− √[1 + (√−3)²] ≤ (sinx − √3 cosx) ≤ √[1 + (√−3)²]

−2 ≤ (sinx − √3cosx) ≤ 2

−2 + 1 ≤ (sinx − √3 cosx + 1) ≤ 2 + 1

−1 ≤ (sinx − √3 cosx + 1) ≤ 3 i.e.,

range = [−1, 3]

For f to be onto S = [−1, 3].

f (x) = cos (log x)

Now let y = f (x) * f (4) − [1 / 2] * [f (x / 4) + f (4x)]

y = cos (log x) * cos (log 4) − [1 / 2] * [cos log (x / 4) + cos (log 4x)]

y = cos (log x) cos (log 4) − [1 / 2] * [cos (log x −log 4) + cos (log x + log 4)]

y = cos (log x) cos (log 4) − [1 / 2] * [2 cos (log x) cos (log 4)]

y = 0

If A = [(x, y) : x² + y² = 25] and B = [(x, y) : x² + 9y² = 144], then A ∩ B contains _______ points.

f (x) = a cos (bx + c) + d ..(i)

For minimum cos (bx + c) = −1

From (i), f (x) = −a + d = (d − a)

For maximum cos (bx + c) = 1

From (i), f (x) = a + d = (d + a)

Range of f (x) = [d − a, d + a]

Let the number of newspapers be x.

If every student reads one newspaper, the number of students would be x (60) = 60x

Since every student reads 5 newspapers, the number of students = [x * 60] / [5] = 300

x = 25

f (x) = cos [π²] x + cos[−π²] x

f (x) = cos (9x) + cos (−10x)

= cos (9x) + cos (10x)

= 2 cos (19x / 2) cos (x / 2)

f (π / 2) = 2 cos (19π / 4) cos (π / 4);

f (π / 2) = 2 * −1 / √2 * 1/ √2

= −1

Let f (x) = y ⇒ x = f⁻¹ (y).

Hence, f (x) = y = 3x − 5

⇒ x = frac{y + 5}/{3}​

⇒f⁻¹ (y) = x = frac{y + 5}/{3}

f⁻¹ (x) = frac{x+ 5}/{3}​

Also, f is one-one and onto, so f⁻¹ exists and is given by f⁻¹ (x) = [x + 5] / [3

Let f (x) = \frac{x^2-1}/{x^2+1}​

= \frac{x^2+1-2}/{x^2+1}​

= 1 − (2 / [x² + 1]) [Because [x² + 1] > 1 also (2 / [x² + 1]) ≤ 2]

So 1 − [2 / [x² + 1]] ≥ 1 − 2;

−1 ≤ f (x) < 1

Thus, f (x) has the minimum value equal to 1.

f (x) = cos² x + sin⁴x

y = f (x) = cos² x + sin²x (1 − cos²x)

y = cos² x + sin²x − sin²x cos²x

y = 1 − sin²x cos²x

y = 1 − [1 / 4] * [sin²2x]

3 / 4 ≤ f (x) ≤ 1, (Because 0 ≤ sin²2x ≤ 1)

f (R) ∈ [3/4, 1]

Function f: R → R is defined by f (x) = e^x.

Let x₁, x₂ ∈ R and f (x₁) = f (x₂) or e^x1 = e^x2 or x₁ = x₂.

Therefore, f is one-one.

Let f (x) = e^x = y.

Taking log on both sides, we get x = logy.

We know that negative real numbers have no pre-image or the function is not onto and zero is not the image of any real number.

Therefore, function f is into.