TS EAMCET · Maths · Application of Derivatives

Let $a, b, c \in ℝ$ be such that $2 a + 3 b + 6 c = 0$ and $g (x)$ be the antiderivative of $f (x) = a x^{2} + b x + c .$ If the slopes of the tangents drawn to the curve $y = g (x)$ at $(1, g (1))$ and $(2, g (2))$ are equal, then

A $\frac{a}{3} = \frac{b}{- 8} = \frac{c}{3}$
B $\frac{a}{6} = \frac{b}{- 18} = \frac{c}{7}$
C $\frac{a}{3} = \frac{b}{- 6} = \frac{c}{2}$
D $a = b = c = - 1$

Verified Solution

Answer & Solution

Correct Answer

(B) $\frac{a}{6} = \frac{b}{- 18} = \frac{c}{7}$

Step-by-step Solution

Detailed explanation

Slopes of tangent of the function gx at x=1 & x=2 implies that f1=f2. ⇒a+b+c=4a+2b+c ⇒3a+b=0 ...i And 2a+3b+6c=0 ...ii (Given) Solving equations i & ii by cross-multiplication method we get, a6=b-18=c7.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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Let a, b, c∈ℝ be such that 2a+3b+6c=0 and g(x) be the antiderivative of f(x)=ax2+bx+c. If the slopes of the tangents drawn to the curve y=g(x) at (1,g(1)) and (2,g(2)) are equal, then

Step-by-step Solution

Let $a, b, c \in ℝ$ be such that $2 a + 3 b + 6 c = 0$ and $g (x)$ be the antiderivative of $f (x) = a x^{2} + b x + c .$ If the slopes of the tangents drawn to the curve $y = g (x)$ at $(1, g (1))$ and $(2, g (2))$ are equal, then