If f(x) = \ cases 2x^ 2 + 3 & ; & x > 3 ax^ 2 + bx + 1 & ; & x ≤ 3 cases ) is differentiable everywhere, then a b^ 2 is equal to

Correct answer: C. 1 ∵ f (x) is continuous at x = 3 ∴ _ x → 3 f (x) = f (3) 2( 3 ^ 2 ) +3 = a ( 3 ^ 2 ) + b (3) + 1 9a + 3b = 20 … (1) Now, f^ ' (x) =\ cases 4x & ; & x > 3 2ax + b & ; & x ≤ 3 cases ) ∵ f (x) is also differentiable at x = 3 ∴ 4 (3) = 2a (3) + b 6a + b = 12 … (2) From (1) & (2) , we get, 9a = 16 ⇒ a = (16)/(9) , b = (4)/(3)

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Continuity and Differentiability

NTA Abhyas 2020 · Mathematics · Continuity and Differentiability

If f(x) = \ cases 2x^ 2 + 3 & ; & x > 3 ax^ 2 + bx + 1 & ; & x ≤ 3...

f\left(x\right) = \left\{

\begin{cases}2x^{2} + 3 & ; & x > 3 \\ ax^{2} + bx + 1 & ; & x \le 3\end{cases}

\right) is differentiable everywhere, then

\dfrac{a}{b^{2}} is equal to

Options

A $5$
B $\dfrac{7}{3}$
C $1$
D $\dfrac{16}{9}$

Correct answer

C. $1$

Step-by-step solution

∵ f $\left(x\right)$ is continuous at $x = 3$
$∴ \lim_{x \to 3}f \left(x\right) = f \left(3\right)$
$2\left({3}^{2}\right) +3 = a \left({3}^{2}\right) +$ b \left(3\right) $+ 1$
9a + 3b = 20 $… \left(1\right)$
Now, $f^{'}\left(x\right) =\left\{$ \begin{cases}4x & ; & x > 3 \\ 2ax + b & ; & x ≤ 3\end{cases} $\right) $ ∵ f \left(x\right) $is also differentiable at$ x = 3 $ $ ∴ 4 \left(3\right) = 2a \left(3\right) + b $ $ 6a + b = 12 $… \left(2\right) From \left(1\right)$ & \left(2\right) $, we get,$ 9a = 16 ⇒ a = $\dfrac{16}{9}$ , $b =$ $\dfrac{4}{3}$

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