not forget the excluded point x = 0.

Answer: (c) \( (-e,-1) \)

Explanation:
1. \( f(e^x) \): \( 0 < e^x < 1 \), so \( x < 0 \).
2. \( f(\ln|x|) \): \( 0 < \ln|x| < 1 \), so \( 1 < |x| < e \).
3. With \( x < 0 \) this gives \( -e < x < -1 \).

Teacher's Note:
1. Put each inner expression into the interval (0, 1).
2. Intersect the two resulting sets.

Answer: (b) \( [-1,1]-\{0\} \)

Explanation:
1. We need \( -1\le[2-3x^2]\le1 \).
2. \( [t]\ge-1 \) gives \( 2-3x^2\ge-1 \), so \( x^2\le1 \).
3. \( [t]\le1 \) gives \( 2-3x^2 < 2 \), so \( x\ne0 \).
4. Domain: \( [-1,1]-\{0\} \).

Teacher's Note:
1. The greatest integer must be -1, 0 or 1.
2. The strict upper bound removes x = 0.

Answer: (c) {-1, 1}

Explanation:
1. For \( x\ne0 \) we have \( \left|\dfrac{1+x^2}{2x}\right|\ge1 \), because \( 1+x^2\ge2|x| \) (AM-GM).
2. We need it to be at most 1, so equality must hold: \( |x|=1 \).
3. Domain: \( \{-1,1\} \).

Teacher's Note:
1. AM-GM says x + 1/x has absolute value at least 2.
2. Only the equality case is allowed here.

Answer: (c) \( (-\infty,\infty) \)

Explanation:
1. The domain is \( x>0 \), so \( \log x \) takes every real value.
2. \( \tan \) takes every real value on the real numbers where it is defined (infinitely many branches).
3. So the range is \( (-\infty,\infty) \).

Teacher's Note:
1. Check what values the inner function takes.
2. The tangent covers all reals on each of its branches.

Answer: (b) \( [2\sqrt2,\infty) \)

Explanation:
1. By AM-GM, \( \dfrac1{|\sin x|}+\dfrac1{|\cos x|}\ge\dfrac{2}{\sqrt{|\sin x\cos x|}}=\dfrac2{\sqrt{\frac12|\sin2x|}}\ge\dfrac2{\sqrt{1/2}}=2\sqrt2 \).
2. Equality holds at \( x=\tfrac\pi4 \).
3. As \( x\to0 \) the function grows without bound, so the range is \( [2\sqrt2,\infty) \).

Teacher's Note:
1. Use AM-GM with the product of the two terms.
2. The minimum is at the angle where sine and cosine are equal in modulus.

Answer: (a) \( \left[0,\tfrac12\right) \)

Explanation:
1. With \( \{x\}=x-[x]\in[0,1) \) we get \( f=\dfrac{\{x\}}{1+\{x\}} \).
2. The function \( \dfrac t{1+t} \) is increasing on \( [0,1) \), from 0 (attained) to \( \tfrac12 \) (not attained).
3. Range: \( \left[0,\tfrac12\right) \).

Teacher's Note:
1. Rewrite using the fractional part.
2. The fractional part never reaches 1.

Answer: (c) \( \lambda=\tfrac14 \)

Explanation:
1. The range of \( \tan^{-1} \) must be \( \left(0,\tfrac\pi2\right) \), so \( x^2+x+\lambda \) must take every value in \( (0,\infty) \) and no negative value.
2. Its minimum value is \( \lambda-\tfrac14 \), so the minimum must be 0.
3. Hence \( \lambda=\tfrac14 \).

Teacher's Note:
1. For onto, the inner expression must have range exactly the positive reals.
2. The minimum of a quadratic is c - b^2/4a.