JEE Mathematics Functions MCQs Set 04

Answer: (b) \( (0, 1) \cup (1, +\infty) \)

Explanation:
1. \( \log_{x^2} 25 = \frac{\log 25}{2\log|x|} = \frac{\log 5}{\log |x|} \) and \( \log_x 5 \) needs \( x > 0 \), \( x \neq 1 \).
2. They agree on the common domain, \( (0, 1) \cup (1, +\infty) \).

Teacher's Note:
1. Equal formulas are equal functions only on the common domain.
2. A log base must be positive and not 1.

Answer: (c) \( R - \left\{x \mid x = (2n+1)\frac{\pi}{2}, n \in Z\right\} \)

Explanation:
1. \( f(x) = 1 \) for all real \( x \).
2. \( g(x) = 1 \) wherever \( \cos x \neq 0 \), i.e. except \( x = (2n+1)\frac\pi2 \).
3. They are equal on \( R \) minus those points.

Teacher's Note:
1. The identity sec² - tan² = 1 hides a domain restriction.
2. Always compare domains before declaring functions equal.

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

Explanation:
1. Write \( f(x) = (x^2+1) + \frac{1}{x^2+1} - 1 \).
2. By AM-GM, \( t + \frac1t \geq 2 \), so \( f(x) \geq 1 \), with equality at \( x = 0 \).
3. As \( x \to \infty \), \( f \to \infty \). Range: \( [1, \infty) \).

Teacher's Note:
1. Add and subtract 1 to create t + 1/t.
2. AM-GM gives the minimum when t = 1.

Answer: (d) none of these

Explanation:
1. \( \frac{x^2}{1+x^2} \) takes values in \( [0, 1) \) (it is 0 at \( x=0 \) and never reaches 1).
2. \( \cos^{-1} \) is decreasing, so \( f \) takes values in \( \left(0, \frac\pi2\right] \).
3. This is not any of the closed intervals listed, so the answer is none of these.

Teacher's Note:
1. Check whether the endpoint 1 is attained.
2. A missing endpoint changes the range.

Answer: (a) [0, 3]

Explanation:
1. Domain: \( -3 \leq x \leq 3 \), so \( 9 - x^2 \in [0, 9] \).
2. Its square root lies in \( [0, 3] \).

Teacher's Note:
1. The square root is never negative.
2. It is the upper half of a circle of radius 3.

Answer: (b) [1, 5]

Explanation:
1. On \( [1, 2] \) the value is constant: \( f = 1 \).
2. At \( x = -1 \), \( f = 2 + 3 = 5 \); at \( x = 3 \), \( f = 2 + 1 = 3 \).
3. The minimum is 1 and the maximum is 5.

Teacher's Note:
1. The sum of distances to 1 and 2 is minimum between them.
2. Check both end points for the maximum.

Answer: (b) \( (-\infty, 2] \)

Explanation:
1. \( 5 + 4x - x^2 = 9 - (x-2)^2 \), which lies in \( (0, 9] \) on the domain.
2. \( \log_3 \) is increasing, so \( y \in (-\infty, \log_3 9] = (-\infty, 2] \).

Teacher's Note:
1. Complete the square to get the max of the quadratic.
2. The log goes to minus infinity as its argument approaches 0.

Answer: (c) {(x, b), (y, a), (z, c)}

Explanation:
1. Test each option for the number of true conditions.
2. (a): (i) and (ii) true. (b): (i) and (iii) true. (d): (i) and (ii) true.
3. (c): (i) false, (ii) false, (iii) true, so exactly one condition holds.

Teacher's Note:
1. Check each option against all conditions.
2. Exactly one true means the other two must be false.

Answer: (d) f is many-one and onto

Explanation:
1. \( f(x) = (x-1)(x-2)(x-3) \), so \( f(1) = f(2) = f(3) = 0 \): many-one.
2. A cubic takes every real value, so \( f \) is onto.

Teacher's Note:
1. Equal values at different inputs mean many-one.
2. Odd-degree polynomials are always onto R.

Answer: (b) f is one-one and onto

Explanation:
1. \( f'(x) = 3x^2 + 2x + 3 + \cos x \geq 3x^2 + 2x + 2 > 0 \) (discriminant \( 4 - 24 \lt 0 \)).
2. So \( f \) is strictly increasing: one-one. It tends to \( \pm\infty \), so it is onto.

Teacher's Note:
1. Positive derivative everywhere means one-one.
2. Use cos x ≥ -1 to bound the derivative.

Answer: (d) many-one and into

Explanation:
1. For \( x \geq 0 \), \( f = 2 \cdot 6^x \geq 2 \). For \( x \lt 0 \), \( f = 6^x + 6^{-x} > 2 \).
2. For example \( f(-1) = 6 + \frac16 \), which is also attained at some \( x > 0 \): many-one.
3. The minimum is 2, so the range is \( [2, \infty) \neq R \): into.

Teacher's Note:
1. Show many-one with one repeated value.
2. Find the least value to show it is into.

Answer: (d) none of these

Explanation:
1. \( f \) is bijective only if it is strictly monotonic: \( f'(x) = p + \cos x \) must not change sign.
2. This needs \( |p| \geq 1 \) (at \( |p| = 1 \) the derivative is zero only at isolated points).
3. So \( p \in (-\infty, -1] \cup [1, \infty) \), which is none of the options.

Teacher's Note:
1. Monotonic functions on R with infinite range are bijections.
2. cos x ranges over [-1, 1].

Answer: (b) one-one and onto

Explanation:
1. \( f'(x) = 2 \mp \sin x > 0 \) wherever it exists, so \( f \) is strictly increasing (continuous): one-one.
2. As \( x \to \pm\infty \), \( f \to \pm\infty \), so it is onto.

Teacher's Note:
1.
2. cos x
3. is at most 1, while 2x grows without bound.
4. A continuous increasing function with infinite limits is onto.

Answer: (c) many-one and into

Explanation:
1. \( f(-x) = f(x) \), so \( f \) is many-one.
2. \( f(x) = 1 - \frac{5}{x^2+1} \), which lies in \( [-4, 1) \), not all of R: into.

Teacher's Note:
1. Even functions are many-one.
2. Rewrite as 1 minus a fraction to find the range.

Answer: (c) \( \frac14 \)

Explanation:
1. \( f \) is onto \( [0, \pi/2) \) if \( x^2 + x + k \) takes every value in \( [0, \infty) \).
2. Its minimum is \( k - \frac14 \), so we need \( k - \frac14 = 0 \), i.e. \( k = \frac14 \).

Teacher's Note:
1. tan⁻¹ of [0, ∞) gives [0, π/2).
2. The minimum of x² + x + k occurs at x = -1/2.

Answer: (d) none of these

Explanation:
1. \( f(x) = x + |x| \) equals \( 2x \) for \( x \geq 0 \) and \( 0 \) for \( x \lt 0 \).
2. It is not injective (all negatives map to 0) and not surjective (no negative values).

Teacher's Note:
1. Write the function piecewise.
2. A flat piece destroys one-one.

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

Explanation:
1. Let \( h(x) = f(x) + f(-x) \). Then \( h(-x) = f(-x) + f(x) = h(x) \).
2. \( f(x) - f(-x) \) is odd, and \( [x] - x \) is not even (fractional part).

Teacher's Note:
1. f(x)+f(-x) is always even; f(x)-f(-x) is always odd.
2. Use this to split any function into even and odd parts.

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

Explanation:
1. \( g(x) = |x-1| + |x-2| + |x-3| \).
2. \( g(1) = 3 \) but \( g(-1) = 2 + 3 + 4 = 9 \); so it is not even, and not odd (\( g(1) \neq -g(-1) \)).
3. It grows without bound, so it is not periodic.

Teacher's Note:
1. Test with one pair x and -x.
2. A shifted modulus sum is symmetric about its median, not about 0.

Answer: (a) an odd function

Explanation:
1. \( f(-x) = \log\left(-x + \sqrt{x^2+1}\right) = \log\frac{1}{x + \sqrt{x^2+1}} = -f(x) \).

Teacher's Note:
1. Rationalise to see the reciprocal.
2. A log of a reciprocal flips the sign.

Answer: (d) none of these

Explanation:
1. Symmetry about the y-axis means the function is even.
2. (a) is odd, (b) forces \( f \) to be odd, and \( \cos x + \sin x \) is neither even nor odd.
3. So none of the given functions is even.

Teacher's Note:
1. Symmetric about the y-axis means even.
2. Check f(-x) for each option.

Answer: (c) \( f(x+y) = f(x) + f(y) \)

Explanation:
1. Symmetry about the origin means \( f(-x) = -f(x) \), i.e. an odd function.
2. Put \( y = -x \): \( f(0) = f(x) + f(-x) \) and \( f(0) = 0 \), so \( f(-x) = -f(x) \).
3. (a) is even and (b) is not defined for negative x.

Teacher's Note:
1. Additive functions are always odd.
2. Put y = -x and use f(0) = 0.

Answer: (a) \( 4x \) for \( 0 \lt x \leq 1 \) and \( 4 \) for \( x > 1 \)

Explanation:
1. For \( 0 \lt x \leq 1 \): \( f(x) = f(-x) = -4(-x) = 4x \).
2. For \( x > 1 \): \( -x \lt -1 \), so \( f(x) = f(-x) = 4 \).

Teacher's Note:
1. Evenness lets you reflect the given pieces.
2. Reflect the interval limits too.

Answer: (b) an odd function

Explanation:
1. For \( |x| \lt 1 \): \( f(-x) = x^2\sin\left(-\frac{\pi x}{2}\right) = -f(x) \).
2. For \( |x| \geq 1 \): \( (-x)|-x| = -x|x| \).
3. Hence \( f(-x) = -f(x) \) everywhere: odd.

Teacher's Note:
1. Check each piece separately.
2. x
3. x
4. is a standard odd function.

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

Explanation:
1. \( \left|\sin\frac x2\right| \) has period \( 2\pi \); \( |\cos x| \) has period \( \pi \).
2. The LCM is \( 2\pi \).

Teacher's Note:
1. Absolute value halves the period of sin and cos.
2. Take the LCM of component periods.

Answer: (c) \( \pi \)

Explanation:
1. The period of \( \cos(bx + c) \) is \( \frac{2\pi}{|b|} \). Here \( b = 2 \), so the period is \( \pi \).

Teacher's Note:
1. Period of cos(bx+c) is 2π/
2. b
3. .
4. The constants 4 and 3 do not matter.