SSC CGL 20201)\({(4a + 3b +2 c)^2}\) का विस्तार करें।
\({16a^2 + 9b^2 +4c^2+24ab +12bc + 16ca}\)
put a, b, c = 1 in the options
opton C will come out to be 81 as in the \({(4a + 3b +2 c)^2} = {(4 + 3 + 2)^2} = 81\)
SSC CGL 20202)यदि A + B = 12 और AB = 17, तो \({A^3 + B^3}\) का मान क्या है ?
1116
\({(A + B)^3} = {A^3} +B^3 + 3AB(A+B) => {(12)^3 = A^3 +B^3 +3(17)(12)}\)
\(A^3 + B^3 = 1116\)
SSC CGL 20203)\((3a - 4b)^3 \) निम्न में से किसके बराबर है ?
\(27a^3-64b^3-108a^2b+144ab^2\)
instead of solving the question with derivative formula, Solve by elimination method.
put a=1 and b=1 and put the value in question \((3a - 4b)^3 = -1\)
Now check options, this will eliminate option A and D, now check for a = 1 and b =0, \((3a - 4b)^3 = 27\)
this will eliminate option C, the correct answer will be B
SSC CGL 20204)यदि a = 2b = 8c और a + b + c = 13 है, तो \(\sqrt{a^2+b^2+c^2}\over2c\) का मान है |
\(9\over2\)
a = 2b = 8c; \({a\over8}={2b\over8}={8c\over8}\); \({a\over8}={b\over4}={c\over1}\); a : b : c = 8 : 4 : 1; and a+b+c=13; \(a={8\over13}\times13=8\), b = 4, c = 1; \(\sqrt{a^2+b^2+c^2}\over2c\) = \(\sqrt{8^2+4^2+1^2}\over2\times1\) = \(\sqrt{81}\over2\) = \(9\over2\)
SSC CGL 20205)यदि x, y, z तीन संख्याएँ इस प्रकार हैं कि x + y = 13, y + z = 15 और z + x = 16 है, तो \(xy +xz\over xyz\) का मान है |
\(5\over18\)
x+y=13 ---(1);
y+z=15 ---(2);
z+x=16$$ ---(3);
By (1) + (2) + (3),
2(x + y + z) = 13 + 15 + 16;
x + y + z = 44/2 = 22;
put the value from eq(1),
13 + z = 22;
z = 9;
From eq(3),
9 + x =16;
x = 7;
From eq(3),
7 + y = 13;
y = 6;
Now,\({xy +xz\over xyz}={(7)(6)+(7)(9)\over(7)(6)(9)} ={5\over18}\)
SSC CGL 20206)यदि \(a^2+b^2-c^2=0\) है, तो \(2(a^6+b^6-c^6)\over3a^2b^2c^2\) का मान ज्ञात कीजिए ।
2
\(a^2+b^2-c^2=0\); \(a^2+b^2=c^2\); {cubing both sides}, \((a^2+b^2)^3=(c^2)^3\); ⇒ \(a^6+b^6+3a^2b^2(a^2+b^2)=c^6\); ⇒ \(a^6+b^6-c^6=-3a^2b^2c^2\);
\({2(a^6+b^6-c^6)\over3a^2b^2c^2}={2\times(-3a^2b^2c^2)\over3a^2b^2c^2}= -2\)
SSC CGL 20207)यदि p + q = 7 और pq = 5 है, तो \((p^3+q^3)\) का मान ज्ञात कीजिए |
238
p + q = 7; cubing both sides \((p^3+q^3)=7^3\); ⇒ \(p^3+q^3+3pq(p+q) =343\); ⇒ \(p^3+q^3=238\)
SSC CGL 20208)यदि x + 3y + 2 = 0 है, तो \(x^3 +27y^3+8-18xy\) का मान ज्ञात कीजिए |
0
x + 3y + 2 = 0; ⇒ x + 3y = -2; cubing both sides, \((x+3y)^3=(-2)^3\); ⇒ \(x^3+(3y)^3+3x(3y)(x+3y)=-8\); ⇒ \(x^3+27y^3+9xy(-2)=-8\); ⇒ \(x^3+27y^3+8-18xy=0\)
SSC CGL 20209)यदि \(a^4+\frac{1}{a^4}=50\), a > 0, तो \((a^3+\frac{1}{a^3})\) का मान ज्ञात कीजिए ।
\(\sqrt{2(1+\sqrt{13})}(-1+2\sqrt{13})\)
\(a^4+\frac{1}{a^4}={(a^2+\frac{1}{a^2})}^2-2\times a^2\times\frac{1}{a^2}=50\); \(a^2+\frac{1}{a^2}=2\sqrt{13}\); Similarly \({(a+\frac{1}{a})^2}=2(\sqrt{13}+1)\); \(a+\frac{1}{a}=\sqrt{2(\sqrt{13}+1)}\); Calculate \(a^3+\frac{1}{a^3}=\sqrt{2(\sqrt{13}+1)}(2\sqrt{13}-1)\)
SSC CGL 202010)\(25a^2-9\) का गुणनखंड इस प्रकार है |
(5a + 3)(5a - 3)
\(25a^2-9 =(5a)^2-(3)^2=(5a+3)(5a-3)\)