Match the pairs relating to Governor – General and Viceroys of India and work done by them or event during their tenure-

1-C, 2-A, 3-D, 4-B

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1-B, 2-C, 3-A, 4-D

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1-B, 2-A, 3-C, 4-D

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1-C, 2-D, 3-A, 4-B

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Solution

The correct option is B
1-B, 2-C, 3-A, 4-D

(B)

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\int_{- 1}^{4} f (x) d x = 4

\int_{2}^{4} [3 - f (x)] d x = 7 t h e n \int_{- 1}^{2} f (x) d x =

$I f \int_{2}^{4} f (x) d x = \frac{1}{2} a n d f (6 - x) = f (x), t h e n \int_{2}^{4} x f (x) d x =$

List I	List II
I. $\int_{- 1}^{1} x \| x \| d x$	(a) $\frac{π}{2}$
II. $\int_{0}^{\frac{π}{2}} (1 + log (\frac{4 + 3 sin x}{4 + 3 cos x})) d x$	(b) $\int_{0}^{\frac{a}{2}} f (x) d x$
III. $\int_{0}^{a} f (x) d x$	(c) $\int_{0}^{a} [f (x) + f (- x)] d x$
IV. $\int_{- a}^{a} f (x) d x$	(d) $0$
	(e) $\int_{0}^{a} f (a - x) d x$

I f I_{1} = \int_{0}^{1} 2^{x^{2}} d x, I_{2} = \int_{0}^{1} 2^{x^{3}} d x, I_{3} = \int_{1}^{2} 2^{x^{2}} d x a n d I_{4} = \int_{1}^{2} 2 x^{3} d x t h e n

\int_{0}^{π / 2} \frac{d x}{1 + {tan}^{5} x} = \frac{π}{4}

\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a + x) d x = \int_{0}^{π / 2} \frac{d x}{1 + {tan}^{3} x} = \int_{0}^{π / 2} \frac{d_{X}}{1 + {cot}^{3} x} = \frac{π}{4}

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