Verified Solution • 2025–26 Session moderate

QJEEM16Ma640d: JEE Main 2016 Mathematics - Class 12 JEEM16Ma640d Solution: If the function f(x) = { − x x < 1 a + cos − 1 ( x + b ) , 1 ≤…

If the function

f(x) = {−xx<1a+cos−1(x+b),1≤x≤2

is differentiable at x = 1, then ab{a \over b} is equal to :

Multiple Choice Options

  • A π−22{{\pi - 2} \over 2}
  • B −π−22{{ - \pi - 2} \over 2}
  • C π+22{{\pi + 2} \over 2}
  • D −1−cos⁡−1(2)- 1 - {\cos ^{ - 1}}\left( 2 \right)

Step-by-Step Verified Solution

As   f(x) is differentiable at x = 1

∴\therefore     lim⁡x→1−(−x)=lim⁡x→1+(a+cos⁡−1(x+b))=f(1)\mathop {\lim }\limits_{x \to {1^ - }} \left( { - x} \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {a + {{\cos }^{ - 1}}\left( {x + b} \right)} \right) = f(1)

⇒\Rightarrow     −- 1 == a + cos −- 1(1 + b)

⇒\Rightarrow    cos −- 1 (1 + b) == −- 1 −- a . . . . . .(1)

As   f(x) is differentiable, so,

2 . H . D == R . H . D

Here, L . H . D == lim⁡h→0\mathop {\lim }\limits_{h \to 0} f(1−h)−f(1)−h{{f\left( {1 - h} \right) - f\left( 1 \right)} \over { - h}}

=lim⁡h→0−(1−h)−(−1)−h= \mathop {\lim }\limits_{h \to 0} {{ - \left( {1 - h} \right) - \left( { - 1} \right)} \over { - h}}

=lim⁡x→0−1+h+1−h= \mathop {\lim }\limits_{x \to 0} {{ - 1 + h + 1} \over { - h}}

== =lim⁡x→0h−h= \mathop {\lim }\limits_{x \to 0} {h \over { - h}}

== −- 1

R. H. D =lim⁡x→0f(1+h)−f(1)h= \mathop {\lim }\limits_{x \to 0} {{f\left( {1 + h} \right) - f\left( 1 \right)} \over h}

=lim⁡h→0a+cos⁡−1(1+h+b)−[a+cos⁡−1(1+b)]h= \mathop {\lim }\limits_{h \to 0} {{a + {{\cos }^{ - 1}}\left( {1 + h + b} \right) - \left[ {a + {{\cos }^{ - 1}}\left( {1 + b} \right)} \right]} \over h}

=lim⁡h→0cos⁡−1(1+h+b)−cos⁡−1(1+b)h[00  form ]= \mathop {\lim }\limits_{h \to 0} {{{{\cos }^{ - 1}}\left( {1 + h + b} \right) - {{\cos }^{ - 1}}\left( {1 + b} \right)} \over h}\left[ {{0 \over 0}\,\,form\left. \, \right]} \right.

=lim⁡h→0−11−(1+h+b)2= \mathop {\lim }\limits_{h \to 0} {{ - 1} \over {\sqrt {1 - {{\left( {1 + h + b} \right)}^2}} }} [ Using L' Hospital Rule]

=−11−(1+b)2= {{ - 1} \over {\sqrt {1 - {{\left( {1 + b} \right)}^2}} }}

∴\therefore     −1=−11−(1+b)2- 1 = {{ - 1} \over {\sqrt {1 - {{\left( {1 + b} \right)}^2}} }}

⇒1−(1+b)2=1\Rightarrow 1 - {\left( {1 + b} \right)^2} = 1

⇒\Rightarrow    (1+b)2=0{\left( {1 + b} \right)^2} = 0

⇒\Rightarrow     b=−1b = - 1

putting value of b in equation (1), we get,

cos⁡−1(1−1)=−1−a{\cos ^{ - 1}}\left( {1 - 1} \right) = - 1 - a

⇒\Rightarrow     π2=−1−a{\pi \over 2} = - 1 - a

⇒\Rightarrow     a=−1−π2a = - 1 - {\pi \over 2}

∴\therefore     ab=−1−π2−1=1+π2=π+22{a \over b} = {{ - 1 - {\pi \over 2}} \over { - 1}} = 1 + {\pi \over 2} = {{\pi + 2} \over 2}

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