JEE Mathematics Functions MCQs Set 06

Answer: (d) 30

Explanation:
1. Periods of the three terms: \( \frac{2\pi}{2\pi}=1,\ \frac{2\pi}{\pi/3}=6,\ \frac{2\pi}{\pi/5}=10 \).
2. Period of the sum is the LCM of 1, 6 and 10, which is 30.

Teacher's Note:
1. The period of a sum of periodic functions is the LCM of the individual periods, provided one exists.
2. Find each period using 2π divided by the coefficient of x.

Answer: (a) periodic with period \( \pi/2 \)

Explanation:
1. \( f(x)=\sqrt{\frac{16}{1-x^2}}=\frac{4}{\sqrt{1-x^2}} \).
2. \( f(\sin x)=\frac{4}{|\cos x|},\ f(\cos x)=\frac{4}{|\sin x|} \).
3. So \( g(x)=|\cos x|+|\sin x| \). Replacing x by x + π/2 swaps |sin x| and |cos x|, so the period is π/2.

Teacher's Note:
1. Simplify the given function first; here it reduces to a sum of absolute values of sine and cosine.
2. Check that x + π/2 interchanges the two terms, which keeps the sum unchanged.

Answer: (a) \( \frac{1}{g(x)} \) and f(x) are identical functions

Explanation:
1. Domain of f: x > 0 and x ≠ 1. Domain of g: x > 0.
2. 1/g(x) = x/ln x needs g(x) ≠ 0, so its domain is x > 0, x ≠ 1, the same as f, and the rule is the same. So (a) is true.
3. 1/f(x) = ln x / x has domain x > 0, x ≠ 1, while g(x) is defined at x = 1, so (b) is false. In (c) and (d) the product f(x)g(x) is not defined at x = 1, so the statements do not hold for all x > 0.

Teacher's Note:
1. Two functions are identical only if both the rule and the domain agree.
2. Always list the points excluded from the domain before cancelling.

Answer: (a) 15

Explanation:
1. Put x = 1: \( f(y)=\frac{f(1)}{y} \), so f(x) = c/x.
2. From f(30) = 20, c = 600.
3. Hence \( f(40)=\frac{600}{40}=15 \).

Teacher's Note:
1. Setting one variable equal to 1 often reveals the form of a functional equation.
2. Once f(x) = c/x is known, one given value fixes c.

Answer: (c) T1 = T2

Explanation:
1. \( f(x)=\cos^4x-\cos^2x+3 \) and \( f\left(x+\tfrac{\pi}{2}\right)=\sin^4x-\sin^2x+3=f(x) \) because \( \sin^4x-\sin^2x=\cos^4x-\cos^2x \). So T1 = π/2.
2. \( g\left(x+\tfrac{\pi}{2}\right)=\cos(-\sin x)+\cos(\cos x)=g(x) \), so T2 = π/2 (π/4 does not work).
3. Hence T1 = T2.

Teacher's Note:
1. Both functions are unchanged when sin and cos are interchanged, and a shift by π/2 does exactly that.
2. Test a smaller candidate like π/4 to confirm it is not a period.

Answer: (d) 1.5

Explanation:
1. \( g(-1,-1.5)=-1-(-1.5)=0.5 \) and \( g(-4,-1.75)=-1.75+4=2.25 \).
2. \( f(0.5,2.25)=2.25^{0.5}=1.5 \).

Teacher's Note:
1. Evaluate the inner functions g first, then apply f to the two results.
2. max is 2.25 and min is 0.5, so the value is the square root of 2.25.

Answer: (d) \( (-\infty,0]\cup[5,\infty) \)

Explanation:
1. \( (f)^2-f-6\ge0\Rightarrow (f-3)(f+2)\ge0 \), so f(x) ≥ 3 or f(x) ≤ −2.
2. f(x) ≥ 3 is the complement of (0, ∞), i.e. (−∞, 0].
3. f(x) ≤ −2 is the complement of (−∞, 5), i.e. [5, ∞).
4. Union: \( (-\infty,0]\cup[5,\infty) \).

Teacher's Note:
1. Factorise the quadratic in f(x) and convert each piece into the complement of the given solution sets.
2. Complement of a strict inequality becomes a non-strict one.

Answer: (c) \( \frac{28}{5} \)

Explanation:
1. The graph of f meets the y-axis at (0, b); its inverse meets the x-axis at (b, 0). The two graphs are mirror images in y = x and meet on that line.
2. Intersection: \( x=\frac{9}{25}x+b\Rightarrow x=\frac{25b}{16} \).
3. R is the quadrilateral O, (b, 0), P, (0, b) with P = (25b/16, 25b/16). It is made of two congruent triangles, so \( \text{Area}=2\times\frac12\times b\times\frac{25b}{16}=\frac{25b^2}{16}=49 \).
4. \( b=\frac{28}{5} \).

Teacher's Note:
1. A function and its inverse are reflections in the line y = x, so the region splits into two equal triangles.
2. Find the intersection point on y = x first.

Answer: (d) I, II and III

Explanation:
1. I is false: f(x) = x and g(x) = −x are one-one but f + g = 0 is constant.
2. II is false: f(x) = g(x) = x gives x², which is not one-one.
3. III is false: f(x) = sin x is odd but not one-one.
4. All three are incorrect.

Teacher's Note:
1. A single counter-example is enough to disprove a general statement.
2. Sums and products of one-one functions need not be one-one.

Answer: (a) odd

Explanation:
1. For rational x: (f − g)(x) = 0 − x = −x. For irrational x: (f − g)(x) = x − 0 = x.
2. −x is rational exactly when x is, so (f − g)(−x) = x for rational x and −x for irrational x.
3. Thus (f − g)(−x) = −(f − g)(x) in both cases: the function is odd.

Teacher's Note:
1. Rationality is preserved under x → −x, so the two cases can be treated separately.
2. Compare (f − g)(−x) with (f − g)(x) in each case.

Answer: (b) 1/n

Explanation:
1. \( f(x)=nx+n-[nx+n]=\{nx+n\}=\{nx\} \) because n is an integer.
2. The function {t} has period 1 in t, so {nx} has period 1/n.

Teacher's Note:
1. Adding an integer inside a fractional part does not change it.
2. If t = nx has period 1, then x has period 1/n.

Answer: (c) [4, 5)

Explanation:
1. Period of \( \sin(kx) \) is \( \frac{2\pi}{k} \) with \( k=\sqrt{[a]} \).
2. \( \frac{2\pi}{\sqrt{[a]}}=\pi\Rightarrow [a]=4 \), so 4 ≤ a < 5.
3. Hence a ∈ [4, 5); a = 5 is excluded because it gives [a] = 5.

Teacher's Note:
1. Equate the period 2π/k to π to find k, and then convert [a] into an interval for a.
2. The greatest integer function equals 4 on the half-open interval [4, 5).

Answer: (a) 10

Explanation:
1. Put y = 1: \( f(x)f(1)=f(x)+f(1)+f(x)-2 \). This gives f(1) = 2.
2. Polynomial solutions have the form f(x) = 1 + x^n. From f(2) = 5, 2^n = 4, so n = 2.
3. f(x) = 1 + x², and f(3) = 10.

Teacher's Note:
1. Substitute y = 1 to get f(1), then use the standard polynomial solution f(x) = 1 ± x^n.
2. The given value f(2) = 5 fixes the sign and the power.

Answer: (a) 5

Explanation:
1. The graph of the inverse is the reflection of the graph of f in y = x.
2. Reflection swaps the x-intercept and the y-intercept, so the total of the intercepts is unchanged: 5.

Teacher's Note:
1. Reflection in y = x interchanges the coordinates of every point.
2. The sum of intercepts therefore stays the same.

Answer: (c) \( h(x)=\cos(\cos^{-1}x) \)

Explanation:
1. (a) \( f(x)=e^{-x}+2+e^{x} \) is even.
2. (b) and (d) are neither odd nor even (they are periodic zigzag graphs not symmetric about the origin).
3. (c) cos(cos⁻¹x) = x on [−1, 1], which is odd.

Teacher's Note:
1. Simplify each function before testing f(−x).
2. A function defined only on [−1, 1] can still be odd if the domain is symmetric.

Answer: (b) f(2) < f(1) < f(4)

Explanation:
1. The condition says the parabola is symmetric about x = 2, so the vertex is at x = 2 (minimum, since the coefficient of x² is positive).
2. The value of f increases with the distance |x − 2|: |2 − 2| = 0, |1 − 2| = 1, |4 − 2| = 2.
3. So f(2) < f(1) < f(4).

Teacher's Note:
1. Symmetry about x = 2 locates the vertex.
2. For an upward parabola, points farther from the vertex have larger values.

Answer: (d) \( \left\{m+\frac1m\ /\ m\in N-\{1\}\right\} \)

Explanation:
1. Write x = m + f with m = [x] and f = {x} ∈ [0, 1). Then m·f = 1, so f = 1/m.
2. f ∈ [0, 1) with f = 1/m requires m > 1, i.e. m is a natural number other than 1.
3. So x = m + 1/m with m ∈ N − {1}.

Teacher's Note:
1. Split x into integer and fractional parts and use 0 ≤ f < 1.
2. m = 1 would give f = 1, which is not allowed.

Answer: (b) 5

Explanation:
1. \( f(f(f(x)))=p^3x+q(p^2+p+1) \).
2. Comparing: \( p^3=8\Rightarrow p=2 \) and \( 7q=21\Rightarrow q=3 \).
3. p + q = 5.

Teacher's Note:
1. Compose a linear function three times and compare coefficients.
2. The constant term is q(p² + p + 1).

Answer: (a) \( (\sec^2 3x+\operatorname{cosec}^2 3x)\tan^2 3x \)

Explanation:
1. \( \sqrt{1-\cos6x}=\sqrt2\,|\sin3x| \). Both \( \tan3x \) and \( |\sin3x| \) have period π/3, so f has period π/3.
2. (a) simplifies to \( \sec^2 3x\,(\tan^2 3x+1)=\sec^4 3x \), with period π/3.
3. (b), (c) and (d) all have period 2π/3 (they contain sin 3x, cos 3x or cosec 3x without absolute values or squares).

Teacher's Note:
1. Simplify f first and find its period before testing the options.
2. A squared or absolute-value trigonometric term has half the period of the plain function.

Answer: (d) \( (-\infty,0)\cup\left[\frac89,\infty\right) \)

Explanation:
1. \( x^2\in[0,\infty),\ x^2\ne 9\Rightarrow 9-x^2\in(-\infty,9]-\{0\} \).
2. Taking reciprocals: for 9 − x² in (0, 9], 1/(9 − x²) ≥ 1/9, so y ≥ 8/9. For 9 − x² < 0, y < 0.
3. Range: \( (-\infty,0)\cup\left[\frac89,\infty\right) \).

Teacher's Note:
1. Find the range of the denominator first, remembering that it cannot be zero.
2. The reciprocal of a positive number up to 9 is at least 1/9.

Answer: (c) \( \left[0,\frac12\right) \)

Explanation:
1. Let t = {x} ∈ [0, 1). Then \( \frac{t}{1+t}=1-\frac{1}{1+t} \) is increasing in t.
2. At t = 0 it is 0, and as t → 1 it approaches 1/2 without reaching it.
3. Range: \( \left[0,\frac12\right) \).

Teacher's Note:
1. Rewrite t/(1 + t) as 1 − 1/(1 + t) to see that it increases with t.
2. The upper end t = 1 is never attained by a fractional part.