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Can’t you do the following problem?

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

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Solution

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

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Similar questions

Solve each of the following quadratic equations:
(i) $\frac{x}{x - 1} + \frac{x - 1}{x} = 4 \frac{1}{4}, x \neq 0, 1$
(ii) $\frac{x - 1}{2 x - 1} + \frac{2 x - 1}{x - 1} = 2, x \neq \frac{1}{2}, 1$

Q. Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

.

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

3 {s i n}^{- 1} \frac{2 x}{1 + x^{2}} - 4 {c o s}^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 {t a n}^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

s i n [(1 / 5) {c o s}^{- 1} x] = 1

{t a n}^{- 1} \sqrt{x^{2} + x} + {s i n}^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2}

y = s e c^{- 1} [\frac{x + 1}{x - 1}] + s i n^{- 1} [\frac{x - 1}{x + 1}]

z = c o s e c^{- 1} [\frac{2 x + 3}{3 x + 2}] + c o s^{- 1} [\frac{3 x + 2}{2 x + 3}]

Now can’t you simplify these?

(i) 2x + (3x + 1)

(ii) x + (2x + 3)

(iii) (x + 2) + (2x + 1)

(iv) (x + 1) + (x + y)

(v) (3x + 2y) + (x + y)

(vi) (3x + 2y) + (2x + 3y)

Q. Prove the following :

∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{3} & {3 x}^{2} & 3 x & 1 x^{2} & x^{2} + 2 x & 2 x + 1 & 1 x & 2 x + 1 & x + 2 & 1 1 & 3 & 3 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ = {(x - 1)}^{6} .

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