Let f (x) = \left[\right. \left(αx\right)^{2} - x + \beta , - 1 \leq x < 1 \\ \int_{- 1}^{x} \left[\right. \frac{t^{4} + 2 t^{2} + 1}{t^{4} + t^{2} + 1} \left]\right. \textrm{ } dt , 1 \leq x \leq 2 \\ x^{3} + px + q , 2 < x \leq 3 If LMVT is applicable for f (x) in [–1, 3] and 'c' is the prescribed value of x for LMVT then [Note: [y] denotes greatest integer less than or equal to y.]

Engineering

Mathematics

Lagranges Mean Value Theorem

Question

Let f (x) = $[\begin{array}{l} {αx}^{2} - x + β, - 1 \leq x < 1 \\ \int_{- 1}^{x} [\frac{t^{4} + 2 t^{2} + 1}{t^{4} + t^{2} + 1}] dt, 1 \leq x \leq 2 \\ x^{3} + px + q, 2 < x \leq 3 \end{array}$
If LMVT is applicable for f (x) in [–1, 3] and 'c' is the prescribed value of x for LMVT then

[Note: [y] denotes greatest integer less than or equal to y.]

α – p = 12

β + q = – 15

[4c²] + p = 6

[c] + b = 5

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$f (x) = [\begin{array}{l} {αx}^{2} - x + β, - 1 \leq x < 1 \\ \int_{- 1}^{x} 1 dt = x + 1, 1 \leq x \leq 2 \\ x^{3} + px + q, 2 < x \leq 3 \end{array}$

continuity at x = 1
α – 1 + β = 2   ⇒   α + β = 3
derivability at x = 1
2αx – 1|_x=1 = 1   ⇒ α = 1
and β = 2
continuity at x = 2
3 = 8 + 2p + q   ⇒   2p + q = – 5
derivability at x = 2
3x² + p |_{x = 2} = 1
p = – 11, q = 17
f (x) = $[\begin{array}{l} x^{2} - x + 2, - 1 \leq x < 1 \\ x + 1, 1 \leq x \leq 2 \\ x^{3} - 11 x + 17, 2 < x \leq 3 \end{array}$
slope of the chord joining end points, (– 1, 4) and (3, 11) = $\frac{7}{4}$
$∴$ 3x² – 11 = $\frac{7}{4}$ ⇒   3x² = $\frac{51}{4}$ ⇒   4c² = 17.