कक्षा 12 गणित अध्याय 5 सांतत्य तथा अवकलनीयता समाधान

बोर्डस्टडि एक्सपेर्ट द्वारा तैयार किया गया कक्षा 12 गणित अध्याय 5 सांतत्य तथा अवकलनीयता समाधान एनसीईआरटी के नवीनतम पाठ्यक्रम के अनुसार तैयार किया गया है। इसमें उन सभी प्रश्नों के उत्तर दिया गया हैं जो एनसीईआरटी कक्षा 12 गणित अध्याय 5 में दिया गया हैं|

इस पोस्ट में अध्याय 5 सांतत्य तथा अवकलनीयता के सभी अध्यायों का समाधान सरल एवं आसान भाषा में दिया गया हैं| परीक्षा के समय अन्य किताबों की तुलना में एनसीईआरटी समाधान छात्रो के लिय बहुत ही मददगार साबित होगा। बोर्डस्टडि पर आपको कक्षा 12 से संबधित और भी अध्ययन समाग्री मिल जायगी।

सांतत्य तथा अवकलनीयता समाधान

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अध्याय 4 : सारणिक
अध्याय 6 : अवकलज के अनुप्रयोग

अध्याय को अच्छे से समझने के लिय अध्याय मे दिय गय सभी प्रश्नों एवं अतिरिक्त प्रश्नों का अभ्यास बहुत जरूरी हैं। पिछले वर्ष का प्रश्न पत्र हल करना भी एक्जाम की तैयारी मे काफी मददगार साबित होता हैं।

एनसीईआरटी अध्याय 5 सांतत्य तथा अवकलनीयता के महत्वपूर्ण प्रश्न

प्रश्न . क्या f (x) = x2 – sinx + 5 द्वारा परिभाषित फलन x=π पर सतत् है

हल यहाँ, f(x) = x2 – sinx + 5

अब,

LHL=limx→π−⁡f(x)=limx→π−⁡(x2−sin⁡x+5)\text{LHL} = \lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} (x^2 – \sin x + 5)

तथा

RHL=limx→π+⁡f(x)=limx→π+⁡(x2−sin⁡x+5)\text{RHL} = \lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} (x^2 – \sin x + 5)

x = π + h रखने पर, x → π– ⇒ h → 0

⇒limh→0⁡[(π−h)2−sin⁡(π−h)+5]=limh→0⁡(π2+h2−2πh−sin⁡h+5)\Rightarrow \lim_{h \to 0} [(\pi – h)^2 – \sin(\pi – h) + 5] = \lim_{h \to 0} (\pi^2 + h^2 – 2\pi h – \sin h + 5)

पुनः,

f(π)=π2−sin⁡π+5=π2+5f(\pi) = \pi^2 – \sin \pi + 5 = \pi^2 + 5
∴LHL=RHL=f(π)\therefore \text{LHL} = \text{RHL} = f(\pi)

अतः फलन x = π पर सतत् फलन है। ⇒

प्रश्न . दर्शाइए कि g(x) = x- (x) द्वारा परिभाषित फलन समस्त पूर्णांक बिंदुओं पर असतत् है, यहाँ (x) वह महत्तम पूर्णांक निरूपित करता है, जो x के बराबर या x से कम है।

हल यहाँ,

g(x)=x−[x]g(x) = x – [x]

मान लीजिए a कोई धन पूर्णांक है, तब

[a−h]=a−1,[a+h]=a तथा [a]=a[a – h] = a – 1, \, [a + h] = a \text{ तथा } [a] = a

x=a पर,

LHL=limx→a−⁡g(x)=limx→a−⁡(x−[x])\text{LHL} = \lim_{x \to a^-} g(x) = \lim_{x \to a^-} (x – [x])

x = a – h रख ने पर, x → a– ⇒ h → 0

⇒limh→0⁡(a−h−[a−h])=limh→0⁡[a−h−(a−1)]\Rightarrow \lim_{h \to 0} (a – h – [a – h]) = \lim_{h \to 0} [a – h – (a – 1)]
=limh→0⁡[−h+1]=1{∵[a−h]=a−1}= \lim_{h \to 0} [-h + 1] = 1 \quad \{\because [a – h] = a – 1\}

तथा

RHL=limx→a+⁡g(x)=limx→a+⁡(x−[x])\text{RHL} = \lim_{x \to a^+} g(x) = \lim_{x \to a^+} (x – [x])

x = a + h रखने पर, x → a+ ⇒ h → 0

⇒limh→0⁡(a+h−[a+h])=limh→0⁡(a+h−a)=limh→0⁡h=0{∵[a+h]=a}\Rightarrow \lim_{h \to 0} (a + h – [a + h]) = \lim_{h \to 0} (a + h – a) = \lim_{h \to 0} h = 0 \quad \{\because [a + h] = a\}
∴LHL≠RHL\therefore \text{LHL} \neq \text{RHL}

अतः g (x) प्रत्येक पूर्णांक के लिए असतत् फलन है।

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