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JEE Advanced 2019 Paper 2 Offline

Numerical
The value of

$${\sec ^{ - 1}}\left( \matrix{ {1 \over 4}\sum\limits_{k = 0}^{10} {\sec \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)} \sec \left( {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2}} \right) \hfill \cr} \right)$$

in the interval $$\left[ { - {\pi \over 4},\,{{3\pi } \over 4}} \right]$$ equals ..........
Your Input ________

Answer

Correct Answer is 0

Explanation

$$\sum\limits_{k = 0}^{10} {\sec \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)} \sec \left( {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2}} \right)$$

= $$\sum\limits_{k = 0}^{10} {{1 \over {\cos \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)\cos \left( {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2}} \right)}}} $$

= $$\sum\limits_{k = 0}^{10} {{{\sin \left[ {\left( {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2}} \right) - \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)} \right]} \over {\cos \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)\cos \left( {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2}} \right)}}} $$

$$ \because $$ $$\left[ {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2} - \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right) = {\pi \over 2}\,and\,sin{\pi \over 2} = 1} \right]$$

$$\sum\limits_{k = 0}^{10} {{\matrix{ \sin \left( {{{7\pi } \over {12}} + {{\left( {k + 1} \right)\pi } \over 2}} \right)\cos \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right) \hfill \cr - \sin \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)\cos \left( {{{7\pi } \over {12}} + {{\left( {k + 1} \right)\pi } \over 2}} \right) \hfill \cr} \over {\cos \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)\cos \left( {{{7\pi } \over {12}} + {{\left( {k + 1} \right)\pi } \over 2}} \right)}}} $$

= $$\sum\limits_{k = 0}^{10} {\left[ {\tan \left( {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2}} \right) - \tan \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)} \right]} $$

= $$\tan \left( {{{7\pi } \over {12}} + {\pi \over 2}} \right) - \tan \left( {{{7\pi } \over {12}}} \right) + \tan \left( {{{7\pi } \over {12}} + {{2\pi } \over 2}} \right) - \tan \left( {{{7\pi } \over {12}} + {\pi \over 2}} \right)$$ ....$$ + \tan \left( {{{7\pi } \over {12}} + {{11\pi } \over 2}} \right) - \tan \left( {{{7\pi } \over {12}} + {{10\pi } \over 2}} \right)$$

= $$\tan \left( {{{7\pi } \over {12}} + {{11\pi } \over 2}} \right) - \tan {{7\pi } \over {12}} = \tan {{7\pi } \over {12}} + \cot {\pi \over {12}}$$

= $${1 \over {\sin {\pi \over {12}}\cos {\pi \over {12}}}} = {2 \over {\sin {\pi \over 6}}} = 4$$

So, $${\sec ^{ - 1}}\left( {{1 \over 4}\sum\limits_{k = 0}^{10} {\sec \left( {{{7\pi } \over {12}} + {{k\pi } \over 2}} \right)\sec \left( {{{7\pi } \over {12}} + {{(k + 1)\pi } \over 2}} \right)} } \right)$$

= $${\sec ^{ - 1}}$$ (1) = 0
2

JEE Advanced 2018 Paper 1 Offline

Numerical
The number of real solutions of the equation $$\eqalign{ & {\sin ^{ - 1}}\left( {\sum\limits_{i = 1}^\infty {} {x^{i + 1}} - x\sum\limits_{i = 1}^\infty {} {{\left( {{x \over 2}} \right)}^i}} \right) \cr & = {\pi \over 2} - {\cos ^1}\left( {\sum\limits_{i = 1}^\infty {} {{\left( {{{ - x} \over 2}} \right)}^i} - \sum\limits_{i = 1}^\infty {} {{\left( { - x} \right)}^i}} \right) \cr} $$ lying in the interval $$\left( { - {1 \over 2},{1 \over 2}} \right)$$ is ........... .

(Here, the inverse trigonometric functions sin$$-$$1 x and cos$$-$$1 x assume values in $${\left[ { - {\pi \over 2},{\pi \over 2}} \right]}$$ and $${\left[ {0,\pi } \right]}$$, respectively.)
Your Input ________

Answer

Correct Answer is 2

Explanation

We have,

$${\sin ^{ - 1}}\left( {\sum\limits_{i = 1}^\infty {} {x^{i + 1}} - x\sum\limits_{i = 1}^\infty {} {{\left( {{x \over 2}} \right)}^i}} \right)$$

$$ = {\pi \over 2} - {\cos ^1}\left( {\sum\limits_{i = 1}^\infty {} {{\left( {{{ - x} \over 2}} \right)}^i} - \sum\limits_{i = 1}^\infty {} {{\left( { - x} \right)}^i}} \right)$$

$$ \Rightarrow {\sin ^{ - 1}}\left[ {{{{x^2}} \over {1 - x}} - {{x.{x \over 2}} \over {1 - {x \over 2}}}} \right]$$

$$ = {\pi \over 2} - {\cos ^{ - 1}}\left[ {{{{{ - x} \over 2}} \over {1 + {x \over 2}}} - {{( - x)} \over {1 + x}}} \right]$$

$$ \because $$ $$\left[ {\sum\limits_{i = 1}^\infty {} {x^{i + 1}} = {x^2} + {x^3} + {x^4} + .... = {{{x^2}} \over {1 - x}}} \right]$$ using sum of infinite terms of GP

$$ \Rightarrow {\sin ^{ - 1}}\left[ {{{{x^2}} \over {1 - x}} - {{{x^2}} \over {2 - x}}} \right] = {\pi \over 2} - {\cos ^{ - 1}}\left[ {{x \over {1 + x}} - {x \over {2 + x}}} \right]$$

$$ \Rightarrow {\sin ^{ - 1}}\left[ {{{{x^2}} \over {1 - x}} - {{{x^2}} \over {2 - x}}} \right] = {\sin ^{ - 1}}\left[ {{x \over {1 + x}} - {x \over {2 + x}}} \right]$$

$$ \because $$ $$\left[ {{{\sin }^{ - 1}}x = {\pi \over 2} - {{\cos }^{ - 1}}x} \right]$$

$$ \Rightarrow {{{x^2}} \over {1 - x}} - {{{x^2}} \over {2 - x}} = {x \over {1 + x}} - {x \over {2 + x}}$$

$$ \Rightarrow {x^2}\left( {{{2 - x - 1 + x} \over {(1 - x)(2 - x)}}} \right) = x{{(2 + x - 1 - x)} \over {(1 + x)(2 + x)}}$$

$$ \Rightarrow {x \over {2 - 3x + {x^2}}} = {1 \over {2 + 3x + {x^2}}}$$ or x = 0

$$ \Rightarrow {x^3} + 3{x^2} + 2x = {x^2} - 3x + 2$$

$$ \Rightarrow {x^3} + 2{x^2} + 5x - 2 = 0$$ or x = 0

Let $$f(x) = {x^3} + 2{x^2} + 5x - 2$$

$$f'(x) = 3{x^2} + 4x + 5$$

$$f'(x) > 0,\,\forall x \in R$$

$$ \therefore $$ $${x^3} + 2{x^2} + 5x - 2$$ has only one real roots

Therefore, total number of real solution is 2.
3

JEE Advanced 2014 Paper 1 Offline

Numerical
Let f : [0, 4$$\pi$$] $$\to$$ [0, $$\pi$$] be defined by f(x) = cos$$-$$1 (cos x). The number of points x $$\in$$ [0, 4$$\pi$$] satisfying the equation $$f(x) = {{10 - x} \over {10}}$$ is
Your Input ________

Answer

Correct Answer is 3

Explanation

Concept :

The number of solutions of equations involving trigonometric functions and algebraic functions are found using graphs of the curves.

We know, $${\cos ^{ - 1}}(\cos x) = \left\{ \matrix{ x,\,if\,x \in [0,\pi ] \hfill \cr 2\pi - x,\,if\,x \in [\pi ,2\pi ] \hfill \cr - 2\pi + x,\,if\,x \in [2\pi ,3\pi ] \hfill \cr 4\pi - x,\,if\,x \in [3\pi ,4\pi ] \hfill \cr} \right.$$


$$y = {{10 - x} \over {10}} = 1 - {x \over {10}}$$

From above figure, it is clear that $$y = {{10 - x} \over {10}}$$ and $$y = {\cos ^{ - 1}}(\cos x)$$ intersect at three distinct points, so number of solutions is 3.

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