If f(x) = \ cases px + q & : & x ≤ 2 x ^ 2 - 5x + 6 & : & 2 then |p| + |q| + |(1)/(a)| + |(1)/(b)| is equal to

Correct answer: 5.85 Continuity at x = 2 ⇒ p(2) + q = 2 ^ 2 - 5(2) + 6 ⇒ q = - 2p Continuity at x = 3 ⇒ a(9) + b(3) + 1 = 0 Differentiable at x = 2 ⇒ p = 2(2) - 5 ⇒ p = - 1 Differentiable at x = 3 ⇒ 2a(3) + b = 2(3) - 5 ⇒ 6a + b = 1 ∴ p = - 1, q = 2, a = (4)/(9), b = - (5)/(3) |p| + |q| + |(1)/(a)| + |(1)/(b)| = 1 + 2 + (9)/(4) + (3)/(5)

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 px + q & : & x ≤ 2 x ^ 2 - 5x + 6 & : & 2 then...

If $f\left(x\right) = \left\{$ \begin{cases}px + q & : & x \le 2 \\ {x}^{2} - 5x + 6 & : & 2 < x < 3 \\ a{x}^{2} + bx + 1 & : & x \ge 3\end{cases} $\right) is differentiable everywhere,</p> <p>then$ \left|p\right| + \left|q\right| + \left|\dfrac{1}{a}\right| + \left|\dfrac{1}{b}\right| is equal to

Correct answer

5.85

Step-by-step solution

Continuity at

x = 2 ⇒ p\left(2\right) + q = {2}^{2} - 5\left(2\right) + 6 ⇒ q = - 2p

Continuity at

x = 3 ⇒ a\left(9\right) + b\left(3\right) + 1 = 0

Differentiable at

x = 2 ⇒ p = 2\left(2\right) - 5 ⇒ p = - 1

Differentiable at

x = 3 ⇒ 2a\left(3\right) + b = 2\left(3\right) - 5 ⇒ 6a + b = 1

∴ p = - 1, q = 2, a = \dfrac{4}{9}, b = - \dfrac{5}{3}

\left|p\right| + \left|q\right| + \left|\dfrac{1}{a}\right| + \left|\dfrac{1}{b}\right| = 1 + 2 + \dfrac{9}{4} + \dfrac{3}{5}

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