Class 11 Maths NCERT Solutions Chapter 2 Miscellaneous Exercise

The two pieces of f meet at x = 3. Check both formulas agree there.

From x² at x = 3: f(3) = 3² = 9

From 3x at x = 3: f(3) = 3(3) = 9

Both pieces give the same value at the boundary, so every x in [0, 10] has exactly one image.

The two pieces of g meet at x = 2. Check both formulas at that point.

From x² at x = 2: g(2) = 2² = 4

From 3x at x = 2: g(2) = 3(2) = 6

The two pieces disagree at x = 2, giving two different images (4 and 6) for the same input.

f(1.1) = (1.1)² = 1.21

f(1) = 1² = 1

(f(1.1) − f(1))/(1.1 − 1) = (1.21 − 1)/(0.1) = 0.21/0.1

The function is undefined wherever the denominator is zero.

x² − 8x + 12 = (x − 6)(x − 2)

This is zero when x = 6 or x = 2.

The expression under the square root must be non-negative: x − 1 ≥ 0 ⟹ x ≥ 1.

As x ranges over [1, ∞), x − 1 ranges over [0, ∞), so √(x − 1) ranges over [0, ∞).

The modulus is defined for every real number.

|x − 1| ≥ 0 always, and as x ranges over R, x − 1 ranges over R, so |x − 1| takes every value in [0, ∞).

Let y = x²/(1 + x²). Rewrite it as:

y = \dfrac{x^2}{1+x^2} = \dfrac{(1+x^2)-1}{1+x^2} = 1-\dfrac{1}{1+x^2}

Since x² ≥ 0, we have 1 + x² ≥ 1, so 0 < 1/(1+x²) ≤ 1.

At x = 0: 1/(1+x²) = 1, so y = 0.

As x → ∞: 1/(1+x²) → 0, so y → 1, but y never actually reaches 1 (since 1+x² is never infinite).

(f + g)(x) = f(x) + g(x) = (x + 1) + (2x − 3) = 3x − 2

(f − g)(x) = f(x) − g(x) = (x + 1) − (2x − 3) = x + 1 − 2x + 3 = 4 − x

(f/g)(x) = f(x)/g(x) = (x + 1)/(2x − 3), defined for x ≠ 3/2

From (0, −1): f(0) = a(0) + b = b = −1, so b = −1.

From (1, 1): f(1) = a(1) + b = a + b = 1 ⟹ a − 1 = 1 ⟹ a = 2.

Check against (2, 3): f(2) = 2(2) − 1 = 3 ✓

Check against (−1, −3): f(−1) = 2(−1) − 1 = −3 ✓

This would require a = a², i.e. a(a − 1) = 0, so a = 0 or a = 1.

Take a = 2: (2,2) ∈ R would need 2 = 2² = 4, which is false.

Take a = 4, b = 2, since 4 = 2². So (4,2) ∈ R.

Is (2,4) ∈ R? This needs 2 = 4² = 16, which is false.

Take c = 2, b = c² = 4, a = b² = 16.

(a,b) = (16,4) ∈ R, since 16 = 4². (b,c) = (4,2) ∈ R, since 4 = 2².

Is (a,c) = (16,2) ∈ R? This needs 16 = 2² = 4, which is false.

Every first element of f (1, 2, 3, 4, 2) belongs to A, and every second element (5, 9, 1, 5, 11) belongs to B.

The first element 2 appears twice: (2,9) and (2,11), giving two different images for the same input.

Check whether the same first element (ab) can arise from different pairs (a,b), giving different values of a + b.

Take a = 1, b = 6: ab = 6, a + b = 7 → (6, 7) ∈ f

Take a = 2, b = 3: ab = 6, a + b = 5 → (6, 5) ∈ f

The same first element, 6, is paired with two different second elements, 7 and 5.

f(9): 9 = 3², highest prime factor = 3

f(10): 10 = 2 × 5, highest prime factor = 5

f(11): 11 is prime, highest prime factor = 11

f(12): 12 = 2² × 3, highest prime factor = 3

f(13): 13 is prime, highest prime factor = 13