Q.25 Find k , if f(x) = kx²(1 - x) , for 0 lt x lt 1 , = 0 otherwise is the p.d.f. of random variable X .

Answer: k = 12 For a p.d.f., ∫_(-∞)^(∞) f(x) dx = 1 , so ∫₀¹ kx²(1 - x) dx = 1 . k∫₀¹ (x² - x³) dx = k[x³/3 - x⁴/4]₀¹ = k(1/3 - 1/4) = k/12 . k/12 = 1 ⇒ k = 12 . (With k = 12 , f(x) ≥ 0 on (0, 1) , so it is a valid p.d.f.)

Find \(k\), if\(f(x) = kx^2(1 - x)\), for \(0 \lt x \lt 1\),\(= 0\) otherwiseis the p.d.f. of random variable \(X\).

Q.25 Find \(k\), if
\(f(x) = kx^2(1 - x)\), for \(0 \lt x \lt 1\),
\(= 0\) otherwise
is the p.d.f. of random variable \(X\).

Doubtly Asked 29 Oct 2026 Board exam · March 2024 hsc 12th Save WhatsApp Report

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Answer: \(k = 12\)

For a p.d.f., \(\int_{-\infty}^{\infty} f(x)\,dx = 1\), so \(\int_0^1 kx^2(1 - x)\,dx = 1\).

\(k\int_0^1 (x^2 - x^3)\,dx = k\left[\frac{x^3}{3} - \frac{x^4}{4}\right]_0^1 = k\left(\frac{1}{3} - \frac{1}{4}\right) = \frac{k}{12}\).

\(\frac{k}{12} = 1 \Rightarrow k = 12\). (With \(k = 12\), \(f(x) \ge 0\) on \((0, 1)\), so it is a valid p.d.f.)

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report

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Find \(k\), if\(f(x) = kx^2(1 - x)\), for \(0 \lt x \lt 1\),\(= 0\) otherwiseis the p.d.f. of random variable \(X\).

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