Question

Previous year Question of jee mains
Chapter:- Function​

Answer 1

Answer:

$f(x) = { \sin}^{ - 1} ( \frac{ |x| + 5}{ {x}^{2} + 1 } )$

For Domain,,

x²+1 ≠ 0 and

$- 1 \leqslant ( \frac{ |x| + 5}{ {x}^{2} + 1 } ) \leqslant 1$

$→\frac{ |x| + 5}{ {x}^{2} + 1 } \geqslant - 1$

$→ |x| + 5 \geqslant - {x}^{2} - 1$

$→ {x}^{2} + |x| +6 = 0$

let |x| = t

$→ {t}^{2} + t + 6 = 0$

$discriminant = 1 - 24 < 0$

here, x is not real.

also,

$→\frac{ |x| + 5}{ {x}^{2} + 1 } \leqslant 1$

$→ |x| + 5 \leqslant {x}^{2} + 1$

$→ {x}^{2} - |x| - 4 \geqslant 0$

let |x| = t

$→ {t}^{2} - t - 4 \geqslant 0$

$→t = \frac{ - ( - 1) ± \sqrt{ {( - 1)}^{2} - 4 \times 1 \times ( - 4)} }{2 \times 1} \\ = \frac{1 ± \sqrt{1 + 16} }{2} \\ = \frac{1± \sqrt{17} }{2}$

→|x| = (1+√17)/2 , (1-√17)/2

since,

$\frac{1 - \sqrt{17} }{2} \: is \: negative \: and \: |x| \: does \: not \: give \: negative \: value.$

So, it will be neglected.

$→ |x| = \frac{ \sqrt{17} + 1}{2}$

$x = ± \frac{ \sqrt{17} + 1 }{2}$

by wavy curve method,

we will get ,

Domain = ( - ∞ , -(√17 + 1)/2 ] U [ (√17 + 1)/2,∞ )

then,

a = (√17 + 1)/2

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = {sin}^{ - 1}{\bigg(\dfrac{ |x| + 5}{ {x}^{2} + 1} \bigg) }$

We know,

$\boxed{ \sf{ \: {sin}^{ - 1}x \: is \: defined \: when \: - 1 \leqslant x \leqslant 1}}$

So,

For domain of f(x) to be defined,

$\rm :\longmapsto\: - 1 \leqslant \dfrac{ |x| + 5}{ {x}^{2} + 1} \leqslant 1$

$\rm :\longmapsto\:As \: {x}^{2} + 1 > 0 \: \forall \: x \: \in \: real$

So,

$\rm :\longmapsto\: - ( {x}^{2} + 1) \leqslant |x| + 5 \leqslant {x}^{2} + 1$

Now, Subtracting 5 from each term, we get

$\rm :\longmapsto\: - {x}^{2} - 1 - 5 \leqslant |x| + 5 - 5 \leqslant {x}^{2} + 1 - 5$

$\rm :\longmapsto\: - {x}^{2} - 6 \leqslant |x| \leqslant {x}^{2} - 4$

So, 2 cases arises.

Case - 1

$\rm :\longmapsto\: - {x}^{2} - 6 \leqslant |x|$

$\rm :\longmapsto\: 0 \leqslant {x}^{2} + |x| + 6$

$\rm :\longmapsto\: {x}^{2} + |x| + 6 \geqslant 0$

can be rewritten as

$\rm :\longmapsto\: { |x| }^{2} + |x| + 6 \geqslant 0$

Now, its a quadratic in | x |.

So, Let we check its Discriminant.

$\rm :\longmapsto\:Discriminant \: = {b}^{2} - 4ac = 1 - 24 = - 23$

As Discriminant, D < 0 and coefficient of x^2 > 0.

$\bf\implies \:\: {x}^{2} + |x| + 6 > 0$

Hence, true for all real values of x.

Now,

Consider,

Case - 2

$\rm :\longmapsto\: |x| \leqslant {x}^{2} - 4$

$\rm :\longmapsto\: {x}^{2} - |x| - 4 \geqslant 0$

can be rewritten as

$\rm :\longmapsto\: { |x| }^{2} - |x| - 4 \geqslant 0$

Let we first find the roots of

$\rm :\longmapsto\: { |x| }^{2} - |x| - 4 = 0 \: using \: quadratic \: formula$

$\rm :\longmapsto\: |x| = \dfrac{ - ( - 1) \: \pm \: \sqrt{ {( - 1)}^{2} - 4( - 4)(1)} }{2(1)}$

$\rm :\longmapsto\: |x| = \dfrac{1 \: \pm \: \sqrt{ 17} }{2}$

So,

$\rm :\longmapsto\: { |x| }^{2} - |x| - 4 \geqslant 0$

can be rewritten in factorization form as

$\rm :\longmapsto\:{\bigg( |x| - \dfrac{1 + \sqrt{17} }{2} \bigg) }{\bigg( |x| - \dfrac{1 - \sqrt{17} }{2} \bigg) } \geqslant 0$

We know, if a and b are two positive real numbers, such that a < b and

$\boxed{ \sf{ \:(x - a)(x - b) \geqslant 0 \implies \: x \leqslant a \: or \: x \geqslant b}}$

So, using this

$\rm :\longmapsto\: |x| \leqslant \dfrac{1 - \sqrt{17} }{2} \: \: or \: \: |x| \geqslant \dfrac{1 + \sqrt{17} }{2}$

Now,

$\rm :\longmapsto\: |x| \leqslant \dfrac{1 - \sqrt{17} }{2} \: \: rejected \: as \: |x| < 0$

So, we have

$\rm :\longmapsto\: |x| \geqslant \dfrac{1 + \sqrt{17} }{2}$

$\rm :\implies\:x \leqslant - \: \dfrac{1 + \sqrt{17} }{2} \: \: or \: \: x \geqslant \dfrac{1 + \sqrt{17} }{2}$

$\rm\implies\:x \in \: ( - \infty ,- \: \dfrac{1 + \sqrt{17} }{2}\bigg] \cup \: \bigg[\dfrac{1 + \sqrt{17} }{2}, \infty )$

As it is given that

$\rm :\longmapsto\:x \: \in \: ( - \infty , - \: a] \: \cup \: [a, \: \infty )$

So, On comparing above two, we get

$\bf :\longmapsto\:a = \dfrac{1 + \sqrt{17} }{2}$

Therefore,

• Option (c) is correct.