If the function fx= a√(x+7) ; 0≤x<2 bx+5 ; x≥2 is differentiable ∀x≥0, then 2a+4b is equal to

Correct answer: (B) 5. fx must be continuous and differentiable at x=2 For continuity at x=2 f2^(-)=f2^(+)=f2 a√(2+7)=b2+5⇒3a=2b+5…i For differentiability at x=2 f^(′) x = a/(2√(x+7)) ; 0≤x<2 b ; x≥2 f^(′) 2^(−) =f^(′) 2^(+) a/(2√(2+7))=b⇒a=6b…ii From i⇒18b=2b+5⇒b=5/16 So, 2a+4b=12b+4b=16b=5

Q14 / 93 · NTA Abhyas (JEE Main) 2020 · Mathematics · Continuity and Differentiability

If the function $f (x) = \{\begin{matrix} a \sqrt{x + 7} & ; & 0 \leq x < 2 \\ b x + 5 & ; & x \geq 2 \end{matrix}$ is differentiable $\forall x \geq 0,$ then $2 a + 4 b$ is equal to

A
$\frac{240}{16}$
Correct answer
B
$5$
C
$\frac{85}{16}$
D
$\frac{250}{16}$

Solution

f (x)

must be continuous and differentiable at

x = 2

For continuity at

x = 2

f (2^{-}) = f (2^{+}) = f (2)

a \sqrt{2 + 7} = b (2) + 5 \Rightarrow 3 a = 2 b + 5 \dots (i)

For differentiability at

x = 2

f^{'} (x) = \{\begin{matrix} \frac{a}{2 \sqrt{x + 7}} & ; & 0 \leq x < 2 \\ b & ; & x \geq 2 \end{matrix}

f^{'} (2^{-}) = f^{'} (2^{+})

\frac{a}{2 \sqrt{2 + 7}} = b \Rightarrow a = 6 b \dots (i i)

From

(i) \Rightarrow 18 b = 2 b + 5 \Rightarrow b = \frac{5}{16}

So,

2 a + 4 b = 12 b + 4 b = 16 b = 5

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

If the function fx=ax+7;0≤x<2bx+5;x≥2 is differentiable ∀x≥0, then 2a+4b is equal to

Solution

More Continuity and Differentiability Questions

If the function $f (x) = \{\begin{matrix} a \sqrt{x + 7} & ; & 0 \leq x < 2 \\ b x + 5 & ; & x \geq 2 \end{matrix}$ is differentiable $\forall x \geq 0,$ then $2 a + 4 b$ is equal to