कक्षा 11 गणित अध्याय 12 सीमा और अवकलज समाधान

कक्षा 11 छात्रों के लिए यहाँ पर कक्षा 11 गणित अध्याय 12 सीमा और अवकलज समाधान दिया गया हैं। परीक्षा के दृष्टिकोण से ये सॉल्यूशंस बहुत ही महत्वपूर्ण हैं। यहाँ पर हमने एनसीईआरटी पाठ्यक्रम में दिय गय प्रश्नावली के साथ साथ विविध प्रश्नावली के भी समाधान दिया हैं|

इस पेज पर सीमा और अवकलज अध्याय में दिय गय सभी प्रश्नो का समाधान दिया हैं जो एनसीईआरटी के नवीनतम पाठ्यक्रम में शामिल हैं| परीक्षा के समय अन्य किताबों की तुलना में एनसीईआरटी समाधान छात्रो के लिय बहुत ही मददगार साबित होगा। बोर्डस्टडि पर आपको कक्षा 11 से संबधित और भी अध्ययन समाग्री मिल जायगी।

सीमा और अवकलज समाधान

प्रश्नावली 12.1

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प्रश्नावली 12.2

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विविध प्रश्नावली

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अध्याय 11: त्रिविमीय ज्यामिति का परिचय समाधान

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।

महत्वपूर्ण प्रश्न

प्रश्न.

limx→2⁡3x2−x−10x2−4\lim_{x \to 2} \frac{3x^2 – x – 10}{x^2 – 4}

हल :

limx→2⁡3x2−x−10x2−4=limx→2⁡(x−2)(3x+5)(x−2)(x+2)\lim_{x \to 2} \frac{3x^2 – x – 10}{x^2 – 4} = \lim_{x \to 2} \frac{(x – 2)(3x + 5)}{(x – 2)(x + 2)}
amp;=limx→2⁡(3x+5)(x+2)amp;=3×2+52+2=114\begin{aligned} &= \lim_{x \to 2} \frac{(3x + 5)}{(x + 2)} \\[10pt] &= \frac{3 \times 2 + 5}{2 + 2} = \frac{11}{4} \end{aligned}

प्रश्न.

limx→0⁡sin⁡axbx.\lim_{x \to 0} \frac{\sin ax}{bx}.

हल :

amp;limx→0⁡sin⁡axbx=limx→0⁡sin⁡axax⋅ab=ab.amp;[∵limx→0⁡sin⁡axax=1]\begin{aligned} & \lim_{x \to 0} \frac{\sin ax}{bx} = \lim_{x \to 0} \frac{\sin ax}{ax} \cdot \frac{a}{b} = \frac{a}{b}. \\[10pt] & \left[ \because \lim_{x \to 0} \frac{\sin ax}{ax} = 1 \right] \end{aligned}

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