If fx= a+cos^(-1)x+b : x≥1 -x : x<1 is differentiable at x=1, then the value of b-a is equal to

Correct answer: (D) (π)/2. Since, f(x) is differentiable, it must be continuous at x=1 ∴f1^(-)=f1^(+)=f1 -1=a+cos^(-1)1+b ... 1 f^(')x= (−1√(1−x+b^2):x≥1−1:x<1Now, for fx to be differentiable at x=1 f^(′) 1^(−) =f^(′) 1^(+) -1)/(√(1-1+b^2))=-1⇒1-1+b^2=1⇒b=-1 ... 2 From equation 1 & 2 -1=a+(π)/2⇒a=-(π)/2-1

Q12 / 93 · NTA Abhyas (JEE Main) 2020 · Mathematics · Continuity and Differentiability

If $f (x) = \{\begin{cases} a + c o s^{- 1} (x + b) & : & x \geq 1 \\ - x & : & x < 1 \end{cases}$ is differentiable at $x = 1$ , then the value of $b - a$ is equal to

A
$0$
B
$1$
C
$- 1$
Correct answer
D
$\frac{π}{2}$

Solution

Since,

f (x)

is differentiable, it must be continuous at

x = 1

∴ f (1^{-}) = f (1^{+}) = f (1)

- 1 = a + c o s^{- 1} (1 + b)

...

(1)

f^{'} (x) = \{\begin{cases} \frac{- 1}{\sqrt{1 - {(x + b)}^{2}}} & : & x \geq 1 \\ - 1 & : & x < 1 \end{cases}

Now, for

f (x)

to be differentiable at

x = 1

f^{'} (1^{-}) = f^{'} (1^{+})

\frac{- 1}{\sqrt{1 - {(1 + b)}^{2}}} = - 1 \Rightarrow 1 - {(1 + b)}^{2} = 1 \Rightarrow b = - 1

...

(2)

From equation

(1)

(2)

- 1 = a + \frac{π}{2} \Rightarrow a = - \frac{π}{2} - 1

Hence, the correct answer is option: 'D'.

If fx=a+cos-1x+b:x≥1-x:x<1 is differentiable at x=1, then the value of b-a is equal to

Solution

More Continuity and Differentiability Questions

If $f (x) = \{\begin{cases} a + c o s^{- 1} (x + b) & : & x \geq 1 \\ - x & : & x < 1 \end{cases}$ is differentiable at $x = 1$ , then the value of $b - a$ is equal to