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^{log49(4x^2+4x+1)}= under root 3^{log3(x}^{^2-2)}^{2}+ 1/x >(or)= 1, for all x>0 is???6-(1+4*9

^{4-2logunder root 33})*log_{7}x = log_{x}7, x belongs to Q^{2}- 2xsec(theta) + 1=0 and a2 and b2 are the roots of the equation x^{2}+ 2xtan(theta) - 1=0. If a1>b1 and a2>b2, then a1 + b2 is???log

_{x}_{-1}3 = 2_{1}x^{2}+b_{1}x+c_{1}=0 are alpha ,beta and that of a_{2}x^{2}+b_{2}x+c_{2}=0 are gamma and delta such that alpha*gamma=beta*delta =1,thena.a

_{1}/a_{2}=b_{1}/b_{2}=c_{1}/c_{2}b.a

_{1}/c_{2}=b_{1}/b_{2}=c_{1}/a_{2}c.a

_{1}a_{2}=b_{1}b_{2}=c_{1}c_{2}d.none of these

the value of log

_{5}9*log_{7}5*log_{3}7/1/2log_{3}6 + 1/1/2log_{4}6 is co prime with^{x√x}= (x√x)^{x}?Options :

A) 1 B) 2 C) 4 D) infinite

I don't know to which chapter it belongs, it's in a book for KVPY

$\mathbf{10}\mathbf{.}If\alpha ,\beta arerootsof375{x}^{2}-25x-2=0and{s}_{n}={\alpha}^{n}+{\beta}^{n},thenfindthevalueof\underset{n\to \infty}{\mathrm{lim}}\sum _{r=1}^{n}{s}_{r}.$

^{log93}+9^{log24}= 10^{logx83 then x = ?}Illustration 2.79 In equation x

^{4}-2x^{3}+4x^{2}+6x-21=0 if two of its roots are equal in magnitude but opposite in sign. find the roots.^{x-1}- 1/ 2^{x+1}+ 1 < 2._{a}(1+ log_{b}[ 1 + log_{c}(1+ log_{p}x) ] ) = 0 is :1) 4

2) 3

3) 2

4) 1

mod(2x-3)<mod(x+2) is ?

_{a}b = 4 , log_{c}d=2, a,b,c,d are natural numbers if b-d = 7 then find c-d? ( plz do it fast) ( KVPY 2009 question)^{2}-15 -m*(2x-8) = 0?5.Let f(x), g(x), and h(x) be the quadratic polynomials having positive leading coefficients and real and distinct roots. If each pair of them has a common root, then find the roots off(x) + g(x) + h(x) = 0.

^{y}=y^{x}and x^{2}=y^{3 }.

log

_{2}^{24}/ log_{96}^{2}-log_{2}^{192 }/log_{12}^{2}^{2}-x+a-3<0 for atleast one negative value of x then complete set of values for a area.(-infinity,4) b.(-infinity,2) c.(-infinity,3) d.(-infinity,1)

^{2})/(x^{2}- 5x + 6) < 0_{0.3}(x-1)<log_{0.09}(x-1) then x lies in the intervalI can understand up to ---2log

_{0.3}(x-1)<log_{0.3}(x-1)(x-1)

^{2}>(x-1) --- next step sign changed from < to >. Can you please explain sir[1] x

^{2}- 2ax + a^{2 }- b^{2}- c^{2}= 0[2] (a - b + c)*x

^{2}+ 4(a- b)*x +(a- b -c) = 0?

^{2}+ bx + c has always the same sign as c ifa) b

^{2 }>4ac b) c>0 c) c<0 d) none of theseIf a

_{1}, a_{2}and b_{1}, b_{2}are roots of the equation2x

^{2}+3x+k_{1}=0and x^{2}+2x+k_{2}=0 respectively, then the system of equation a_{1}y+a_{2}z=0 and b_{1}y+b_{2}z=0 has a non zero solution. What will be the value of k_{1}:k_{2}?

$5.\mathrm{If}{\mathrm{log}}_{10}2,{\mathrm{log}}_{10}\left({2}^{\mathrm{x}}-1\right)\mathrm{and}{\mathrm{log}}_{10}\left({2}^{\mathrm{x}}+3\right)\mathrm{are}\mathrm{three}\mathrm{consecutive}\mathrm{terms}\mathrm{of}\mathrm{an}\phantom{\rule{0ex}{0ex}}\mathrm{AP},\mathrm{then}\mathrm{which}\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{is}\mathrm{correct}?\phantom{\rule{0ex}{0ex}}\left(\mathrm{a}\right)\mathrm{x}=0\phantom{\rule{0ex}{0ex}}\left(\mathrm{b}\right)\mathrm{x}=1\phantom{\rule{0ex}{0ex}}\left(\mathrm{c}\right)\mathrm{x}={\mathrm{log}}_{2}5\phantom{\rule{0ex}{0ex}}\left(\mathrm{d}\right)\mathrm{x}={\mathrm{log}}_{5}2$

|x-1| ≥ |x-3|

Find the values of m for which the equation sin^2x-(m-3)sinx+m=0 has real roots.

I can assume sinx to be t and reduce the equation to simple for. now t is binded in the interval -1,1. Now how to do sir? please clear!

1/x-2 + 1/x-1 > 1/x

I am having difficulty with this cubic:

t^{3}- 2t^{2}+3t - 6 > 0