The area under the curve y = x on [-(π/2), (π/2)] is given by : / वक्र y = x के अंतर्गत अंतराल [-(π/2), (π/2)] पर क्षेत्रफल होगा :

(1/2) square units / (1/2) वर्ग इकाई

π square units / π वर्ग इकाई

(π/4) square units / (π/4) वर्ग इकाई

Mathematics & Numerical Analysis Practice Questions (Q111-115) - Statistical Sciences

111

ICAR PG Statistical Sciences 29 Jun 2024Shift 1

MCQ (Single Correct) +4 −1

Evaluate $\iint_R (4xy - y^2)\, dx\, dy$, where $R$ is the rectangle bounded by $x = 1, x = 2, y = 0$ and $y = 3$:

$\iint_R (4xy - y^2)\, dx\, dy$ का मूल्यांकन कीजिए, जहां $R$ आयत $x = 1, x = 2, y = 0$ और $y = 3$ से परिबद्ध है:

18 18 16 16 15 15 14 14

112

ICAR PG Statistical Sciences 29 Jun 2024Shift 1

MCQ (Single Correct) +4 −1

Evaluate $\int_0^1 \frac{1}{\sqrt{-\log_e x}}\, dx$:

$\int_0^1 \frac{1}{\sqrt{-\log_e x}}\, dx$ का मान ज्ञात कीजिए:

$\sqrt{\frac{\pi}{2}}$ $\sqrt{\frac{\pi}{2}}$ $2\pi^2$ $2\pi^2$ $\frac{\sqrt{\pi}}{2}$ $\frac{\sqrt{\pi}}{2}$ $\sqrt{\pi}$ $\sqrt{\pi}$

113

ICAR PG Statistical Sciences 29 Jun 2024Shift 1

MCQ (Single Correct) +4 −1

Find the value of $\int_{-1}^2 |x|\, dx$:

$\int_{-1}^2 |x|\, dx$ का मान ज्ञात कीजिए:

2/5 2/5 3/5 3/5 1/2 1/2 5/2 5/2

114

ICAR PG Statistical Sciences 09 Jul 2023Shift 1

MCQ (Single Correct) +4 −1

The value of $\int \cos^{-1}(2x)\,dx$ is given by :

$\int \cos^{-1}(2x)\,dx$ का मान होगा :

$x\cos^{-1}(2x) - \frac{1}{2}\sqrt{1-4x^2} + C$ $x\cos^{-1}(2x) - \frac{1}{2}\sqrt{1-4x^2} + C$ $x\sin^{-1}(2x) - \frac{1}{2}\sqrt{1-4x^2} + C$ $x\sin^{-1}(2x) - \frac{1}{2}\sqrt{1-4x^2} + C$ $x\cos^{-1}(2x) + \frac{1}{2}\sqrt{1-4x^2} + C$ $x\cos^{-1}(2x) + \frac{1}{2}\sqrt{1-4x^2} + C$ $x\sin^{-1}(2x) + \frac{1}{2}\sqrt{1-4x^2} + C$ $x\sin^{-1}(2x) + \frac{1}{2}\sqrt{1-4x^2} + C$

Using integration by parts: $\int u\,dv = uv - \int v\,du$.
Let $u = \cos^{-1}(2x)$ and $dv = dx$.
Then $du = -\frac{2}{\sqrt{1-(2x)^2}}dx = -\frac{2}{\sqrt{1-4x^2}}dx$, and $v = x$.

Thus: $\int \cos^{-1}(2x)\,dx = x\cos^{-1}(2x) - \int x\left(-\frac{2}{\sqrt{1-4x^2}}\right)dx = x\cos^{-1}(2x) + \int \frac{2x}{\sqrt{1-4x^2}}dx$. To evaluate the remaining integral, let $w = 1 - 4x^2$, so $dw = -8x\,dx \implies 2x\,dx = -\frac{1}{4}dw$. Then $\int \frac{2x}{\sqrt{1-4x^2}}dx = -\frac{1}{4}\int w^{-1/2}dw = -\frac{1}{4}(2w^{1/2}) = -\frac{1}{2}\sqrt{1-4x^2}$. Combining terms gives: $\int \cos^{-1}(2x)\,dx = x\cos^{-1}(2x) - \frac{1}{2}\sqrt{1-4x^2} + C$. Detailed analysis of options:

$x\cos^{-1}(2x) - \frac{1}{2}\sqrt{1-4x^2} + C$ is the correct result.
Uses $\sin^{-1}$ instead of $\cos^{-1}$.
Has a plus sign instead of minus sign.
Uses $\sin^{-1}$ and plus sign.

खंडशः समाकलन (integration by parts) का प्रयोग करने पर: $\int u\,dv = uv - \int v\,du$.
माना $u = \cos^{-1}(2x)$ तथा $dv = dx$.
तब $du = -\frac{2}{\sqrt{1-4x^2}}dx$ और $v = x$.

अतः: $\int \cos^{-1}(2x)\,dx = x\cos^{-1}(2x) - \int x\left(-\frac{2}{\sqrt{1-4x^2}}\right)dx = x\cos^{-1}(2x) + \int \frac{2x}{\sqrt{1-4x^2}}dx$. शेष समाकल हेतु, माना $w = 1 - 4x^2 \implies dw = -8x\,dx \implies 2x\,dx = -\frac{1}{4}dw$. अतः $\int \frac{2x}{\sqrt{1-4x^2}}dx = -\frac{1}{4}(2\sqrt{w}) = -\frac{1}{2}\sqrt{1-4x^2}$. अंतिम मान: $x\cos^{-1}(2x) - \frac{1}{2}\sqrt{1-4x^2} + C$. विकल्पों का विस्तृत विश्लेषण:

$x\cos^{-1}(2x) - \frac{1}{2}\sqrt{1-4x^2} + C$ सही उत्तर है।
$\cos^{-1}$ के स्थान पर $\sin^{-1}$ लिखा है।
ऋण चिन्ह के स्थान पर धन चिन्ह लिखा है।
गलत है।

Explanation not matching this question? क्या व्याख्या इस प्रश्न से मेल नहीं खा रही? Report

115

ICAR PG Statistical Sciences 09 Jul 2023Shift 1

MCQ (Single Correct) +4 −1

The area under the curve $y = \cos x$ on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ is given by :

वक्र $y = \cos x$ के अंतर्गत अंतराल $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ पर क्षेत्रफल होगा :

$\frac{1}{2}$ square units $\frac{1}{2}$ वर्ग इकाई 2 square units 2 वर्ग इकाई $\pi$ square units $\pi$ वर्ग इकाई $\frac{\pi}{4}$ square units $\frac{\pi}{4}$ वर्ग इकाई

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