| Class XII |
Informatics Practices |
Informatics Practices |
7 |
any other similar questions, and solving them step-by-step, with detailed explanation and documentation. As an example, in the following pages, we ... |
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| Class XII |
Informatics Practices |
Informatics Practices |
7 |
Worker - Household Industry - Males' 'Scheduled Tribe Marginal Worker - Household Industry - Females' 'Scheduled Tribe Marginal Worker - Other Work... |
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| Class XII |
Informatics Practices |
Informatics Practices |
7 |
have noticed that the Category column contains data under six different categories — '10-14', '15-19', '20-24', 'Adolescent (10-19)', 'All Ages', '... |
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| Class XII |
Informatics Practices |
Informatics Practices |
7 |
a barchart for the DataFrame obtained in Step 7. d.plot(kind='bar') plt.show() Reprint 2026-27 208 InformatIcPractIces The barchart shown at Figure... |
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| Class XII |
Mathematics |
Mathematics Part-II |
1 |
INTEGRALS 225 Chapter 7 INTEGRALS ❖ Just as a mountaineer climbs a mountain – because it is there, so a good mathematics student studies new materi... |
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| Class XII |
Mathematics |
Mathematics Part-II |
1 |
Such a process is called integration oranti differentiation. Letusconsiderthefollowingexamples: d We know that (sin x) = cos x ... (1) dx d x 3 2 d... |
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| Class XII |
Mathematics |
Mathematics Part-II |
1 |
+1) +C 2 2 1 2 1 2 (C) −log x + log (x +1) + C (D) 2 log x +log (x +1) +C 7.6 IntegrationbyParts Inthissection,wedescribeonemoremethodofintegration... |
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| Class XII |
Mathematics |
Mathematics Part-II |
1 |
f ′(x) = g ′(x) – ′(x) ∀ x ∈ I or f′ (x) = 0, ∀ x ∈ I by hypothesis, i.e., the rate of change of f with respect to x is zero on I and hence f is co... |
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| Class XII |
Mathematics |
Mathematics Part-II |
1 |
=sin x + C dx d (iii) (– cos x )sin x ; ∫sin x dx = –cos x + C dx d 2 2 (iv) (tan x)=sec x ; ∫sec x dx = tan x + C dx d 2 2 (v) (– cot x)= cosec x ... |
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| Class XII |
Mathematics |
Mathematics Part-II |
1 |
arbitrary constant. ∫ Proof Let F be any anti derivative of f, i.e., d dx F(x) = f(x) Then f (x) dx= F(x) + C ∫ Therefore d f (x) dx = d (F(x) + C ... |
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| Class XII |
Mathematics |
Mathematics Part-II |
1 |
Thus, in view of Property (II), it follows by (1) and (2) that ∫(f (x) + g(x))dx = ∫ f (x) dx +∫g(x) dx . k f (x) dx = k f (x) dx (IV) For any real... |
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| Class XII |
Mathematics |
Mathematics Part-II |
1 |
+ x 4 2 3 dx ( )= 3x + 4x . Therefore, an anti derivative of 3x + 4x is x + x . 4 Reprint 2026-27 232 MATHEMATICS (iii) We know that d 1 d 1 1 (log... |
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| Class XII |
Mathematics |
Mathematics Part-II |
1 |
x 2 + 2e – log x +C Example3Findthefollowingintegrals: (i) (sin x + cos x) dx (ii)cosec x (cosec x + cot x) dx ∫ ∫ 1– sin x (iii)∫ cos x dx Solutio... |
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| Class XII |
Mathematics |
Mathematics Part-II |
1 |
it is often necessary to satisfy an additional condition which then determines a specific value of C giving unique antiderivativeofthegivenfunction... |
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| Class XII |
Mathematics |
Mathematics Part-II |
1 |
(D) 2x + x2+C d 3 3 22. If dx f (x) = 4x − x4 such that f(2) = 0. Then f(x) is 4 1 129 3 1 129 (A) x + 3 − (B) x + 4 + x 8 x 8 4 1 129 3 1 129 (C) ... |
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| Class XII |
Mathematics |
Mathematics Part-II |
1 |
make the substitution mx = t so that mdx = dt. 1 1 1 Therefore, ∫ sin mx dx = ∫sin t dt= – cos t + C = – cos mx + C m m m 2 2 (ii) Derivative of x ... |
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| Class XII |
Mathematics |
Mathematics Part-II |
1 |
3sin x 1 cos x 24. 25. 2 2 26. 6cos x + 4sin x cos x (1– tan x) x cos x 27. sin 2x cos 2x 28. 1+ sin x 29. cot x log sin x sin x sin x 1 30. 1+ cos... |
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| Class XII |
Mathematics |
Mathematics Part-II |
1 |
2 5 1 1 = – 10 cos5x + 2 cos x + C (iii) From the identity sin 3x = 3 sin x – 4 sin x, we find that 3sin x – sin 3x sin x = Therefore, ∫sin x d... |
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| Class XII |
Mathematics |
Mathematics Part-II |
1 |
sin2 x− cos2 x 23. ∫ 2 2 dx is equal to sin xcos x (A) tan x + cot x + C (B) tan x + cosec x + C (C) – tan x + cot x + C (D) tan x + sec x + C x 24... |
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| Class XII |
Mathematics |
Mathematics Part-II |
1 |
= 1 (a+ x)+(a− x) 1 1 1 a – x 2 2a (a + x) (a − x) = + 2a − x a+ x Reprint 2026-27 INTEGRALS 245 dx 1 dx + dx Therefore, ∫... |
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| Class XII |
Mathematics |
Mathematics Part-II |
1 |
2 = log x + x + a − log|a|+C 1 = log x + x +a +C 2 , where C = C – log |a| Applying these standard formulae, we now obtain some more formulae which... |
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| Class XII |
Mathematics |
Mathematics Part-II |
1 |
dx (ii) dx ∫ x −16 ∫ 2x− x 2 Solution (i) We have dx = dx = 1 log x –4 +C [by 7.4 (1)] ∫ x2 −16 ∫ x – 4 2 8 x+ 4 (ii) Put x – 1 = t. Then dx = dt. ... |
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| Class XII |
Mathematics |
Mathematics Part-II |
1 |
6 6 1 6x −4 = log +C 1 17 6x +30 1 3x−2 1 1 = 17 log x 5 +C 1+17 log3 1 log 3x −2 +C C + 1 log 1 = 17 x +5 , where C = 1 17 3 Reprint 2026-27 INTEG... |
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| Class XII |
Mathematics |
Mathematics Part-II |
1 |
1 dx = ∫ 2 2 2 x+ 3 + 2 Put x + =t , so that dx = dt, we get 1 dt 1 tan 2 tC I2= 2∫ 1 2 = 1 2 [by7.4(3)] t + 2× 2 –1 3 –1 = t... |
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| Class XII |
Mathematics |
Mathematics Part-II |
1 |
dt. Therefore, I = dt = sin–1 t+ C2 [by 7.4 (5)] 2 ∫ 3 −t 2 3 – 1x +2 = sin +C 2 ... (3) Substituting (2) and (3) in (1), we obtain x+3 2 –1x + C 1... |
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