"11 \$ \\quad \$ (i) \$ \\frac { x - 1 } { x + 1 } = \\frac { 2 x - 5 } { 3 x - 7 } \$"

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Assertion (

A

): Derivative of

\frac{{tan}^{- 1} x}{1 + {tan}^{- 1} x}

with respect to

{tan}^{- 1} x

\frac{1}{(1 + {tan}^{- 1} x)^{2}}

Reason (

R

): Derivative of

\frac{x}{1 + x}

with respect to

x

\frac{1}{(1 + x)^{2}}

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Solve:-

(i) $5 (3 x - 4) - 8 (6 x - 7) = 9 x - 8$

(ii) $2 x - \frac{3 x}{5} = 7$

(iii) $\frac{2 z}{3} - 5 = \frac{3 z}{4} - 1$

(iv) $\frac{4 x}{7} - 12 = 7$

View Solution

Assertion :

{cosec}^{- 1} (3 / 2) + {cos}^{- 1} (2 / 3) - 2 {cot}^{- 1} (1 / 7) - {cot}^{- 1} 7

is equal to

{cot}^{- 1} 7

Reason:

{sin}^{- 1} x + {cos}^{- 1} x = π / 2, {tan}^{- 1} x + {cot}^{- 1} x = π / 2, {cosec}^{- 1} x = {sin}^{- 1} (1 / x),

{cot}^{- 1} (1 / x) = {tan}^{- 1} (x), {cot}^{- 1} (x) = {tan}^{- 1} (1 / x)

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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

.
Prove that
(a)

2 {t a n}^{- 1} \frac{1}{5} + {s e c}^{- 1} \frac{5 \sqrt{2}}{7} + 2 {t a n}^{- 1} \frac{1}{8} = \frac{π}{4}

(b)

{c o s}^{- 1} \frac{12}{13} + 2 {c o s}^{- 1} \sqrt{(\frac{64}{65})} + {c o s}^{- 1} \sqrt{(\frac{49}{50})} = {c o s}^{- 1} \frac{1}{\sqrt{2}}

View Solution

Study the following pattern:

1 + 3 = 2 × 2
1 + 3 + 5 = 3 × 3
1 + 3 + 5 + 7 = 4 × 4
1 + 3 + 5 + 7 + 9 = 5 × 5

By observing the above pattern, find

(i) 1 + 3 + 5 + 7 + 9 + 11
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
(iii) 21 + 23 + 25 + ........ + 51

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"11 \\( \\quad \\) (i) \\( \\frac { x - 1 } { x + 1 } = \\frac { 2 x - 5 } { 3 x - 7 } \\)"